**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}**