Students finding difficulties in their maths exam preparations can refer these NCERT solutions. We at BYJU’S provide free NCERT solutions for class 12 Maths chapter 8. These solutions are available in PDF format which can be easily downloaded and used by the students in their preparations.

NCERT Solutions for class 12 maths chapter 8 Application of Integrals includes a complete set of solved questions from the NCERT maths textbook. Students can practice more questions from these materials as it includes various solved MCQ, and both short and long type questions. NCERT Solutions for class 12 maths chapter 8 pdf covers solutions for various questions, including finding the area between two curves, determining the average value of a function, etc.

### NCERT Solutions Class 12 Maths Chapter 8 Exercises

- NCERT Solutions Class 12 Maths Chapter 8 Application of Integrals Exercise 8.1
- NCERT Solutions Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2
- NCERT Solutions Class 12 Maths Chapter 8 Application of Integrals Exercise 8.3

**Area between two curves**

**CASE – 1:**

For finding the **Area** bounded by the curve **y = f(x),** **x-axis** and the lines **x = a**, **x = b** let us consider a very thin vertical strip of length **y** and width **dx**. Therefore, **Area** of the strip **(dA) = y dx [Since, y = f(x)]**

Hence the total **Area** enclosed by the curve **y = f(x), x- axis** and the lines **x = a, x = b**:

**\(\boldsymbol{A =\int_{a}^{b} dA=\int_{a}^{b}y\;dx}\)**

**Therefore, A = \(\int_{a}^{b}f(x)\;dx\)**

** **

**CASE – 2:**

** **

Similarly the **Area** bounded by the curve **x = g(y), y-axis** and the **lines y = c, y = d **is given by:

**\(A=\int_{c}^{d} dA=\int_{c}^{d} x\;dy\)**

**Therefore, A = \(\int_{c}^{d}g(y)\;dy\)**

** **

**CASE – 3:**

** **

If the curve lies below **x-axis**, then the **Area** bounded by the curve **y = f(x)**, **x-axis** and the lines **x = a**, **x = b** will come negative.

Since, the **area** cannot be negative therefore we will neglect the negative sign and considering its **absolute value** only.

**\(A = \left |\int_{a}^{b}f(x)\;dx \right|\)**

** **

**CASE – 4:**

As shown above, some portion of the curve lies above the x-axis and some portion of the curve lies below the x-axis.

Here in this case, **Area-1 > 0** and **Area-2 < 0**

Therefore, **total** Area bounded by the **curve y= f(x)**, **x-axis** and the lines **x=a**, **x=b** is given by: **A = |A|+B**

** **

**NOTE:**

**\(\boldsymbol{\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}}\)**

** **

**Example 1: Find the area enclosed by equation: x ^{2 }+ y^{2} = 9**

** **

**Sol:**

Equation **x ^{2 }+ y^{2} = 3^{2}** represents the

**circle**with

**centre (0,0)**and

**radius 3 units.**

**Therefore, y = \(\sqrt{3^{2}-x^{2}}\)**

** **

**From the figure, the Area enclosed by the circle = 4 ****×**** (Area enclosed by the curve ABOA)**

**Now, Area enclosed by the curve** **ABOA**:

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{3}y\;dx = \int_{0}^{3} \sqrt{9-x^{2}}\;dx\)

**Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin^{-1}\frac{x}{3} \right | _{0}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2} \sqrt{9-9}+\frac{9}{2} \sin^{-1}(1) – \frac{0}{2} \sqrt{9-0}-\frac{0}{2} \sin^{-1}\frac{0}{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9}{2}\times \frac{\pi }{2}=\frac{9\pi }{4}}\) **unit ^{2}**

**Therefore, the Area enclosed by the circle =** **4 ****×**** (area enclosed by curve ABOA)**

**\(\boldsymbol{\Rightarrow }\) \(4\times \frac{9\pi }{4}=9\pi\)**

**Hence the Area enclosed by the circle = 9****π**** unit ^{2}**

**Example 2: Find the area enclosed by the curve \(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)**

** **

**Sol:**

**Equation** **\(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)** **represents an** **ellipse with major axis = 3 units and minor axis = 2 units**

**Since, \(\frac{x^{2}}{9}+\frac{y^{2}}{4} = 1\)**

**\(\boldsymbol{\Rightarrow }\) 4x ^{2} + 9y^{2} = 36**

**\(\boldsymbol{\Rightarrow }\) 9y ^{2} = 6^{2 }– (2x)^{2}**

**Therefore, \(y=\sqrt{\frac{6^{2}-(2x)^{2}}{9}}\)**

**Therefore, the Area of region enclosed by the ellipse = 4 ****× ****(Area enclosed by the curve ABOA)**

Now, **Area enclosed by the curve ABOA = \(\int_{0}^{3} y\;dx\)**

**Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\times \int_{0}^{3}\sqrt{6^{2}-(2x)^{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\int_{0}^{3}\;\frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}\left | \frac{2x}{2}\sqrt{6^{2}-(2x)^{2}}+\frac{36}{2}\sin^{-1}\frac{2x}{6} \right |_{0}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3}[\frac{6}{2}\sqrt{36-36}+18\sin^{-1}(1)]=3\pi }\)

**Therefore, the Area enclosed by the** **curve ABOA = 3****π**** unit ^{2}**

**Hence, the total Area enclosed by an ellipse ABA’B’ = ****4****×****3****π** **= 12****π unit ^{2}.**

** **

**Area of region bounded by the curve and line:**

** **

**Example 3: Find the area of the region enclosed by the curve y = x ^{2} and the line y = 3.**

** **

**Sol:**

** **

**y = x ^{2} **represents a

**parabola**, symmetrical about y-axis as shown in the above figure. By substituting

**y =9**in equation of parabola

**y = x**we will get coordinates of point

^{2}**M**.

i.e. **x ^{2} = 9**

Therefore, **x = +3 or -3**

**Hence the coordinates of point M = (3, 9)**

The area of the region bounded by the curve **y = x ^{2}** and the line

**y = 9**is the area enclosed by curve

**POMP.**

Now, **Area of region POMP = 2(area of region AOMA)**

Since, **x ^{2} = y**

Therefore, **x = \(\sqrt{y}\)**

Thus, **the Area of region bounded by the curve AOMA:**

\(\boldsymbol{\Rightarrow }\) \(\int_{0}^{9}x\;dy\) = \(\int_{0}^{9}\sqrt{y}\;dy\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{0}^{9}=\frac{2}{3}\times (9^{\frac{3}{2}})}}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{2}{3}\times 27 = 18}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve AOMA = 18 unit ^{2}**

**Hence, the Area of region bounded by the curve POMP ****= 2 ****×**** (area of region bounded by curve AOMA) ****= 36 unit ^{2}**

** **

** **

**Example 4: Find the area enclosed by the curve x ^{2 }+ y^{2 }= 50, x-axis and the line y = x in the 1^{st} quadrant.**

** **

**Sol:**

Equation **x ^{2 }+ y^{2 }= 50** represents a

**circle**with

**radius \(\sqrt{50}\) units**.

Since y = x

Therefore x^{2 }+ x^{2 }= 50 [for points of intersection of both the curves]

Hence, **x = +5 and -5**

Similarly, ** y = +5 and -5**

**Thus the coordinates of point B are (5, 5)**

** **

Form the figure, the area of region bounded by the curve **x ^{2 }+ y^{2 }= 50**,

**x = y**and

**x – axis in the 1**is the

^{st}quadrant**Area enclosed by the curve OMABO**

**.**

**i.e. Area of triangle OMB + Area under the curve MBAM.**

**Now, the Area of triangle = \(\frac{1}{2}\times Base \times Altitude\)**

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times OM \times BM\;\;=\;\frac{1}{2}\times 5\times 5 =\frac{25}{2}\) unit^{2}.

Now, the Area under the curve **MBAM** = \(\int_{5}^{5\sqrt{2}}y\;dx\)

Since, **x ^{2} +y^{2} = 50**. Therefore,

**y**

^{2}= 50 – x^{2}\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\sqrt{50-x^{2}}}\)

Therefore, the Area under curve **MBAM [x ^{2 }+ y^{2 }= 50]**:

**Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{5}^{5\sqrt{2}}\sqrt{50-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{50-x^{2}}+\frac{50}{2}\;\sin^{-1}\frac{x}{5\sqrt{2}} \right |_{5}^{5\sqrt{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{5\sqrt{2}}{2}\times\ \sqrt{50-50}]+[\frac{50}{2}\;\sin^{-1}(1)]-[\frac{5}{2}\times \sqrt{50-25}]-[\frac{50}{2}\times \sin^{-1}\frac{1}{\sqrt{2}}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{0+\frac{25\pi }{2}-\frac{25}{2}-\frac{25\pi }{4}=\frac{25}{4}(\pi -2)}\)**unit ^{2}**

Therefore, the **Area under the curve MBAM = \(\boldsymbol{\frac{25}{4}(\pi -2)}\) unit ^{2}**

**Now, the total Area under the shaded region** **=** **Area of triangle OMB + Area under the curve MBAM**

\(\boldsymbol{\Rightarrow }\) \(\frac{25}{2}+\frac{25}{4}(\pi -2)=\frac{25}{2}+\frac{25\pi }{4}-\frac{25}{2}\boldsymbol{=\frac{25\pi }{4}}\)**unit ^{2}**

**Hence, the Area of shaded region OMABO = \(\boldsymbol{\frac{25\pi }{4}}\)unit ^{2}**

** **

** ****Exercise 8.1**

** **

**Q.1: Find the area enclosed by the curve y = x ^{2} and the lines y = 2, y = 4 and the y-axis.**

** **

**Sol:**

Equation **y = x ^{2}** represents a

**parabola**symmetrical about y-axis.

** **

The Area of the region bounded by the curve **y = x ^{2}**,

**y = 2**, and

**y**

**=**

**4**, is the Area enclosed by the curve AA’B’BA.

Now, the **Area of region AA’B’BA = 2 (Area of region ABNMA)**

Since, x^{2} = y

Therefore, **x = \(\sqrt{y}\)**

Thus, **the Area of region bounded by the curve ABNMA [****y = x ^{2}**

**]:**

\(\boldsymbol{\Rightarrow }\) \(\int_{2}^{4}x\;dy\) = \(\int_{2}^{4}\sqrt{y}\;dy\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{2}^{4}=\frac{2}{3}\times [(4^{\frac{3}{2}})-(2)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{2}{3}\times (8-2\sqrt{2}) = \frac{4}{3}(4-\sqrt{2})}\) **unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve ABNMA = \(\boldsymbol{\frac{4}{3}(4-\sqrt{2})}\)unit ^{2}**

**Hence, the ****Area of region bounded by the curve AA’B’BA ****=**** 2(Area of region bounded by the curve ABNMA)****= \(\boldsymbol{\frac{8}{3}(4-\sqrt{2})}\)unit ^{2}**

** **

**Q.2: Find the area enclosed by the curve y ^{2} = 4x and lines x = 1, x = 3 and the x- axis in the first quadrant.**

** **

**Sol:**

** **

**Equation y ^{2} = 4x **represents a

**parabola**, symmetrical about the

**x-axis**as shown in the above figure.

The area of the region bounded by the curve **y ^{2} = 4x**,

**x = 1**,

**x**

**=**

**3**and the

**x-axis**is the area enclosed by the curve

**ABCDA**.

Now, the **Area of region bounded by the curve ABCDA:**

Since, **y ^{2} = 4x**

Therefore, **y = \(2\sqrt{x}\)**

Hence, **the area of region bounded by the curve ABCDA [y ^{2} = 4x]:**

\(\boldsymbol{\Rightarrow }\) \(\int_{1}^{3}y\;dx\) = \(\int_{1}^{3}2\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=\frac{4}{3}\times [(3^{\frac{3}{2}})-(1)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4}{3}\times (3\sqrt{3}-1) = \frac{-4\;+\;12\sqrt{3}}{3}}\) **unit ^{2}**

**Therefore, the ****area**** of region ****bounded ****by the ****curve ABCDA \(\boldsymbol{= \frac{-4\;+\;12\sqrt{3}}{3}}\)unit ^{2}**

** **

** **

**Q.3: Find the value of k if the line x = k divides the area enclosed by the curve y ^{2} = 9x and the line x = 4 in to two equal parts.**

** **

**Sol:**

**Equation y ^{2} = 9x **represents a

**parabola**, symmetrical about the

**x-axis**as shown in the above figure.

Since, the line **x = k** divides the Area OCBB’O in to two **equal halves** and the curve is symmetrical to **x-axis**. Therefore**, Area of the region OADO = Area of the region ABCDA.**

The Area of the region bounded by the curve **y ^{2} = 9x** and the line

**x = k**is the Area of region enclosed by the curve

**OADO**.

Since, **y ^{2} = 9x**

Therefore, **y = \(3\sqrt{x}\)**

Hence, **the Area of region bounded by the curve OADO:**

\(\boldsymbol{\Rightarrow }\) **\(\int_{0}^{k}y\;dx\) = \(\int_{0}^{k}3\sqrt{x}\;dx\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{1}^{3}=2\times [(k^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{2k^{\frac{3}{2}}}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve OADO = \(\boldsymbol{2k^{\frac{3}{2}}}\)unit ^{2}**

The Area of the region bounded by the curve **y ^{2} = 9x** and the lines

**x = k and x = 4**is the Area under the curve

**ABCDA**

**Now, the Area of region bounded by the curve ABCDA [****y ^{2} = 9x**

**]:**

\(\boldsymbol{\Rightarrow }\) ** \(\int_{k}^{4}y\;dx\) = \(\int_{k}^{4}3\sqrt{x}\;dx\)**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 3\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{k}^{4}=2\times [(4^{\frac{3}{2}})-(k)^{\frac{3}{2}}}]}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})}\) **unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve ABCDA = ****\(\boldsymbol{\Rightarrow (16-2k^{\frac{3}{2}})}\) unit ^{2}**

**Since, the Area of region OADO = Area of region ABCDA [GIVEN]**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{(16-2k^{\frac{3}{2}})=2k^{\frac{3}{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k^{\frac{3}{2}}=4}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{k = 4^{\frac{2}{3}}}\)

**Therefore, the value of k = \(\boldsymbol{4^{\frac{2}{3}}}\)**

** **

** **

**Q.4: Find the area enclosed by the curve y ^{2} = 16x, y-axis and the line y = 2.**

** **

**Sol:**

**Equation y ^{2} = 16x **represents a

**parabola**, symmetrical about the

**x-axis**as shown in the above figure.

The Area of the region bounded by the curve **y ^{2} = 16x**,

**y = 2**and the

**y-axis**is the Area of region enclosed by the curve

**OABO**.

Now, the **Area of region **enclosed by the curve **OABO:**

Since, **y ^{2} = 16x**

Therefore, **x = \(\frac{y^{2}}{16}\)**

Thus, **the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{2}x\;dy\) = \(\int_{0}^{2}\frac{y^{2}}{16}\;dy\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | \frac{1}{16}\times \frac{y^{3}}{3}\right |_{0}^{2}=\frac{1}{16}\times [\frac{8}{3}-0}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}}\) **unit ^{2}**

**Therefore, ****Area**** of region ****bounded ****by the ****curve OABO \(\boldsymbol{=\frac{1}{6}}\)unit ^{2}**

** **

** **

**Q.5: Find the area enclosed by the curve y ^{2} = 25x and the line x = 3**

** **

**Sol:**

** **

**Equation y ^{2} = 25x **represents a

**parabola**, symmetrical about x-axis as shown in the above figure.

The area of the region bounded by the curve **y ^{2} = 25x** and

**x = 3**is the Area enclosed by the curve

**BOCAB**.

Now, the** Area of region BOCAB = 2(Area of region OABO)**

Since, y^{2} = 25x

Therefore, **y = \(5\sqrt{x}\)**

Hence, **the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3}y\;dx\) = \(\int_{0}^{3}5\sqrt{x}\;dx\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 5\times \frac{2}{3}\times (x)^{\frac{3}{2}} \right |_{0}^{3}=\frac{10}{3}\times [(3^{\frac{3}{2}})-(0)}]}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{10}{3}\times (3\sqrt{3})=10\sqrt{3}}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve OABO\(\boldsymbol{=10\sqrt{3}}\) unit ^{2}**

**Now, the Area of region BOCAB = ****2 ****×**** (Area of region OABO) ****\(\boldsymbol{=20\sqrt{3}}\) unit ^{2}**

** **

** **

**Q.6: Find the area enclosed by the curve x ^{2 }= 8y, y = 1, y = 9 and the y-axis in the first quadrant.**

** **

**Sol:**

** **

**Equation x ^{2} = 8y **represents a

**parabola**, symmetrical about the y-axis as shown in the above figure.

The area of the region bounded by the curve **x ^{2} = 8y**,

**y = 1, y = 9 and the first quadrant**is the area enclosed by the curve ABDCA.

Now, the **Area of the region ABDCA:**

Since, **x ^{2} = 8y**

Therefore, **x = \(2\sqrt{2y}\)**

Hence, **the Area of region bounded by the curve ABDCA:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{1}^{9}x\;dy\) = \(\int_{1}^{9}2\sqrt{2y}\;dy\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{{\left | 2\sqrt{2}\times \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{9}=\frac{4\sqrt{2}}{3}\times [(9^{\frac{3}{2}})-(1)}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{4\sqrt{2}}{3}\times (27-1)=104\times \frac{\sqrt{2}}{3}}\) **unit ^{2}**

Therefore, **Area** of region **bounded** by the **curve ABDCA ****\(\boldsymbol{=104\times \frac{\sqrt{2}}{3}}\)unit ^{2}**

** **

** **

**Q.7: Find the area bounded by the curve whose equation is x ^{2} = 9y and the line x = 6y – 3.**

** **

**Sol:**

** **

**Equation x ^{2} = 9y **represents a

**parabola**, symmetrical about the y-axis as shown in the above figure.

The Area of the region bounded by parabola **x ^{2} = 9y** and the line

**x =6y – 3**is the Area enclosed under the curve

**ABC0A**.

Since, the **parabola x ^{2} = 9y** and the

**line x = 6y – 3**intersect each other at points

**A**and

**C**, hence the coordinates of

**points**

**A**and

**C**are given by:

(6y-3)^{2} = 9y

**\(\boldsymbol{\Rightarrow }\) 36y ^{2} – 45y + 9 = 0**

By **Hit and Trial method** solutions of this quadratic equation are:

**y = 1 and y = \(\frac{1}{4}\)**

Hence, the **co-ordinates** of **point A** and **point C** are** (3,1)** and** \((\frac{-3}{2},\frac{1}{4})\)** respectively

Since, **x ^{2} = 9y**

Therefore, **x = \(3\sqrt{y}\)**

The **Area** of region bounded by the curve **ABCOA** = [**Area** of region **OBCbO** – **Area** of region **OCbO**] + [**Area** of region **ABOaA** – **Area** of region **OAaO**]

**The Area enclosed by the curve OBCbO:**

\(\\\boldsymbol{\Rightarrow }\) \(\int_{0}^{3} y\;dx = \boldsymbol{\int_{0}^{3}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(\frac{9}{2}+9)=\frac{9}{4}}\) **unit ^{2}**

**The Area enclosed by the curve ABOaA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx} = \boldsymbol{\int_{\frac{-3}{2}}^{0}\frac{x+3}{6}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{2}}{2}+ 3x \right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}(0-\frac{9}{8}+\frac{9}{2})=\frac{9}{16}}\) **unit ^{2}**

**The Area ****enclosed by the curve ****OAaO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{-3}{2}}^{0}\;y\;dx\; =\;\int_{\frac{-3}{2}}^{0} \frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{\frac{-3}{2}}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{27}(0-\frac{-27}{8})=\frac{1}{8}}\)**unit ^{2}**

**The Area ****enclosed by the curve ****OCbO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3} y\;dx\; = \;\int_{0}^{3}\frac{x^{2}}{9}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}\left | \frac{x^{3}}{3}\right |_{0}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{9}(\frac{27}{3}-0)=1}\)** unit ^{2}**

**Since,** **the Area of region bounded by the curve ABCOA = [Area of region OBCbO – Area of region OCbO] + [Area of region ABOaA – Area of region OAaO]**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{9}{4}-1]+[\frac{9}{16}-\frac{1}{8}]=\frac{27}{16}}\)

**Therefore, the Area of region bounded by the curve ABCOA \(=\frac{27}{16}\) unit ^{2}**

**Q.8: Find the area enclosed by the curve 4y = x ^{2} and y = |x|**

** **

**Sol:**

** **

**Equation x ^{2} = 4y **represents a

**parabola**, symmetrical about the y-axis as shown in the above figure.

The area of the region bounded by the curve **x ^{2} = 4y** and

**y = |x|**is

**2(OAEO**) i.e.

**(area OCFO+ area OAEO)**

Now, **Area of region OAEO = OABO – OEABO**

Since, **x ^{2} = 4y**

\(\boldsymbol{\Rightarrow }\) \(y=\frac{x^{2}}{4}\)

Now, **the Area of region bounded by the curve OEABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}y\;dx\;=\;\int_{0}^{4}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{4}\left | \frac{x^{3}}{3} \right |_{0}^{4}=\frac{16}{3}}\) **unit ^{2}**

**Therefore, the area of region bounded by the curve OEABO = \(=\frac{16}{3}\) unit ^{2}**

**Now, the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}y\;dx\;\Rightarrow \;\int_{0}^{4}x\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{4}=8}\) **unit ^{2}**

**Therefore, the Area of the region bounded by the curve OABO = 8 unit ^{2}**

**Now, ****Area of region OAEO = Area of region (OABO – OEABO)**

\(\boldsymbol{\Rightarrow }\) \(8-\frac{16}{3}\) = **\(\frac{8}{3}\) unit ^{2}**

**Therefore, the total Area of shaded region = ****2****×\(\frac{8}{3}\)**** = \(\frac{16}{3}\)unit ^{2}**

** **

** **

**Q.9: ****Find the area enclosed by the curve \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)**

** **

**Sol:**

Equation **\(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)** represents an **ellipse.**

**Since, \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\\\)**

**\(\\\boldsymbol{\Rightarrow }\) \(\frac{y^{2}}{9}=1-\frac{x^{2}}{4}\\\)**

**\(\\\boldsymbol{\Rightarrow }\) \(y=\frac{3}{2}\sqrt{4-x^{2}}\)**

Therefore, the Area of region enclosed by the ellipse = 4 × (Area enclosed by the curve ABOA)

Now, the **Area enclosed by the curve ABOA:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2} y\;dx\;=\; \frac{3}{2}\times \int_{0}^{2}\sqrt{2^{2}-(x)^{2}}\;dx}\)

**Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}\) **unit ^{2}**

Therefore, the **Area** enclosed by the **curve ABOA \(\boldsymbol{=\frac{3\pi}{2}}\)**** unit ^{2}**

**Hence, the total Area enclosed by the ellipse ABA’B’ = ****4 ****× \(\boldsymbol{\frac{3\pi}{2}}\)**** unit ^{2}**

**= 6****π unit ^{2}**

** **

** **

**Q.10: Find the area enclosed by the curve x ^{2 }+ y^{2} = 9, line x = \(2\sqrt{2}y\) and the first quadrant.**

** **

**Sol:**

** **

Equation **x ^{2 }+ y^{2 }= 9** represents a

**circle**with

**radius equal to 3units**.

Since, **x = \(2\sqrt{2}y\)**

Therefore (\(2\sqrt{2}y\))^{2 }+ y^{2} = 9 (for points of intersection of both the curves)

Hence, the coordinates of point **B** **= (****\(2\sqrt{2}y\)****,1)**

Now, the area of region bounded by the curve **x ^{2 }+ y^{2 }= 9**,

**x =**

**\(2\sqrt{2}y\)**, and the

**first quadrant**is the

**area enclosed by the curve OMABO.**

**i.e. Area of triangle OMB + Area under the curve MBAM.**

**Now, the Area of triangle = \(\frac{1}{2}\times Base \times Altitude\)**

\(\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times OM \times BM\;=\;\frac{1}{2}\times2\sqrt{2}y\times 1 =\sqrt{2}\;y\;\)unit^{2}

**Now, the Area under the curve MBAM:**

\(\boldsymbol{\Rightarrow }\) \(\int_{2\sqrt{2}}^{3}\;y\;dx\)

**Since, x ^{2} +y^{2} =9**

**Therefore, y ^{2} = 9 – x^{2}**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\sqrt{9-x^{2}}}\)

**Since, \(\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2\sqrt{2}}^{3}\sqrt{3^{2}-x^{2}}\;dx=\left | \frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\;\sin^{-1}\frac{x}{3} \right |_{2\sqrt{2}}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\;\boldsymbol{[\frac{3}{2}\times\ \sqrt{9-9}]+[\frac{9}{2}\;\sin^{-1}(1)]-[\frac{2\sqrt{2}}{2}\times \sqrt{9-8}]-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\\\) **unit ^{2}**

**Therefore, the Area under the curve MBAM:**

**= \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit ^{2}**

Now, total Area under the shaded region =** Area under the curve MBAM + Area of the triangle OMB**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-\sqrt{2}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]+\sqrt{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)** unit ^{2}**

**Hence, the required Area is given by the region OMABO:**

**\(\boldsymbol{=\frac{9\pi }{4}-[\frac{9}{2}\times \sin^{-1}\frac{2\sqrt{2}}{3}]}\)unit ^{2}**

** **

** **

**Q.11: Find the area of larger part of circle x ^{2 }+ y^{2 }= 16 cut off by the line x = \(\frac{4}{\sqrt{2}}\) in the first quadrant.**

** **

**Sol:**

Equation **x ^{2 }+ y^{2 }= 16 **represents a

**circle**with

**radius**=

**4**

**units**.

For coordinates of point B:

\(\boldsymbol{\Rightarrow }\) \((\frac{4}{\sqrt{2}})^{2}+y^{2}=16\)

\(\boldsymbol{\Rightarrow }\) 8 + y^{2} = 16

\(\boldsymbol{\Rightarrow }\) y = \(2\sqrt{2}\)

Hence, the **coordinates** of **point** **B** are: **\(\boldsymbol{(\frac{4}{\sqrt{2}},2\sqrt{2})}\)**

**The required Area is given by the curve OMBCO:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2\sqrt{2}}y\;dx\;=\;\int_{0}^{2\sqrt{2}}\sqrt{16-x^{2}}\;dx}\)

**Since, \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}\sin^{-1} \frac{x}{4} \right |_{0}^{2\sqrt{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{2\sqrt{2}}{2}\sqrt{16-8}+8\sin^{-1}\frac{1}{\sqrt{2}}]-0}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{2}\times 2\sqrt{2}+8\times \frac{\pi }{4}=[4+2\pi] }\) **unit ^{2}**

**Therefore, the Area of shaded region OMBCO = [4 + 2****π****]unit ^{2}**

Let us assume two curves represented by the equation **y = f(x)** and **y = g(x) **in **[a, b]** as shown in the above figure. In this case the height of an elementary strip will be **[f(x) – g(x)] **and its width will be **dx**.

Now, Area of the elementary strip **(dA) = [f(x) – g(x)] dx**

Hence, the **total Area **of shaded region** (A) = \(\int_{a}^{b} [f(x)-g(x)]\;dx\)**

**Example – 1: Find the area enclosed between two curves whose equations are: x ^{2 }= 4y and y^{2 }= 4x.**

** **

**Sol:**

The Equation **x ^{2 }= 4y **represents a parabola symmetrical about y-axis and the equation

**y**represents a parabola symmetrical about x-axis.

^{2 }= 4x**On solving both the equations:**

\(\boldsymbol{\Rightarrow }\) \(\left ( \frac{y^{2}}{4} \right )^{2}=4y\)

\(\boldsymbol{\Rightarrow }\) y^{3} = 64 **i.e. y = 4**

Which gives x = 4. Hence, the **coordinates** of **point N** are **(4, 4)**.

Now, The Area of region enclosed by the curve** NAOBN** = Area of region enclosed by the curve **OANMO** – Area of region enclosed by the curve **OBNMO**.

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}2\sqrt{x}\;dx-\int_{0}^{4}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2\times \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{4}-\left [ \frac{x^{3}}{12} \right ]_{0}^{4}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4}{3}\times (4)^{\frac{3}{2}}-0 \right ]-\left [ \frac{64}{12}-0 \right ]= \frac{16}{3}}\) **unit ^{2}**

**Therefore, the Area of shaded region = \(\frac{16}{3}\) unit ^{2}**

**EXAMPLE – 2: Find the area enclosed by the sides of a triangle whose vertices have coordinates (1, 0) (3, 5) and (5, 4).**

** **

**Sol:**

**Form the above figure:**

Let, **A (1, 0), B (3, 5) and C (5, 4)** be the vertices of **triangle ABC**.

**Now, the equation of line AB:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{5\;-\;0}{3\;-\;1}\right]\)

\(\boldsymbol{\Rightarrow }\) **2y = 5x – 5**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{5x\;-\;5}{2}}\)

**The Equation of line BC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-5)=(x-3)\times \left[\frac{4-5}{5-3}\right]\)

\(\boldsymbol{\Rightarrow }\) **2y – 10 = 3 – x**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{13\;-\;x}{2}}\)

**The Equation of line AC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right]\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-1)\times \left[\frac{4-0}{5-1}\right]\)

\(\boldsymbol{\Rightarrow }\) **4y = 4x – 4**

\(\boldsymbol{\Rightarrow }\) ** y = x – 1**

Now, the** Area **of** triangle ABC = Area **under the curve** ABMA + Area **under the curve** MBCN – Area **under the curve** ACNA.**

**The Area under the curve ABMA [2y = 5x – 5]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{3}y\;dx\;=\;\int_{1}^{3}\frac{5x-5}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{2}}{4}-\frac{5x}{2} \right ]_{1}^{3}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{45}{4}-\frac{15}{2}-\frac{5}{4}+\frac{5}{2}=5}\) **unit ^{2}**

Therefore**, Area **under the** curve ABMA = 5 unit ^{2}**

**The Area under the curve MBCN [2y = 13 – x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}\frac{13-x}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{13x}{2}-\frac{x^{2}}{4} \right ]_{3}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{65}{2}-\frac{25}{4}-\frac{39}{2}+\frac{9}{4}=9}\) **unit ^{2}**

**Therefore, Area **under curve** MBCN = 9 unit ^{2}**

**The Area under the curve ACNA [y = x – 1]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{5}y\;dx\;=\;\int_{1}^{5}(x-1)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{2}}{2}-x \right ]_{1}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{25}{2}-5-\frac{1}{2}+1=8}\)** unit ^{2}**

**Therefore, Area under the curve ACNA = 8 unit ^{2}**

Now, **Area** of **triangle** **ABC** = **Area** under curve **ABMA** + **Area** under curve **MBCN** – **Area** under curve **ACNA**.

**Therefore, the Area of triangle ABC = 5 + 9 – 8 = 6** **unit ^{2}**

** **

**Exercise – 8.2**

** **

** Q.1: Find the area lying above the x-axis enclosed between two curves whose equations are given as: x ^{2 }+ y^{2} = 6x and y^{2} = 3x.**

** **

**Sol:**

The Equation **y ^{2} = 3x** represents a

**parabola**

**symmetrical**about

**x-axis**.

The Equation **x ^{2 }+ y^{2} = 6x** i.e.

**(x – 3)**represents a

^{2 }+ y^{2}= 9**circle**with

**centre**

**(3, 0)**and

**radius**

**3**

**units**.

Substituting the equation of parabola in equation of circle:

(x – 3)^{2 }+ 3x = 9 \(\boldsymbol{\Rightarrow}\) **x ^{2 }+ 9 – 6x + 3x = 9**

\(\boldsymbol{\Rightarrow}\) x^{2} – 3x = 0 i.e. x = 0 or x = 3 which gives **y = 0 and y = ****±**** 3**

Therefore, the **coordinates** of **point A** above the **x-axis** are **(3, 3)**

Now, the Area of region bounded by the curve **OQABO** = Area of region bounded by the curve **OQAMO **+ Area of region bounded by the curve **ABMA**

**The Area of region bounded by the curve OQAMO [y ^{2} = 3x]:**

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\int_{0}^{3}y\;dx=\int_{0}^{3}\sqrt{3x}\;dx}\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\sqrt{3}\times\left [ \frac{2x^{\frac{3}{2}}}{3} \right ]_{0}^{3}=\sqrt{3}\times \frac{2}{3}\times 3^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\sqrt{3}\times \frac{2}{3}\times 3\sqrt{3}=6}\)** unit ^{2}**

**Therefore, the Area of region bounded by the curve OQAMO = 6 unit ^{2}**

**Now, the Area of region bounded by the curve ABMA [(x – 3) ^{2 }+ y^{2} = 9)]:**

\(\boldsymbol{\Rightarrow}\) \(\boldsymbol{\int_{3}^{6}y\;dx=\int_{3}^{6}\sqrt{9-(x-3)^{2}}\;dx}\\\)

**Since,** \(\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}\\\)

\(\boldsymbol{\Rightarrow}\) \(\boldsymbol{\left [ \frac{x-3}{2}\sqrt{\left (3 \right )^{2}-\left ( x-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{x-3}{3} \right ]_{3}^{6}}\\\)

\(\boldsymbol{\Rightarrow \;\;\;\left [\frac {6-3}{2}\sqrt{{9}-\left (6-3\right )^{2}}+\frac{9}{2}\sin^{-1}\frac{6-3}{3}\right]-\left [ \frac{x-3}{2}\;\sqrt{9-\left (3-3 \right )^{2}}\;+\frac{9}{2}\;\sin^{-1}\frac{3-3}{3}\;\right]}\\\)\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\left [0 +\frac{9}{2}\times \sin^{-1}1-0\right ]=\frac{9}{2}\times \frac{\pi }{2}}\\\)

\(\\\boldsymbol{\Rightarrow}\) \(\boldsymbol{\frac{9\pi }{4}}\) **unit ^{2}**

**Therefore, the Area of region bounded by the curve ABMA \(=\frac{9\pi }{4}\) unit ^{2}**

**The Area of region bounded by the curve OQABO = Area of region bounded by the curve OQAMO + Area of region bounded by the curve ABMA**

\(\boldsymbol{\Rightarrow}\) \(6+\frac{9\pi }{4}=\frac{24+9\pi }{4}\) unit^{2}

**Therefore, the Area of shaded region (OQABO)\(=\frac{24+9\pi }{4}\) unit ^{2}**

** **

** **

**Q.2: Find the area enclosed between two curves: 9x ^{2 }+ 9y^{2 }= 4 and (x –**

**\(\frac{2}{3}\))**

^{2 }+ y^{2 }= \(\frac{4}{9}\)** **

**Sol:**

Equation **9x ^{2 }+ 9y^{2 }= 4 . . . . . . . (1)**, represents a circle with

**centre (0, 0)**and

**radius 3 units**.

Equation (**x –** **\(\frac{2}{3}\) ) ^{2 }+ y^{2 }=**

**\(\frac{4}{9}\)**

**. . . . . . . . . . .(2)**, represents a circle with

**centre (3, 0)**and

**radius 3 units**.

**On solving equation (1) and equation (2), we will get:**

\(\\\boldsymbol{\Rightarrow }\) \((x-\frac{2}{3})^{2}+\frac{4-9x^{2}}{9}=\frac{4}{9}\)

\(\\\boldsymbol{\Rightarrow }\) \(x^{2}+\frac{4}{9}-\frac{4x}{3}+\frac{4}{9}-x^{2}=\frac{4}{9}\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{4}{9}=\frac{4}{3}\;x\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{x=\frac{1}{3}}\) which gives \(\boldsymbol{y=\pm \frac{1}{\sqrt{3}}}\)

Therefore, the **coordinates** of **points M** and **N** are: \((x=\frac{1}{3})\;\;(y=\pm \frac{1}{\sqrt{3}})\)

**Now, the Area enclosed by region BMONB = 2 ****×**** [Area enclosed by the curve BMOB].**

And the Area enclosed by the curve **BMOB** = Area of region enclosed by the curve **(MOPM+MPBM).**

**Now, the Area of region enclosed by the curve MPBM [****9x ^{2 }+ 9y^{2 }= 4**

**]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\frac{1}{3}}^{\frac{2}{3}}y\;dx=\int_{\frac{1}{3}}^{\frac{2}{3}}\sqrt{\frac{4}{9}-x^{2}}\;dx}\)

**Since,** \(\\\int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-x^{2}}+\frac{4}{2\times 9}\sin^{-1}\frac{3x}{2} \right ]_{\frac{1}{3}}^{\frac{2}{3}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\frac{1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\sin^{-1}(1) \right ] -\left [ \frac{1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\sin^{-1}\frac{1}{2}\right ] }\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 0+\frac{2}{9}\times \frac{\pi }{2} \right ]-\left [ \frac{1}{6}\times \frac{1}{\sqrt{3}}+\frac{2}{9}\times \frac{\pi }{6} \right ]=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]}\) **unit ^{2}**

**Therefore, area of region enclosed by the curve MPBM \(=\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]\) unit ^{2}**

**Now, the Area of region enclosed by the curve MOPM \(\left [ \left ( x-\frac{2}{3} \right )^{2}+y^{2}=\frac{4}{9} \right ]\) :**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{3}}y\;dx=\int_{0}^{\frac{1}{3}}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}\;dx}\\\)

**Since**, \(\\\int \sqrt{a^{2}-(x-b)^{2}}\;dx = \frac{x-b}{2}\sqrt{a^{2}-(x-b)^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x-b}{a}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x-\frac{2}{3}}{2}\sqrt{\left ( \frac{2}{3} \right )^{2}-\left ( x-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{x-\frac{2}{3}}{\frac{2}{3}} \right ]_{0}^{\frac{1}{3}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{\frac{1}{3}-\frac{2}{3}}{2}\sqrt{\frac{4}{9}-\left ( \frac{1}{3}-\frac{2}{3} \right )^{2}}+\frac{2}{9}\sin^{-1}\frac{\frac{1}{3}-\frac{2}{3}}{\frac{2}{3}} \right]- \left [ \frac{0-\frac{2}{3}}{2}\;\sqrt{\frac{4}{9}-\left ( 0-\frac{2}{3} \right )^{2}}\;+\frac{2}{9}\;\sin^{-1}\frac{0-\frac{2}{3}}{\frac{2}{3}}\; \right ]}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{6}\times \sqrt{\frac{4}{9}-\frac{1}{9}}+\frac{2}{9}\times \sin^{-1}\frac{-1}{2} \right ]- \left [ \frac{-1}{3}\times \sqrt{\frac{4}{9}-\frac{4}{9}}+\frac{2}{9}\times \sin^{-1}(-1) \right ]}\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{6\sqrt{3}}+\frac{2}{9}\times \frac{-\pi }{6} \right ]-\left [ \frac{2}{9}\times \frac{-\pi }{2} \right ]= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]}\) **unit ^{2}**

**Therefore, the Area of region enclosed by the curve MOPM \(= \left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ]\) unit ^{2}**

Now, the Area enclosed by the **curve BMOB** = Area of region enclosed by the curve **(MOPM+MPBM)**

\(\boldsymbol{\Rightarrow }\) \(\left [ \frac{2\pi }{27}-\frac{1}{6\sqrt{3}} \right ] +\left [\frac{2\pi }{27}-\frac{ 1}{6\sqrt{3}} \right ]\\\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]}\) **unit ^{2}**

Now, the Area enclosed by the curve **BMONB** = 2 [Area enclosed by the curve **BMOB**] \(=2\times \left [ \frac{4\pi }{27}-\frac{1}{3\sqrt{3}} \right ]\) unit^{2}

**Therefore, the Area of shaded region = \(\left [ \frac{8\pi }{27}-\frac{2}{3\sqrt{3}} \right ]\) unit ^{2}**

** **

** **

**Q.3: Find the area lying above the x-axis enclosed between two curves whose equations are given as: 4x ^{2 }+ 4y^{2} = 9 and x^{2} = 4y.**

** **

**Sol:**

The Equation **x ^{2} = 4y** represents a

**parabola**

**symmetrical**about

**y-axis**.

The Equation **4x ^{2 }+ 4y^{2} = 9** i.e.

**x**represents a

^{2 }+ y^{2}= \(\frac{3}{2}\)**circle**with

**centre**

**(0, 0)**and

**radius**

**\(\frac{3}{2}\)**

**units**.

Now, on substituting the equation of parabola in the equation of circle we will get:

4(4y) + 4y^{2} =9 i.e.** 4y ^{2 }+ 16y – 9 = 0**

From the above quadratic equation: **a = 4, b = 16 and c = -9**

Substituting the values of a, b and c in **quadratic** **formula**:

\(\\\boldsymbol{\Rightarrow }\) \(\\y=\frac{-16+\sqrt{(16)^{2}-4(4\times-9)}}{2\times 4}\;and\;y=\frac{-16-\sqrt{(16)^{2}-4(4\times -9)}}{2\times 4}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+\sqrt{256+144}}{8}\;and\;y=\frac{-16-\sqrt{256+144}}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+\sqrt{400}}{8}\;and\;y= \frac{-16-\sqrt{400}}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\\ y=\frac{-16+20}{8}\;and\;y=\frac{-16-20}{8}\)

\(\\\boldsymbol{\Rightarrow }\) \(y = \frac{1}{2}\;\;and\;\;y = \frac{-9}{2}\)

Which gives x = \(\pm \;\sqrt{2}\) [Neglecting y = \(\frac{-9}{2}\) as it gives absurd results]

Therefore, the **coordinates** of **point** **B** are \(\left (\sqrt{2},\;\frac{1}{2} \right )\)

Now, Area of region bounded by the curve **ODCBO** = **2 ****×** (Area of region bounded by the curve **OBCO**)

Now, Area of region bounded by curve **OBCO** = (Area of region bounded by the curve **OCBMO** + Area of region bounded by the curve **OBMO**)

**Area of region bounded by the curve OCBMO [4x ^{2 }+ 4y^{2} = 9]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\sqrt{\left ( \frac{3}{2} \right )^{2}-x^{2}}\;\;dx}\)

**Since,** \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{\frac{9}{4}- x^{2}}+\frac{9}{2\times 4}\sin^{-1}\frac{2\times x}{3} \right ]_{0}^{\sqrt{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\frac {\sqrt{2}}{4}\times \sqrt{{\frac{9}{4}-2}}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right ]\\}\) **unit ^{2}**

**Therefore, Area of region bounded by the curve ABMA = \(\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]\) unit ^{2}**

**Area of region bounded by the curve OBMO [x ^{2} = 4y]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{2}}y\;dx=\int_{0}^{\sqrt{2}}\frac{x^{2}}{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^3}{12} \right ]_{0}^{\sqrt{2}}=\frac{2\sqrt{2}}{12}-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{3\sqrt{2}}}\) **unit ^{2}**

**Therefore, the area of region bounded by the curve OBMO = \(\frac{1}{3\sqrt{2}}\) unit ^{2}**

**Now, the Area of region bounded by curve OBCO = [Area of region bounded by the curve OCBMO – Area of region bounded by the curve OBMO]**

\(\\\boldsymbol{\Rightarrow }\) **\(\left [\sqrt{2}\times \frac{1}{4}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3}\right]-\frac{1}{3\sqrt{2}}\)**

\(\\\boldsymbol{\Rightarrow }\) **\(\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit ^{2}**

**Therefore,** **Area of region bounded by the curve** **OBCO**** = ****\(\left [ \frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit ^{2}**

**Now, the Area of region bounded by the curve ODCBO = 2 ****×**** [Area of region bounded by the curve OBCO]**

\(\\\boldsymbol{\Rightarrow }\) \(2\times \left [\frac{\sqrt{2}}{12}+\frac{9}{8}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\)

**Therefore, the Area of shaded region = \(\left [ \frac{1}{2\sqrt{2}}+\frac{9}{4}\sin^{-1}\frac{2\sqrt{2}}{3} \right ]\) unit ^{2}**

** **

** **

**Q.4: Find the area enclosed by the sides of a triangle whose vertices have coordinates (-2, 0) (3, 4) and (5, 2).**

** **

**Sol:**

**Form the above figure:**

Let, **A (1, 0), B (3, 5) and C (5, 4)** be the vertices of **triangle ABC**

**Now, the equation of line AB:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-0)\;=\;(x+2)\times \left[\frac{4\;-\;0}{3\;+\;2}\right]\)

\(\\\boldsymbol{\Rightarrow }\) **5y = 4x + 8**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{y=\frac{4x\;+\;8}{5}}\)

**The Equation of line BC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-4)=(x-3)\times \left[\frac{2\;-\;4}{5\;-\;3}\right]\)

\(\\\boldsymbol{\Rightarrow }\) **2y – 8 = 6 – 2x**

\(\\\boldsymbol{\Rightarrow }\) **y = 7 – x**

**The Equation of line AC:**

Since, \((y-y_{1})=(x-x_{1})\times \left[\frac{y_{2}\;-\;y_{1}}{x_{2}\;-\;x_{1}}\right]\)

\(\\\boldsymbol{\Rightarrow }\) \((y-0)=(x+2)\times \left[\frac{2\;-\;0}{5\;+\;2}\right]\)

\(\\\boldsymbol{\Rightarrow }\) **7y = 2x + 4**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{ y= \frac{2\;x+4}{7}}\)

**Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.**

**The Area under the curve ABMA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{3}y\;dx\;=\;\int_{-2}^{3}\frac{4x\;+\;8}{5}\;dx}\\\) = \(\boldsymbol{\left [ \frac{4x^{2}}{10}+\frac{8x}{5} \right ]_{-2}^{3}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{36}{10}+\frac{24}{5}-\frac{16}{10}-\frac{-16}{5}=10}\) **unit ^{2}**

**Therefore, the Area under the curve ABMA = 10 unit ^{2}**

**The Area under the curve MBCN:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}=(7-x)\;dx}\\\) = \(\boldsymbol{\left [ 7x-\frac{x^{2}}{2} \right ]_{3}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{35-\frac{25}{2}- 21+\frac{9}{2}=6}\) **unit ^{2}**

**Therefore, the Area under the curve MBCN = 6 unit ^{2}**

**The Area under the curve ACNA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{5}y\;dx\;=\;\int_{-2}^{5}\frac{2x\;+\;4}{7}\;dx}\\\) = \(\boldsymbol{\left [ \frac{2x^{2}}{14}+\frac{4x}{7}\right ]_{-2}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{25}{7}+\frac{20}{7}-\frac{4}{7}-\frac{-8}{7}=7}\) **unit ^{2}**

**Therefore, the Area under the curve ACNA = 7 unit ^{2}**

Now, **Area** of triangle **ABC** = Area under curve **ABMA** + Area under curve **MBCN** – Area under curve **ACNA**.

**Therefore, the Area of triangle ABC = 10 + 6 – 7 = 9**** unit ^{2}**

** **

** **

**Q.5: Find the area enclosed by the sides of a triangle whose equations are: y = 4x + 2, y = 3x + 2 and x = 5.**

** **

**Sol:**

** **

**From the above figure,**

The Equation of **side AC: y = 4x + 2 . . . . . . (1)**

The Equation of **side BC: y =3x + 2 . . . . . . (2)**

**And,** **x = 5 . . . . . . . . . . (3)**

From **equation (1) and equation (3):**

y = 4(5) + 2 = 22 **[Since, x = 5]**

Therefore, the **coordinates** of **point** **A** are **(5, 22).**

**From equation (2) and equation (3):**

y = 3(5) + 2 = 17 **[Since, x = 5]**

Therefore, the **coordinates** of **point** **B** are **(5, 17).**

**Now, substituting equation (1) in equation (2):**

3x + 2 = 4x + 2 **i.e. x = 0 and y = 2**

Therefore, the **coordinates** of **point** **C** are **(0, 2).**

Now, the **Area** of triangle **ABC** = Area enclosed by the curve **ACOMA** – Area enclosed by the curve **COMBC**.

**The Area under the curve ACOMA [****y = 4x + 2****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(4x+2)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4x^{2}}{2}+2x \right ]_{0}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{50+10-0=60}\) **unit ^{2}**

**Therefore, the Area under the curve ABMA = 60 unit ^{2}.**

**The Area under the curve MBCN [y = 3x + 2]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{5}y\;dx\;=\;\int_{0}^{5}(3x+2)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3x^{2}}{2}+2x \right ]_{0}^{5}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{75}{2}+10-0=47.5}\) **unit ^{2}**

**Therefore, the Area under the curve MBCN = 47.5 unit ^{2}.**

**Now, the Area of triangle ABC = Area enclosed by the curve ACOMA – Area enclosed by the curve COMBC**

\(\\\boldsymbol{\Rightarrow }\) 60 – 47.5 **= 12.5 unit ^{2}**

**Therefore, the Area of triangle ABC = 12.5 unit ^{2}**

^{ }

^{ }

**Q.6: Find the area enclosed by the curves y = x ^{2 }+ 3, y = 2x, x = 2 and x =0.**

** **

**Sol:**

The Equation **y = x ^{2 }+ 3** represents a

**parabola**

**symmetrical**about

**y-axis**.

On substituting equation of line x = 2 in the equation of parabola we will get the **coordinates** of **point** **C** i.e. **(2, 7).**

On substituting **x = 2** in the equation of line **y = 2x** we will get the **coordinates** of **point B** i.e. **(2, 4).**

From the above figure,

The **Area** of region enclosed by the curve **ODCBO** = **Area** of region enclosed by the curve **ODCAO** – **Area** of region enclosed by the curve **OBAO**

**Now, the Area of region enclosed by the curve ODCAO [****y = x ^{2 }+ 3**

**]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(x^{2}+3)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{3}}{3}+3x \right ]_{0}^{2}=\frac{8}{3}+6=\frac{26}{3}}\)**unit ^{2}**

**Therefore, the Area of region enclosed by the curve ODCAO = \(\frac{26}{3}\) unit ^{2}**

**Now, the Area of region enclosed by the curve OBAO [y = 2x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx = \int_{0}^{2}(2x)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [x^{2} \right ]_{0}^{2}=4}\) **unit ^{2}**

**Therefore, the Area of region enclosed by the curve OBAO = 4 unit ^{2}**

Now, the **Area** of region enclosed by the **curve ODCBO** = Area of region enclosed by the **curve ODCAO** – Area of region enclosed by the **curve OBAO**

\(\Rightarrow \frac{26}{3}-4=\frac{2}{3}\) unit^{2}

**Therefore, The Area of shaded region (ODCBO) = \(\frac{2}{3}\)unit ^{2}**

^{ }

^{ }

**Q.7: Find the area enclosed between the curve y ^{2 }= 6x and line y = 3x.**

** **

**Sol:**

Equation **y ^{2} = 6x** represents a parabola, symmetrical about

**x-axis**.

Now, substituting the Equation of line **y = 3x** in the equation of parabola:

(3x)^{2} = 6x **\(\Rightarrow x = \frac{2}{3}\) **which gives **y = 2**

Hence the **coordinates** of point **A** are **\((\frac{2}{3},2)\)**

**The area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

**Now, the Area enclosed by the curve OMABO [y ^{2 }= 6x]:**

**Since, y ^{2} = 6x**

**Therefore, y = \(\sqrt{6x}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx = \int_{0}^{\frac{2}{3}}\sqrt{6x}\;dx}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{6}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{2}{3}}=\sqrt{6}\times \frac{2}{3}\times \left ( \frac{2}{3} \right )^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{2} \times \sqrt{3}\times \frac{2}{3}\times 2\sqrt{2}\times \frac{1}{3\sqrt{3}}=\frac{8}{9}}\\\) **unit ^{2}**

**Therefore, the Area enclosed by the curve OMABO = \(\frac{8}{9}\) unit ^{2}**

**Now, the Area enclosed by the curve OAMO [y = 3x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{2}{3}} y\;dx\Rightarrow \int_{0}^{\frac{2}{3}}3x\;dx}\)

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{3\left [\frac{x^{2}}{2}\right ]_{0}^{\frac{2}{3}}=\frac{3}{2}\times \left ( \frac{2}{3} \right )^2}\) = \(\boldsymbol{\frac{2}{3}}\)** unit ^{2}**

**Therefore, the Area enclosed by the curve OAMO = \(\frac{2}{3}\) ****unit ^{2}**

**Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

\(\\\boldsymbol{\Rightarrow }\) \(\frac{8}{9}-\frac{2}{3} = \frac{2}{9}\) unit^{2}

**Therefore, the Area enclosed by the curve OABO = \(\frac{2}{9}\) unit ^{2}**

**Miscellaneous Exercise**

**Q.1: Find the area enclosed by the curve whose equations are: y = 2x ^{2}, x = 2, x = 3 and x-axis.**

**Sol:**

The equation **y = 2x ^{2}** represents a

**parabola**symmetrical about

**y-axis**.

**Now, the Area of region enclosed by the curve ABCDA [****y = 2x ^{2}**

**]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2}^{3}y\;dx=\int_{2}^{3}2x^{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{2x^{3}}{3} \right ]_{2}^{3}=\left [ 18-\frac{16}{3} \right ]=\frac{38}{3}}\) **unit ^{2}**

**Therefore, the Area of shaded region \(=\frac{38}{3}\) unit ^{2}**

** **

** **

**Q.2: Find the area enclosed by the curve whose equations are: y = 5x ^{4}, x = 3, x = 7 and x-axis.**

** **

**Sol:**

The equation **y = 2x ^{2}** represents a

**quartic**

**parabola**symmetrical about

**y-axis**.

**Now, the Area of region enclosed by the curve ABCDA [****y = 5x ^{4}**

**]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{7}y\;dx=\int_{3}^{7}5x^{4}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{5}}{5} \right ]_{3}^{7}=\left [ 16807-243 \right ]=16564}\) **unit ^{2}**

**Therefore, the Area of shaded region = 16807 unit ^{2}**

** **

** **

**Q.3****: Find the area enclosed between the curve y ^{2 }= 3x and line y = 6x.**

** **

**Sol:**

Equation **y ^{2} = 3x** represents a parabola, symmetrical about

**x-axis**.

Now, substituting the Equation of line **y = 6x** in the equation of parabola:

**(6x) ^{2} = 3x**

**\(\Rightarrow x = \frac{1}{12}\)**which gives

**y = \(\frac{1}{2}\)**

Hence the **coordinates** of point **A** are **\((\frac{1}{12},\frac{1}{2})\)**

**The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

**Now, the Area enclosed by the curve, OMABO [y ^{2 }= 3x]:**

**Since,** **y ^{2} = 3x**

Therefore,** y = \(\sqrt{3x}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{12}} y\;dx = \int_{0}^{\frac{1}{12}}\sqrt{3x}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{3}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{1}{12}}=\sqrt{3}\times \frac{2}{3}\times \left ( \frac{1}{12} \right )^{\frac{3}{2}}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\sqrt{3} \times \frac{2}{3}\times \frac{1}{24\sqrt{3}}=\frac{1}{36}}\)** unit ^{2}**

**Therefore, the Area enclosed by the curve OMABO = \(\frac{1}{36}\)****unit ^{2}**

**Now, the Area enclosed by the curve OAMO [y = 6x]:**

\(\\\boldsymbol{\Rightarrow }\) \(Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{1}{12}\times \frac{1}{2}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{48}}\)**unit ^{2}**

**Therefore, the Area enclosed by the curve OAMO = \(\frac{1}{48}\)****unit ^{2}**

**Now, the area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{36}-\frac{1}{48} = \frac{1}{144}\)**unit ^{2}**

**Therefore, the Area enclosed by the curve OABO = \(\frac{1}{144}\)****unit ^{2}**

** **

** **

**Q.4 Find the area enclosed by the curve y = 2x ^{2} and the lines y = 1, y = 3 and the y-axis.**

** **

**Sol:**

Equation **y = 2x ^{2}** represents a

**parabola**symmetrical about

**y-axis**.

** **The **Area** of the region bounded by the curve **y = 2x ^{2}**,

**y = 1**, and

**y**

**=**

**3**, is the

**Area**enclosed by the curve

**AA’B’BA**.

Now, the **Area of region AA’B’BA = 2 (Area of region ABNMA)**

**Since,** **2x ^{2} = y**

Therefore, **x = \(\sqrt{\frac{y}{2}}\)**

Thus, **the Area of region bounded by the curve ABNMA [****y = 2x ^{2}**

**]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{3}x\;dy=\int_{1}^{3}\sqrt{\frac{y}{2}}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{\sqrt{2}}\times {\left | \frac{2}{3}\times (y)^{\frac{3}{2}} \right |_{1}^{3}=\frac{\sqrt{2}}{3}\times [(3^{\frac{3}{2}})-(1)^{\frac{3}{2}}}}]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{\sqrt{2}}{3}\times (3\sqrt{3}-1) = \frac{\sqrt{2}}{3}(3\sqrt{3}-1)}\)**unit ^{2}**

Therefore, the **Area** of region **bounded** by the **curve ABNMA = \(\frac{\sqrt{2}}{3}(3\sqrt{3}-1)\)****unit ^{2}**

**Hence, the ****Area of region bounded by the curve AA’B’BA ****=**** 2(Area of region bounded by the curve ABNMA)****\(= \frac{2\sqrt{2}}{3}(3\sqrt{3}-1)\)****unit ^{2}**

**Q.5****: Find the area enclosed between the curve y ^{2 }= 9ax and line y = mx.**

** **

**Sol:**

Equation **y ^{2} = 9ax** represents a parabola, symmetrical about

**x-axis**.

Now, substituting the Equation of line **y = mx** in the equation of parabola:

9ax = (mx)^{2} i.e **x = \(\frac{9a}{m^{2}}\)** which gives **y = \(\frac{9a}{m}\)**

Hence the **co-ordinates** of point **A** are **\(\left ( \frac{9a}{m^{2}},\frac{9a}{m}\right )\)**

**The Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

**Now, the Area enclosed by the curve OMABO [y ^{2 }= 9ax]:**

**Since,** **y ^{2} = 9ax**

Therefore,** y = \(3\sqrt{ax}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{9a}{m^{2}}} y\;dx = \int_{0}^{\frac{9a}{m^{2}}}3\sqrt{ax}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{3\sqrt{a}\left [ \frac{2}{3}\times x^{\frac{3}{2}} \right ]_{0}^{\frac{9\;a}{m^{2}}}=3\sqrt{a}\times \frac{2}{3}\times \left ( \frac{9\;a}{m^{2}} \right )^{\frac{3}{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{2\sqrt{a} \times 27\times a\sqrt{a}\times \frac{1}{m^{3}}=\frac{54\;a^{2}}{m^{3}}}\)** unit ^{2}**

**Therefore, the Area enclosed by the curve OMABO \(=\frac{54\;a^{2}}{m^{3}}\) unit ^{2}**

**Now, the Area enclosed by the curve OAMO [y = mx]:**

\(\\\boldsymbol{\Rightarrow }\) \(Area\;of \;\Delta OAM = \frac{1}{2}\times Base\times Altitude\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{1}{2}\times |OM|\times |AM|=\boldsymbol{\frac{1}{2}\times \frac{9\;a}{m^{2}}\times \frac{9\;a}{m}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{81\;a^{2}}{2\;m^{3}}}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve OAMO =\(\frac{81\;a^{2}}{2\;m^{3}}\) unit ^{2}**

**Now, the Area of region enclosed by the curve OABO = Area enclosed by the curve OMABO – Area enclosed by the curve OAMO**

i.e. \(\frac{54\;a^{2}}{m^{3}}-\frac{81\;a^{2}}{2\;m^{3}} = \frac{27\;a^{2}}{2\;m^{3}}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve OABO = \(\frac{27\;a^{2}}{2\;m^{3}}\)unit ^{2}**

** **

**Q.6: Find the area bounded by the curve whose equation is 3x ^{2} = 4y and the line 2y – 12 = 3x.**

** **

**Sol:**

**Equation 3x ^{2} = 4y **represents a

**parabola**, symmetrical about the

**y-axis**as shown in the above figure.

The **Area** of the region bounded by parabola **3****x ^{2} = 4y** and the line

**2y -12 = 3x**is the

**Area**enclosed under the curve

**ABC0A**.

Since, the **parabola 3x ^{2} = 4y** and the

**line 3x = 2y – 12**intersect each other at points

**A**and

**C**, hence the coordinates of

**points**

**A**and

**C**are given by:

**Since,** \(\;x=\frac{2y-12}{3}\)

\(\\\boldsymbol{\Rightarrow }\) \(3\left ( \frac{2y-12}{3} \right )^{2}=4y\;\;\;i.e. \;\;\;(2y-12)^{2}=12y\)

\(\\\boldsymbol{\Rightarrow }\) 4y^{2} +144 – 48y – 12 y = 0

\(\\\boldsymbol{\Rightarrow }\) y^{2} – 15y + 36 = 0

By **splitting the middle term Method** solutions of this quadratic equation are:

y^{2} – (12+3)y + 36 = 0 \(\Rightarrow\) y(y – 12) –3(y – 12) = 0

\(\Rightarrow\) (y – 3) (y – 12) = 0

Therefore, y = 12 and y = 3 which gives x = 4 and x = -2 respectively.

Hence, the **co-ordinates** of **point A** and **point C** are** (-2, 3)** and** (4, 12) **respectively**.**

**Since,** 3x^{2} = 4y

Therefore, **y = \(\frac{3x^{2}}{4}\)**

**The Area of region bounded by the curve ABCOA = [Area of region aACba] – [Area of region OCbO + Area of region OAaO ]**

**The Area enclosed by the curve aACBa [2y – 12 = 3x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{4} y\;dx \Rightarrow \int_{-2}^{4}\frac{3x+12}{2}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\left | \frac{3x^{2}}{2}+ 12x \right |_{-2}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\left [ 24+48-6-(-24) \right ]=45}\) **unit ^{2}**

**The Area enclosed by the curve OAaO [****3x ^{2} = 4y**

**]:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} \frac{3x^{2}}{4}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{-2}^{0}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{3}{4}(0-\frac{-8}{3}) \right |=2}\) **unit ^{2}**

**The Area enclosed by the curve OCbO [****3x ^{2} = 4y**

**]:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{4}\;y\;dx\;\Rightarrow \int_{0}^{4} \frac{3x^{2}}{4}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{4}\left | \frac{x^{3}}{3}\right |_{0}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{3}{4}(\frac{64}{3}-0) \right |=16}\) **unit ^{2}**

**Now, the Area of region bounded by the curve ABCOA = [ Area of region aACba ] – [ Area of region OCbO + Area of region OAaO ]**

\(\\\boldsymbol{\Rightarrow }\) **45 – [2 + 16] = 27 unit ^{2}**

**Therefore, the Area of shaded region ABCOA = 27 unit ^{2}**

** **

** **

**Q.7: Find the area enclosed by the curves {(x , y) : 6y ****≥ ****x ^{2 }and y = |x|}**

** **

**Sol:**

**Equation x ^{2} = 6y **represents a

**parabola**, symmetrical about the

**y-axis**as shown in the above figure.

The Area of the region bounded by the curve **x ^{2} = 6y** and

**y = |x|**is

**2(OAEO**) i.e.

**(area OCFO+ area OAEO)**

Now, **Area of region OAEO = OABO – OEABO**

**Since,** **x ^{2} = 6y**

Therefore, \(y=\frac{x^{2}}{6}\)

Now, **the Area of region bounded by the curve OEABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}\frac{x^{2}}{6}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{6}\left | \frac{x^{3}}{3} \right |_{0}^{6}=12}\) unit^{2}

**Therefore, the Area of region bounded by the curve OEABO = 12 unit ^{2}**

**Now, the Area of region bounded by the curve OABO:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{6}y\;dx\;\Rightarrow \;\int_{0}^{6}x\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{x^{2}}{2} \right |_{0}^{6}=18}\) **unit ^{2}**

**Therefore, the Area of the region bounded by the curve OABO = 18 unit ^{2}**

**Now, ****Area of region OAEO = Area of region (OABO – OEABO)**

\(\\\boldsymbol{\Rightarrow }\) 18 – 12 = **6 unit ^{2}**

**Therefore, the total Area of shaded region = ****2 ****× 6 ****= 12 unit ^{2}**

** **

** **

**Q.8****: Find the area enclosed by the sides of a triangle whose vertices have coordinates (3, 0) (5, 8) and (7, 5).**

** **

**Sol:**

**Form the above figure:**

**Let,** **A (3, 0), B (5, 8) and C (7, 5)** be the vertices of **triangle ABC**.

**Now, the equation of line AB:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-3)\times \left(\frac{8-0}{5-3}\right)\)

\(\boldsymbol{\Rightarrow }\) 2y = 8x – 24

\(\boldsymbol{\Rightarrow }\) ** y=4x-12**

**The Equation of line BC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-8)=(x-5)\times \left(\frac{5-8}{7-5}\right)\)

\(\boldsymbol{\Rightarrow }\) 2y – 16 = -3x + 15

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\Rightarrow y=\frac{31-3x}{2}}\)

**The Equation of line AC:**

**Since,** \((y-y_{1})=(x-x_{1})\times \left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\)

\(\boldsymbol{\Rightarrow }\) \((y-0)=(x-3)\times \left(\frac{5-0}{7-3}\right)\)

\(\boldsymbol{\Rightarrow }\) 4y=5x-15

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\Rightarrow y=\frac{5x-15}{4}}\)

**Now, the Area of triangle ABC = Area under the curve ABMA + Area under the curve MBCN – Area under the curve ACNA.**

**The Area under the curve ABMA [y = 4x – 12]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{5}y\;dx\;=\;\int_{3}^{5}(4x-12)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{4x^{2}}{2}-12x\right ]_{3}^{5}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[50-60]-[18-36]=8}\)**unit ^{2}**

**Therefore, Area under the curve ABMA = 8 unit ^{2}**

**The Area under the curve MBCN [2y + 3x = 31]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{5}^{7}y\;dx\;=\;\int_{5}^{7}\frac{31-3x}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times \left [ 31x-\frac{3x^{2}}{2}\right ]_{5}^{7}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[\frac{217}{2}-\frac{147}{4}]-[\frac{155}{2}-\frac{75}{4}]=13}\) **unit ^{2}**

**Therefore, Area under the curve MBCN = 13 unit ^{2}**

**The Area under the curve ACNA [4y = 5x – 15]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{3}^{7}y\;dx\;=\;\int_{3}^{7}\left ( \frac{5x-15}{4} \right )dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{5x^{2}}{8}-\frac{15x}{4} \right ]_{3}^{7}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{245}{8}-\frac{105}{4} \right ]-\left [ \frac{45}{8}-\frac{45}{4} \right ]=10}\)** unit ^{2}**

**Therefore, Area under the curve ACNA = 10 unit ^{2}**

**Now, Area of triangle ABC = Area under curve ABMA + Area under curve MBCN – Area under curve ACNA**.

**Therefore, the Area of triangle ABC = 8 + 13 – 10 = 11** **unit ^{2}**

** **

** **

**Q.9: Find the area enclosed by the sides of a triangle whose equations are: 2x – 4 = y, – 2y = -3x + 6 and x – 3y = -5.**

** **

**Sol:**

** **

**From the above figure:**

The Equation of **line AB: 3y = x + 5 . . . . . . (1)**

The Equation of **line BC: y = 4 – 2x . . . . . . (2)**

The Equation of **line AC: 2y = 3x – 6 . . . . . . (3)**

From **equation (1) **and** equation (2):**

3(4 – 2x) = x + 5 i.e. **x = 1**, which gives **y = 2**

Therefore, the **coordinates** of **point** **B** are **(1, 2)**

From **equation (2)** and **equation (3):**

2(4 – 2x) = 3x – 6 i.e. **x = 2** which gives **y = 0**

Therefore, the **coordinates** of **point** **C** are **(2, 0).**

From **equation** **(1)** and **equation (3):**

2y = 3(3y – 5) – 6 i.e. **y = 3** which gives **x = 4**

Therefore, the **coordinates** of **point** **A** are **(4, 3).**

Now, the **Area** of triangle **ABC** = **Area** enclosed by the curve **ABMNA** **–** **Area** enclosed by the curve **BMCB – Area **enclosed by the curve** ACNA**

**The Area under the curve ABMNA [****3y = x + 5****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{4}y\;dx\;=\;\int_{1}^{4}\frac{x+5}{3}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{2}}{6}+\frac{5x}{3} \right ]_{1}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{16}{6}+\frac{20}{3}-\frac{1}{6}-\frac{5}{3}=\frac{15}{2}}\) **unit ^{2}**

**Therefore, the Area under the curve ABMNA \(=\frac{15}{2}\) unit ^{2}**

**The Area under the curve BMCB [****y = 4 – 2x****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{1}^{2}y\;dx\;=\;\int_{1}^{2}(4-2x)\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 4x-x^{2} \right ]_{1}^{2}}\)

\(\\\boldsymbol{\Rightarrow }\) [8 – 4] – [4 – 1] **=1 unit ^{2}**

**Therefore, the Area under the curve MBCN = 1 unit ^{2}**

**The Area under the curve ACNA [****2y = 3x – 6****]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{2}^{4}y\;dx\;=\;\int_{2}^{4}\frac{3x-6}{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3x^{2}}{4}-\frac{6x}{2} \right ]_{2}^{4}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{3\times 16}{4}-\frac{6\times 4}{2} \right ]- \left [ \frac{3\times 4}{4}-\frac{6\times 2}{2} \right ]=3}\)** unit ^{2}**

**Therefore, the Area under the curve ACNA = 3 unit ^{2}**

The **Area** of triangle **ABC** = **Area** enclosed by the curve **ABMNA** **–** **Area** enclosed by the curve **BMCB – Area **enclosed by the curve** ACNA**

**\(\\\boldsymbol{\Rightarrow }\) \(\frac{15}{2}-1-3 = 3.5\) unit ^{2}**

**Therefore, the Area of triangle ABC = 3.5 unit ^{2}**

** **

** **

**Q.10: Find the area enclosed by the curve 2x ^{2 }= y and the line y = 2x + 12 and x – axis.**

** **

**Sol:**

**Equation 2x ^{2} = y **represents a

**parabola**, symmetrical about the

**y-axis**as shown in the above figure.

The **Area** of the region bounded by parabola **2****x ^{2} = y** and the line

**y = 2x + 12 and x-axis**is the

**Area**enclosed under the curve

**ABCOA**.

Since, the **parabola 2x ^{2} = y** and the

**line y = 2x + 12**intersect each other at points

**A and C**, hence the coordinates of

**points**

**A**and

**C**are given by:

**Since, y = 2x + 12**

\(\\\boldsymbol{\Rightarrow }\) 2x^{2 }= (2x+12)

\(\\\boldsymbol{\Rightarrow }\) **x ^{2} – x – 6 = 0**

By **splitting the middle term Method** solutions of this quadratic equation are:

x^{2} – (3 – 2)x – 6 = 0 \(\Rightarrow\) x(x – 3) +2(x – 3) = 0

\(\Rightarrow\) (x – 3) (x + 2) = 0

Therefore, **x = 3** and **x = -2** which gives **y = 18** and **y = 8** respectively.

Hence, the **co-ordinates** of **point E** and **point A** are** (3, 18)** and** (-2, 8) **respectively**.**

**Since,** **2x ^{2} = y**

**Therefore,** **y = 2x ^{2}**

**The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]**

**The Area enclosed by the curve ACOA [ ****2x ^{2 }= y **

**]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-2}^{0}\;y\;dx\;\Rightarrow \int_{-2}^{0} 2x^{2}\;dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | \frac{2\;x^{3}}{3}\right |_{-2}^{0}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left |(0-\frac{-16}{3}) \right |=\frac{16}{3}}\) **unit ^{2}**

**The Area enclosed by the curve ABC [ ****y = 2x + 12 ****] :**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{Area\;of\;\Delta ABC=\frac{1}{2}\times Base\times Altitude}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times |BC|\times |AC|=\frac{1}{2}\times \left | 4 \right |\times |8|=16}\) unit^{2}

**The Area of region bounded by the curve ABCOA = [Area of region ACOA] + [Area of region ABC]**

\(\Rightarrow \frac{16}{3}+16=\frac{64}{3}\) unit^{2}

**Therefore, the Area of shaded region ABCOA \(=\frac{64}{3}\)unit ^{2}**

** **

** **

**Q.11: Plot the curve y = |x + 4| and hence evaluate \(\int_{-9}^{0}|x+4|\;dx\)**

** **

**Sol:**

From the given equation the corresponding values of x and y are given in the following table.

X |
-9 | -8 | -7 | -6 | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |

Y |
5 | 4 | 3 | 2 | 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

Now, on using these values of x and y, we will plot the graph of **y = |x + 4|**

** **

From the above graph,** the required Area = the Area enclosed by the curve ABCA + the Area enclosed by the curve CDOC**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{-9}^{0}|x+4|\;dx+\int_{-9}^{-4}(x+4)\;dx+\int_{-4}^{0}(x+4)\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-9}^{-4}+\left [ \left | \frac{x^{2}}{2}+4x \right | \right ]_{-4}^{0}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left | 8-16-\frac{81}{2}+36 \right |+\left | 0-(8-16) \right |=\left | \frac{-25}{2} \right |+8}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{41}{2}}\)** unit ^{2}**

**Therefore, the area of shaded region = \(\frac{41}{2}\) unit ^{2}**

**Q.12: Find the area enclosed by the curve y = sin x between 0 ****≤ ****x ****≤ 2π**

**Sol:**

From the above figure, the required Area is represented by the curve **OABCD**.

**Now, the Area bounded by the curve OABCD = the Area bounded by the curve OABO + the Area bounded by the curve BCDB**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\pi } sin(x)\;dx+\left | \int_{\pi }^{2\pi } sin(x)\;dx\right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos(x)\right ]_{0}^{\pi }+\left | \left [ -cos (x) \right ] _{\pi }^{2\pi }\right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{[-cos(\pi )+cos(0)]+\left | [-cos(2\pi )+cos(\pi )] \right |}\\\)

\(\boldsymbol{\Rightarrow }\) **[1+1] + [ |– 1 – 1| ] unit ^{2}**

**Therefore, the area of shaded region = 4 unit ^{2}**

** **

** **

**Q.13: ****Find the area of smaller region enclosed by the curve \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) and the line \(\frac{x}{2}+\frac{y}{3}=1\)**

** **

**Sol:**

The Equation **\(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)** represents an **ellipse.**

The Equation **\(\frac{x}{2}+\frac{y}{3}=1\)** represents a **line** with **x and y** **intercepts** as **2 and 3** respectively.

** **

**Since, \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)**

**\(\\\boldsymbol{\Rightarrow }\) \(\frac{y^{2}}{9}=1-\frac{x^{2}}{4}\)**

**\(\\\boldsymbol{\Rightarrow }\) \(y=\frac{3}{2}\sqrt{4-x^{2}}\)**

Therefore, the **Area** of **smaller** **region** enclosed by the Ellipse **\(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\)** and the line **\(\frac{x}{2}+\frac{y}{3}=1\)** is represented by **curve ACBA**

**Now, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – Area enclosed by the curve ABOA**

**Now, the Area enclosed by the curve ACBOA:**

**Since, \(\ \int \sqrt{a^{2}-x^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}\;\left [ \frac{x}{2}\sqrt{2^{2}-(x)^{2}}+\frac{4}{2}\sin^{-1}\frac{x}{2} \right ]_{0}^{2}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{3}{2}[\frac{2}{2}\sqrt{4-4}+2\sin^{-1}(1)-0]=\frac{3\pi }{2}}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve ACBOA = \(\boldsymbol{\frac{3\pi}{2}}\)unit ^{2}**

**Now, the Area enclosed by the curve ABOA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{Area \;of\;\Delta ABO = \frac{1}{2}\times AO\times BO}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{1}{2}\times 2\times 3=3}\) **unit ^{2}**

**Therefore, the Area enclosed by the curve ABOA = 3 unit ^{2}**

**Since, the Area enclosed by the curve ACBA = Area enclosed by the curve ACBOA – the Area enclosed by the curve ABOA**

\(\\\boldsymbol{\Rightarrow }\) \(\frac{3\pi }{2}-3=\frac{3}{2}(\pi -2)\)**unit ^{2}**

**Therefore, the Area of shaded region \(=\frac{3}{2}(\pi -2)\) unit ^{2}**

^{ }

** **

**Q.14: Find the area enclosed by the curve |x| + |y| = 2, by using the method of integration.**

** **

**Sol:**

Equation **|x| + |y| = 2** represent a region bounded by the lines:

**x + y = 2 . . . . . . (1)**

**x – y = 2 . . . . . . (2)**

**-x + y = 2 . . . . . . (3)**

**-x – y = 2 . . . . . . (4)**

**From equations (1), (2), (3) and (4)** we conclude that the curve intersects **x-axis** and **y-axis** axis at points **A (0, 2), B (2, 0), C (0, -2)** and **D (-2, 0) respectively.**

**From the above figure:**

Since, the curve is symmetrical to **x-axis** and **y-axis**. Therefore, the **Area** of region bounded by the curve **ABCDA** = **4** × **Area** of region bounded by the curve **ABOA**

**Now, the Area of region bounded by the curve ABOA:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=4\int_{0}^{2}(2-x)dx}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2x-\frac{x^{2}}{2} \right ]_{0}^{2}=(4-2)=2}\\\)** unit ^{2}**

**Therefore, the Area of region bounded by the curve ABOA = 2unit ^{2}**

**Since, the Area of region bounded by the curve ABCDA = 4 × Area of region bounded by the curve ABOA**

**Therefore, the Area of region bounded by the curve ABCDA =** (2 × 4) **= 8 unit ^{2}**

**Therefore, the Area of shaded region = 8 unit ^{2}**

** **

** **

**Q.15: Find the area which is exterior to curve x ^{2} = 2y and interior to curve x^{2 }+ y^{2} = 15.**

** **

**Sol:**

The Equation **x ^{2} = 2y** represents a

**parabola**

**symmetrical**about

**y-axis**.

The Equation **x ^{2 }+ y^{2} = 15** represents a

**circle**with

**centre**

**(0, 0)**and

**radius**

**units**.

Now, on substituting the equation of parabola in the equation of circle we will get:

(2y) + y^{2} = 15 i.e.** y ^{2 }+ 2y – 15 = 0**

Now, by **splitting the middle term method** solutions of this quadratic equation are:

y^{2} + (5 – 3)y – 15 = 0 \(\Rightarrow\) y(y + 5) – 3(y +5) = 0

\(\Rightarrow\) (y – 3) (y + 5) = 0

Neglecting y = -5 ** [gives absurd values of x]**

Therefore, **y = 3 **which gives **x =** **\(\pm \sqrt{6}\)**

Hence, the **coordinates** of **point** **B and point D** are **(\(\sqrt{6}\), 3) (\(-\sqrt{6}\)****, 3****)** respectively.

**Now, the Area of region bounded by the curve BAC’A’DOB = 2 ****× ****(Area of region bounded by the curve OBMO + Area of region bounded by the curve BAMB+ Area of region bounded by the curve OAC’O)**

**Area of region bounded by the curve BAMB [ x ^{2 }+ y^{2} = 15 ]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{\sqrt{6}}^{\sqrt{15}}y\;dx=\int_{\sqrt{6}}^{\sqrt{15}}\sqrt{\left ( \sqrt{15} \right )^{2}-x^{2}}\;\;dx}\\\)

**Since,** \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{15- x^{2}}+\frac{15}{2}\sin^{-1}\frac{x}{\sqrt{15}} \right ]_{\sqrt{6}}^{\sqrt{15}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac {\sqrt{15}}{2}\times \sqrt{{15-15}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{15}}{\sqrt{15}} \right ]-}\\\) \(\\\boldsymbol{\left [\frac {\sqrt{6}}{2}\times \sqrt{{15-6}}+\frac{15}{2}\sin^{-1}\frac{\sqrt{6}}{\sqrt{15}}\right]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 0+\frac{15\pi }{4} \right ]-\left [\frac{3\sqrt{6}}{2}+\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}\right]}\\\)

**\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{15\pi }{4}=\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5} \right ]}\)unit ^{2}**

**Area of region bounded by the curve OBMO [ x ^{2} = 2y ]:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\sqrt{6}}y\;dx=\int_{0}^{\sqrt{6}}\frac{x^{2}}{2}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^3}{6} \right ]_{0}^{\sqrt{6}}=\frac{6\sqrt{6}}{6}-0}\\\) = \(\boldsymbol{\sqrt{6}}\) **unit ^{2}**

**Area of region bounded by the curve OAC’O:**

\(\boldsymbol{\Rightarrow }\) \(\boldsymbol{\frac{Area\;of\;circle}{4}=\frac{\pi \times \left ( \sqrt{15} \right )^{2} }{4}}\\\) = \(\boldsymbol{\frac{15\pi }{4}}\)unit^{2}

Now, the Area of region bounded by the curve **BAC’A’DOB** = **2 ****×** (**Area** of region bounded by the curve **OBMO + Area **of region bounded by the curve** BAMB +** **Area** of region bounded by the curve **OAC’O**)

**Therefore, the Area of region bounded by the curve BAC’A’DOB:**

\(\boldsymbol{\Rightarrow }\) \(2\left [ \frac{15\pi }{4}-\frac{3\sqrt{6}}{2}-\frac{15}{2}\sin^{-1}\frac{\sqrt{10}}{5}+\sqrt{6} +\frac{15\pi }{4}\right ]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\\\)**unit ^{2}**

**Therefore, the Area of shaded region: \(\boldsymbol{ \left [ 15\pi -\sqrt{6}-15\left ( \sin^{-1}\frac{\sqrt{10}}{5} \right ) \right ]}\)unit ^{2}**

** **

** **

**Q.16: Find the area enclosed by the curve y = x ^{3}, x-axis and the lines x = -2 and x = 2.**

** **

**Sol.**

The Equation **y = x ^{3}** represents a

**cubic parabola**which intersects the

**line x = 2**and

**x = -2**at

**points A**and

**D**respectively

**From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}x^{3}\;dx+\left | \int_{-2}^{0}x^{3}\;dx \right |}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{4}}{4} \right ]_{0}^{2}+\left | \left [ \frac{x^{4}}{4} \right ]_{-2}^{0} \right |}\)

\(\\\boldsymbol{\Rightarrow }\) **4 – 0 + 0 + 4 =16 unit ^{2}**

**Therefore, the Area of shaded region = 16 unit ^{2}**

** **

** **

**Q.17: Find the area enclosed by the curve y = x|x|, y – axis and the lines y = -1 and y = 3.**

** **

**Sol:**

**Now, y = x|x| is equal to [y = x ^{2}] when x > o**

**And y = x|x| is equal to [y = -x ^{2}] when x < o**

** **

**From the above figure, the required Area = Area enclosed by the curve ABOA + Area enclosed by the curve CODC**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{3}x^{2}\;dx+\left | \int_{-1}^{0}-x^{2}\;dx \right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x^{3}}{3}\right ]_{0}^{3}+\left | \left [ \frac{x^{3}}{3} \right ]_{-1}^{0} \right |}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{9+\frac{1}{3}=\frac{28}{3}}\)**unit ^{2}**

**Therefore, the area of shaded region \(=\frac{28}{3}\)unit ^{2}**

** **

** **

**Q.18: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and y – axis, when [0 ****≤ x ≤ \(\frac{\pi }{2}\)****].**

** **

**Sol:**

**y = Cos(x) . . . . . . . . (1)**

**y = Sin(x) . . . . . . . . . (2)**

Now, **from equation (1) and equation (2):**

**Cos (x) = Sin (x)** \(\Rightarrow\) Cos (x) = Cos \(\left [ \frac{\pi }{2}-x \right ]\)

\(\Rightarrow\) x = \(\left [ \frac{\pi }{2}-x \right ]\) \(\Rightarrow\) **x = \(\frac{\pi }{4}\)**

**Therefore, the coordinates of point A are:** **\(\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )\)
**

** **

**Now, from the above figure:**

**The required Area =** **Area** enclosed by the **curve ADMA** **+** **Area** enclosed by the **curve AMOA**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\\\)

**Since, \(\int \sin^{-1}y\;dy= y\sin^{-1}y+\sqrt{1-y^{2}}\\\)**

**And, \(\\\int \cos^{-1}y\;dy= y \cos^{-1}y-\sqrt{1-y^{2}}\)**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{\frac{1}{\sqrt{2}}}sin^{-1}y\;dy+ \int_{\frac{1}{\sqrt{2}}}^{1}cos^{-1}y\;dy}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ y\sin^{-1}y+\sqrt{1-y^{2}} \right ]_{0}^{\frac{1}{\sqrt{2}}}+\left [ y\cos^{-1}y-\sqrt{1-y^{2}} \right ]_{\frac{1}{\sqrt{2}}}^{1}}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{1}{\sqrt{2}}\times \frac{\pi }{4}+\frac{1}{\sqrt{2}}-(0+1) \right ]+\left [ 0-\left ( \frac{1}{\sqrt{2}}\times \frac{\pi }{4}-\frac{1}{\sqrt{2}} \right ) \right ]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [\sqrt{2}-1 \right ]}\)**unit ^{2}**

**Therefore, the Area of shaded region = \(\left [\sqrt{2}-1 \right ]\)unit ^{2}**

^{ }

^{ }

**Q.20: Find the area which is exterior to curve y ^{2} = 6x and interior to curve x^{2 }+ y^{2} = 16.**

** **

**Sol:**

The Equation **y ^{2} = 6x** represents a

**parabola**

**symmetrical**about

**x – axis**.

The Equation **x ^{2 }+ y^{2} = 16** represents a

**circle**with

**centre**

**(0, 0)**and

**radius**

**4**

**units**.

Now, on substituting the equation of parabola in the equation of circle we will get:

x^{2} + (6x) = 16 i.e.** x ^{2 }+ 6x – 16 = 0**

Now, by **splitting of middle term method** solutions of this quadratic equation are:

x^{2} + (8 – 2)x – 16 = 0 \(\Rightarrow\) x(x + 8) – 2(x + 8) = 0

\(\Rightarrow\) (x – 2) (x + 8) = 0

Neglecting x = -8 [gives absurd values of y]

Therefore, **x = 2 **which gives **y = \(\pm 2\sqrt{3}\)**

Hence, the **coordinates** of **point** **B and point D** are **(****2, \(2\sqrt{3}\)****) (2, \(-2\sqrt{3}\))** respectively.

**Now, the Area of region bounded by the curve PBA’B’NOP** **=** **2 ****×** (**Area** of region bounded by the curve **BPMOB – Area **of region bounded by the curve** POMP +** **Area** of region bounded by the curve **BOA’B**)

**Area of region bounded by the curve BPMOB [x ^{2 }+ y^{2} = 16]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{\left ( 4\right )^{2}-x^{2}}\;\;dx}\\\)

**Since,** \(\\\int \sqrt{a^{2}-b^{2}}\;dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{x}{2}\sqrt{16- x^{2}}+\frac{16}{2}\sin^{-1}\frac{x}{4} \right ]_{0}^{2}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{2}{2}\times \sqrt{{16-4}}+8\times \sin^{-1}\frac{1}{2} \right ]-0}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ 2\sqrt{3}+\frac{8\times \pi }{6} \right ]=\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]}\\\)**unit ^{2}**

**Therefore, Area of region bounded by the curve BPMOB = \(\left [ 2\sqrt{3}+\frac{4\pi }{3} \right ]\)unit ^{2}**

**Area of region bounded by the curve POMP [y ^{2} = 6x]:**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\int_{0}^{2}y\;dx=\int_{0}^{2}\sqrt{6x}\;dx}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\left [ \frac{\sqrt{6}\times 2}{3}\times x^\frac{3}{2} \right ]_{0}^{2}=\frac{2\sqrt{6}}{3}\times 2\sqrt{2}-0\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\frac{8\sqrt{3}}{3}\)**unit ^{2}**

**Therefore, Area of region bounded by the curve POMP : \(\frac{8\sqrt{3}}{3}\)unit ^{2}**

**Area** of region bounded by the curve **BOA’B:**

\(\\\boldsymbol{\Rightarrow }\) **4****π**** unit ^{2}**

Therefore, **Area** of region bounded by the curve **BOA’B = 4****π**** unit ^{2}**

Now, the **Area** of region bounded by the curve **PBA’B’NOP** = **2 ****× **(**Area** of region bounded by the curve **BPMOB – Area **of region bounded by the curve** POMP +** **Area** of region bounded by the curve **BOA’B**)

**Therefore, **the **Area** of region bounded by the curve **PBA’B’NOP:**

\(\\\boldsymbol{\Rightarrow }\) \(2\left [2\sqrt{3}+\frac{4\pi }{3}-\frac{8\sqrt{3}}{3}+4\pi \right ]\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{32\pi }{3}-\frac{4\sqrt{3}}{3} \right ]=\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]}\\\)**unit ^{2}**

**Therefore, the Area of shaded region = \(\frac{4}{3}\times \left [ 8\pi -\sqrt{3} \right ]\)unit ^{2}**

** **

** **

**Q.21: Find the area bounded by the curve y = Cos (x) and y = Sin (x) and x – axis, when [0 ****≤ x ≤ **** ].**

** **

**Sol:**

**y = Cos(x) . . . . . . . . (1)**

**y = Sin(x) . . . . . . . . . (2)**

Now, from **equation (1)** and **equation (2):**

**Cos (x) = Sin (x)** \(\Rightarrow\) Cos (x) = Cos \(\left [ \frac{\pi }{2}-x \right ]\\\)

\(\\\Rightarrow\) x = \(\left [ \frac{\pi }{2}-x \right ]\\\) \(\\\Rightarrow\) **x = \(\frac{\pi }{4}\)**

**Therefore, the coordinates of point A are:** **\(\left ( \frac{\pi }{4},\frac{1}{\sqrt{2}} \right )\)**

**Now, from the above figure:**

**The required Area =** **Area** enclosed by the **curve AONA** **+** **Area** enclosed by the **curve ANBA**

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos(x) \right ]_{0}^{\frac{\pi }{4}}+[sin (x)]_{\frac{\pi }{4}}^{\frac{\pi }{2}}}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ -cos\frac{\pi }{4}+cos(0)+sin\frac{\pi }{2}-sin\frac{\pi }{4}\right ]}\\\)

\(\\\boldsymbol{\Rightarrow }\) \(\boldsymbol{\left [ \frac{-1}{\sqrt{2}}+1+1-\frac{1}{\sqrt{2}} \right ]=[2-\sqrt{2}\;]}\)**unit ^{2}**

**Therefore, the Area of shaded region \( = [2-\sqrt{2}\;]\)unit ^{2}**