**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}**