Class 12 Important Questions: CBSE Maths Chapter 8 Applications of Integrals 2024-25

Long Answer Type Questions (6 Marks)

Q.1.Find the area enclosed by a circle \[{x^2} + {y^2} = {a^2}\].

Ans: 

Drawing circle

$ {{x^2} + {y^2} = {a^2}} $

${{\text{ Center }} = (0,0)} $    

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Radius $ = a$

Hence

${{\text{OA}} = {\text{OB}} = {\text{ Radius }} = a} \\ $

$ {\;{\text{A}} = (a,0)} \\ $

${{\text{B}} = (0,a)} $ 

Since, The Circle is Symmetric about X-Axis and Y-Axis.

Area of circle $ = 4 \times $ Area of Region OBAO

$ = 4 \times \int_0^a y dx$

We know that

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  ${{x^2} + {y^2} = {a^2}} $

${{y^2} = {a^2} - {x^2}} $

  ${y =  \pm \sqrt {{a^2} - {x^2}} } $

Since AOBA lies in ${1^{{\text{st }}}}$ Quadrant,

Value of $y$ is positive

$y = \sqrt {{a^2} - {x^2}} $

Now,

${\text{ Area of circle }} = 4 \times \int_0^a {\sqrt {{a^2} - {x^2}} } dx$

$  {{\text{ Using: }}\sqrt {{a^2} - {x^2}} dx = \dfrac{1}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{a} + c} $

$  {\quad  = 4\left[ {\dfrac{x}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{a}} \right]_0^a} $

$ { = 4\left[ {\left( {\dfrac{a}{2}\sqrt {{a^2} - {a^2}}  + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{a}{a}} \right) - \left( {\dfrac{0}{2}\sqrt {{a^2} - 0}  + \dfrac{{{0^2}}}{2}{{\sin }^{ - 1}}(0)} \right)} \right]} $

$  { = 4\left[ {0 + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(1) - 0 - 0} \right]} $

$  { = 4 \times \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(1)} $

$  { = 4 \times \dfrac{{{a^2}}}{2} \times \dfrac{\pi }{2}} $

$  { = \pi {a^2}} $

Q. 2. Find the area of region bounded by:$\left\{ {(x,y):|x - 1| \leqslant y \leqslant \sqrt {25 - {x^2}} } \right\}$

Ans:

We have provided

$\left\{ {(x,y):|x - 1| \leqslant y \leqslant \sqrt {\left. {\left( {25 - {x^2}} \right)} \right\}} } \right.$

Equation of curve is $y = \sqrt {\left( {25 - {x^2}} \right)} $ or ${y^2} + {x^2} = 25$, which is a circle with center at $(0,0)$ and radius 5 

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Equation of line is $y = |x - 1|$ Consider, $y = x - 1$ and $y = \sqrt {25 - {x^2}} $

Eliminating $y$, we get $x - 1 = \sqrt {25 - {x^2}} $

$ \Rightarrow \quad {x^2} + 1 - 2x = 25 - {x^2}$

$ \Rightarrow \quad 2{x^2} - 2x - 24 = 0$

$ \Rightarrow \quad {x^2} - x - 12 = 0$

$ \Rightarrow \quad (x - 4)(x + 3) = 0$

$ \Rightarrow \quad x =  - 3,4$

The Required Area is

${ = \int_{ - 3}^4 {\sqrt {25 - {x^2}} } dx - \int_{ - 1}^1 {( - x + 1)} dx - \int_1^2 {(x - 1)} dx} $

$  { = \left[ {\dfrac{x}{2}\sqrt {25 - {x^2}}  + \dfrac{{25}}{2}{{\sin }^{ - 1}}\dfrac{x}{{\sqrt {25} }}} \right]_{ - 3}^4 - \left[ { - \dfrac{{{x^2}}}{2} + x} \right]_{ - 1}^1 - \left[ {\dfrac{{{x^2}}}{2} - x} \right]_1^2} $

$ = \left( {6 + \dfrac{{25}}{2}{{\sin }^{ - 1}}\dfrac{4}{{\sqrt {25} }}} \right) + 6 - \dfrac{{25}}{2}{\sin ^{ - 1}}\left( { - \dfrac{3}{{\sqrt {25} }}} \right) - \left( {\dfrac{{ - 1}}{2} + 1 + \dfrac{1}{2} + 1} \right) - \left( {2 - 2 - \dfrac{1}{2} + 1} \right)$

${ = \dfrac{{25}}{2}\left( {{{\sin }^{ - 1}}\dfrac{4}{{\sqrt 5 }} + {{\sin }^{ - 1}}\dfrac{3}{{\sqrt 5 }}} \right) + 2 - 2 - \dfrac{1}{2}} \\ $

${ = \dfrac{{25}}{2}{{\sin }^{ - 1}}\left[ {\dfrac{4}{{\sqrt 5 }}\sqrt {1 - \dfrac{1}{5}}  + \dfrac{3}{{\sqrt 5 }}\sqrt {1 - \dfrac{4}{5}} } \right] - \dfrac{1}{2}} $

$ { = \dfrac{{25}}{2}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{4}{5} + \dfrac{1}{5}} \right)} \right] - \dfrac{1}{2}} \\ $

$ { = \dfrac{{25}}{2}{{\sin }^{ - 1}}(1) - \dfrac{1}{2}} \\ $ 

$ { = \left( {\dfrac{{25\pi }}{4} - \dfrac{1}{2}} \right){\text{ sq}}{\text{. units }}} $

Q.3. Find the area enclosed by the ellipse${\text{  }}\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$.

Ans: 

We have to find Area Enclosed by ellipse

Since Ellipse is symmetrical about both $x$ -axis and $y$ -axis

$\therefore $ Area of ellipse $ = 4 \times $ Area of OAB

$ = 4 \times \int_0^a y dx$

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We know that,

$ {\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1} \\ $

$ {\dfrac{{{y^2}}}{{{b^2}}} = 1 - \dfrac{{{x^2}}}{{{a^2}}}} $

$ {\dfrac{{{y^2}}}{{{b^2}}} = \dfrac{{{a^2} - {x^2}}}{{{a^2}}}} \\ $

$ {{y^2} = \dfrac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right)} \\ $

$ {y =  \pm \sqrt {\dfrac{{{b^2}}}{{{a^2}}}\left( {{a^2} - {x^2}} \right)} } \\ $ 

$ {y =  \pm \dfrac{b}{a}\sqrt {\left( {{a^2} - {x^2}} \right)} } $

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Since ${\text{OAB}}$ is in ${1^{{\text{st }}}}$ quadrant,

value of $y$ is positive

$\therefore \quad y = \dfrac{b}{a}\sqrt {{a^2} - {x^2}} $

${\text{Area of ellipse }} = 4 \times \int_0^a y  \cdot dx$

$ { = 4\int_0^a {\dfrac{b}{a}} \sqrt {{a^2} - {x^2}} dx} \\ $

$ { = \dfrac{{4b}}{a}\int_0^a {\sqrt {{a^2} - {x^2}} } dx} $

$ {{\text{ It is of form }}} \\ $

$ {\sqrt {{a^2} - {x^2}} dx = \dfrac{1}{2}x\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{a} + c} $

$ { = \dfrac{{4b}}{a}\left[ {\dfrac{x}{2}\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{a}} \right]_0^a} \\ $

$ { = \dfrac{{4b}}{a}\left[ {\left( {\dfrac{a}{2}\sqrt {{a^2} - {a^2}}  + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{a}{a}} \right) - \left( {\dfrac{0}{2}\sqrt {{a^2} - 0}  - \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(0)} \right)} \right]} \\ $

$ { = \dfrac{{4b}}{a}\left[ {0 + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(1) - 0 - 0} \right]} $ 

$ { = \dfrac{{4b}}{a} \times \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}(1)} \\ $

$ { = 2ab \times {{\sin }^{ - 1}}(1)} \\ $

$ { = 2ab \times \dfrac{\pi }{2}} \\ $

$ { = \pi ab} \\ $

$ {\therefore {\text{ Required Area }} = \pi ab{\text{ square units }}} $

Q. 4. Find the area of region in the first quadrant enclosed by x–axis, the line y = x and the circle \[{x^2} + {y^2} = 32\].

Ans:

Equation of Given Circle is: -

$ {{x^2} + {y^2} = 32} \\ $

$ {{x^2} + {y^2} = 16 \times 2} \\ $

$ {{x^2} + {y^2} = {4^2} \times 2} \\ $

$ {{x^2} + {y^2} = {{(4\sqrt 2 )}^2}} $

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Let point where line and circle intersect be point ${\text{M}}$

Required Area = Area of shaded region

= Area OMA

 First, we find Point ${\text{M}}$

Point $M$ is intersection of line and circle

We know that

$y = x$

Putting this in Equation of Circle, we get

$ {{x^2} + {y^2} = 32} \\ $

$ {{x^2} + {x^2} = 32} \\ $

$ {2{x^2} = 32} \\ $

$ {{x^2} = 16} \\ $ 

$ {\therefore \quad x =  \pm 4} $ 

When x =4 

Y = x =4

So, points are (4, 4)

 When x = -4

Y = x = -4

So, points are (-4, -4)

As, Point M is in 1st Quadrant 

M = (4, 4)

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$\therefore \quad {\text{ Required Area}} = {\text{ Area OMP}} + {\text{ Area PMA}} = \int_0^4 {{y_1}} dx + \int_4^{4\sqrt 2 } {{y_2}} dx$

$ {{x^2} + {y^2} = {{(4\sqrt 2 )}^2}} \\ $

$ {{y^2} = {{(4\sqrt 2 )}^2} - {x^2}} \\ $

$ {y =  \pm \sqrt {{{(4\sqrt 2 )}^2} - {x^2}} } $ 

As Required Area is in first Quadrant

$\therefore {y_2} = \sqrt {{{(4\sqrt 2 )}^2} - {x^2}} $

$ {\therefore {\text{ Required Area }} = \int_0^4 x dx + \int_4^{4\sqrt 2 } {\sqrt {{{(4\sqrt 2 )}^2} - {x^2}} } dx}\\ $

$ {{\text{ Taking }}{{\text{I}}_1}{\text{ i}}{\text{.e}}{\text{. }}}\\ $

$ { = \int_0^4 x  \cdot dx} \\ $

$ { = \left[ {\dfrac{{{x^2}}}{2}} \right]_0^4} \\ $

$ { = \dfrac{{{{(4)}^2} - 0}}{2}} \\ $

$ { = \dfrac{{16}}{2}} \\ $

$ { = 8} $ 

Now solving ${{\text{I}}_2}$

${{\text{I}}_2} = \int_4^{4\sqrt 2 } {\sqrt {{{(4\sqrt 2 )}^2} - {x^2}} } dx$

It is of form

$\int {\sqrt {{a^2} - {x^2}} } dx = \dfrac{1}{2}x\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + c$

 Replacing a by 4√2, we get

${{\text{I}}_2} = \left[ {\dfrac{x}{2}\sqrt {{{(4\sqrt 2 )}^2} - {x^2}}  + \dfrac{{{{(4\sqrt 2 )}^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{{4\sqrt 2 }}} \right]_4^{4\sqrt 2 }$

$ = \dfrac{{4\sqrt 2 }}{2}\sqrt {{{(4\sqrt 2 )}^2} - {{(4\sqrt 2 )}^2}}  + \dfrac{{{{(4\sqrt 2 )}^2}}}{2}{\sin ^{ - 1}}\dfrac{{4\sqrt 2 }}{{4\sqrt 2 }}$

$ - \dfrac{4}{2}\sqrt {{{(4\sqrt 2 )}^2} - {{(4)}^2}}  - \dfrac{{{{(4\sqrt 2 )}^2}}}{2}{\sin ^{ - 1}}\dfrac{4}{{4\sqrt 2 }}$

$ = 0 + \dfrac{{16 \times 2}}{2}{\sin ^{ - 1}}(1) - 2\sqrt {32 - 16}  - \dfrac{{16 \times 2}}{2}{\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$

$ { = 16{{\sin }^{ - 1}}(1) - 2\sqrt {16}  - 16{{\sin }^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \\ $

$ { = 16\left[ {{{\sin }^{ - 1}}(1) - {{\sin }^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right] - 8} \\ $

$ { = 16\left[ {\dfrac{\pi }{2} - \dfrac{\pi }{4}} \right] - 8} \\ $

$ { = 16\left[ {\dfrac{{4\pi  - 2\pi }}{{4 \times 2}}} \right] - 8} \\ $ 

$ { = \dfrac{{16}}{8}[2\pi ] - 8} \\ $ 

$ { = 2[2\pi ] - 8} \\ $

$ { = 4\pi  - 8} $ 

Putting Value of i1 and i2 in (1)

Area = 8 + 4π – 8

          =4π

Therefore, Required Area is 4π Sq. unit.

Q.5. Find the area of region \[\{ \left( {{\mathbf{x}},{\text{ }}{\mathbf{y}}} \right):{\text{ }}{\mathbf{y}}{\text{ }}{\mathbf{2}} \leqslant {\mathbf{4x}},{\text{ }}{\mathbf{4x}}{\text{ }}{\mathbf{2}}{\text{ }} + {\text{ }}{\mathbf{4y}}{\text{ }}{\mathbf{2}} \leqslant {\mathbf{9}}\} .\]

Ans:

x-coordinate of point of intersection is x = 1/2 

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$ {{\text{ Required area }}} \\ $

$ { = 2\left( {\int_0^1 2 \sqrt x dx + \int_{\dfrac{1}{2}}^3 {\sqrt {\dfrac{9}{4} + {x^2}} } dx} \right)} \\ $

 $ { = 2\left[ {\dfrac{4}{3}{x^{3/2}}^{\dfrac{1}{2}} + \dfrac{x}{2}\sqrt {\dfrac{9}{4} - {x^2}}  + \left. {\dfrac{9}{8}\sin \dfrac{{2x}}{3}} \right|_2^3} \right]} \\ $

$ { = \dfrac{{\sqrt 2 }}{6} + \dfrac{9}{4}\left( {\dfrac{\pi }{2} - \sin \dfrac{1}{3}} \right){\text{ or }}\dfrac{{\sqrt 2 }}{6} + \dfrac{9}{4}{{\cos }^{ - 1}}\dfrac{1}{3}} $

Q. 6. Prove that the curve \[{\mathbf{y}}{\text{ }} = {\text{ }}{{\mathbf{x}}^{\mathbf{2}}}\] and, \[{\mathbf{x}}{\text{ }} = {\text{ }}{{\mathbf{y}}^{\mathbf{2}}}\]divide the square bounded by x = 0, y = 0, x = 1, y = 1 into three equal parts.

Ans:

Let OABC be the square bounded by x = 0, y=0, x=1, y=1.

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 $A(OABC){\text{ }} = 1 \times 1 = 1{\text{ sq}}{\text{. units }}$

$ {{\text{ From, }}y = {x^2}{\text{ and x = }}{{\text{y}}^2}} \\ $

$ {x = {{({x^2})}^2}} $

${x^4} = x$

$ {{x^4} - x = 0} \\ $

$x(x^{3}-1)=0$

$x=0, x=\pm 1$

 Point of intersection of the two parabolas is (0, 0) and (1,1).

 ${\text{ Area of part III }} = \int_0^1 y dx\left( {{\text{ parabola }}{x^2} = y} \right)$

$ { = \int_0^1 {\dfrac{{{x^2}}}{{}}} dx = \left[ {\dfrac{{{x^3}}}{3}} \right]_0^1} \\ $

$  { = \dfrac{1}{3}{\text{ sq}}{\text{. units }}} $

${\text{ Area of }}I = {\text{ Area of square  -  Area of II and III }}$

$ { = 1 - \int_0^1 {\sqrt x } dx} \\ $

$ { = 1 - \dfrac{1}{3}\left[ {{x^{3/2}}} \right]_0^1} \\ $

$ { = 1 - \dfrac{1}{3}{\text{ sq}}{\text{. units }}} \\ $

$ { = \dfrac{2}{3}{\text{ sq}}{\text{. units }}} $

Area of II = Area of square - Area of I - Area of III

$ { = 1 - \dfrac{1}{3} - \dfrac{1}{3}{\text{ sq}}{\text{. units }}} \\ $

$  { = \dfrac{1}{3}{\text{ sq}}{\text{. units }}} $ 

The two curves divide the square into three equal parts.

Q.7. Find the area of the smaller region bounded by the ellipse \[\dfrac{{{x^2}}}{a} + \dfrac{{{y^2}}}{b} = 1\]

and straight-line \[\dfrac{x}{a} + \dfrac{y}{b} = 1\]

Ans:

The given ellipse is x2/a2 + y2/b2 = 1 and 

the line is x/a + y/b = 1.

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The sketch is as— The shaded region is the required area

$ {\therefore \quad \operatorname{ar} (ABCA) = \int_0^a {\left( {{y_1} - {y_2}} \right)} dx} \\ $

$ { = \int_0^a {(y{\text{ of the ellipse }})} dx - \int_0^a {(y{\text{ of the line }})} dx} \\ $

$ { = \int_0^a {\dfrac{b}{a}} \sqrt {{a^2} - {x^2}} dx - \int_0^a {\dfrac{{b(a - x)}}{a}} dx} $

$ { = \dfrac{b}{a}\left[ {\dfrac{{x\sqrt {{a^2} - {x^2}} }}{2} + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{a}} \right]_0^a - \dfrac{b}{a}\left[ {ax - \dfrac{{{x^2}}}{2}} \right]_0^a} \\ $

$ { = \dfrac{b}{a}\left[ {0 + \dfrac{{{a^2}}}{2}{{\sin }^{ - 1}}\dfrac{a}{a} - 0} \right] - \dfrac{b}{a}\left( {{a^2} - \dfrac{{{a^2}}}{2} - 0} \right)} \\ $

$ { = \dfrac{{ab}}{2} \cdot {{\sin }^{ - 1}}1 - \dfrac{b}{a} \times \dfrac{{{a^2}}}{2} = \left( {\dfrac{{\pi ab}}{4} - \dfrac{{ab}}{2}} \right)} $

${\text{ So, The required area }} = \left( {\dfrac{{\pi ab}}{4} - \dfrac{{ab}}{2}} \right){\text{ sq}}{\text{. units }}$

Q.8. Find the common area bounded by the circles \[{{\mathbf{x}}^{\mathbf{2}}} + {\text{ }}{{\mathbf{y}}^{\mathbf{2}}} = {\text{ }}{\mathbf{4}}\] and \[{\left( {{\mathbf{x}}{\text{ }}--{\text{ }}{\mathbf{2}}} \right)^{\mathbf{2}}} + {\text{ }}{{\mathbf{y}}^{\mathbf{2}}} = {\text{ }}{\mathbf{4}}.\]

Ans:

Equations of given circles

\[{x^2} + {y^2} = 4..........(1)\]

\[{(x - 2)^2} + {y^2} = 4\]

Centre of the circle from equation –(i) is at origin (0, 0) and radius is 2 units.

Centre of the circle of equation (ii) is (2, 0) and at the x axis and radius is 2 units.

 Required enclosed area is shown by shaded part in figure

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 Solving equation (i) and (ii) 

Obtained intersecting points of circles are p (1, √3) and Q (1 – √3)

 Out of two circles one symmetric about x axis

 ∴ Required area = 2(Area OPACO) = 2 [Area OPCO + Area CPAC] 

= 2[Area OPCO (part of circle (x – 2)2 + y2 = 4) + Area CPAC (part of circle x2 + y2 = 4) 

 ∴ Required Area = 2 ∫ y dx (for circle (x – 2)2 + y2 = 4) + ∫ y dx (for circle x2 + y2 = 4)

 $ { = 2\left[ {\int_0^1 {\sqrt {4 - {{(x - 2)}^2}} } dx + \int_1^2 {\sqrt {4 - {x^2}} } dx} \right]} \\ $

$ { = 2\left[ {\left\{ {\left( {\dfrac{{x - 2}}{2}} \right)\sqrt {4 - {{(x - 2)}^2}}  + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{{(x - 2)}}{2}} \right\}_0^1} \right.} \\ $

$ {\left. { + \left\{ {\dfrac{x}{2}\sqrt {4 - {x^2}}  + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{x}{2}} \right\}_1^2} \right]} $

$ { = 2\left\{ {\dfrac{{1 - 2}}{2}\sqrt {4 - {{(1 - 2)}^2}}  + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{{1 - 2}}{2}} \right\}} \\ $

$ { - \left\{ {\left( {\dfrac{{0 - 2}}{2}} \right)\sqrt {4 - {{(0 - 2)}^2}}  + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{{0 - 2}}{2}} \right\}} \\ $

$ { + \left\{ {\dfrac{2}{2}\sqrt {4 - 4}  + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{2}{2}} \right\}} \\ $ 

$ {\left. {\quad  - \left\{ {\dfrac{{\sqrt 3 }}{2}\sqrt {4 - 3}  + \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{{\sqrt 1 }}{2}} \right\}} \right]} $

$ = 2\left[ { - \dfrac{1}{2}\sqrt {4 + 1}  + \dfrac{4}{2}{{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right. + \sqrt {4 - 4}  - \dfrac{4}{2}{\sin ^{ - 1}}( - 1)\left. { + \dfrac{4}{2}{{\sin }^{ - 1}}1 - \dfrac{{\sqrt 3 }}{2} - \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{1}{2}} \right]$

$ = 2\left[ { - \dfrac{1}{2} \times \sqrt 3  - \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{1}{2} + } \right.\dfrac{4}{2}{\sin ^{ - 1}}1\left. { + \dfrac{4}{2}{{\sin }^{ - 1}}1 - \dfrac{{\sqrt 3 }}{2} - \dfrac{4}{2}{{\sin }^{ - 1}}\dfrac{1}{2}} \right]$

$ { = 2\left[ { - \dfrac{{\sqrt 3 }}{2} - \dfrac{4}{2} \times \dfrac{\pi }{6} + \dfrac{4}{2} \times \dfrac{\pi }{2} + \dfrac{4}{2} \times \dfrac{\pi }{2} - \dfrac{{\sqrt 3 }}{2} - \dfrac{4}{2} \times \dfrac{\pi }{6}} \right]} \\ $

$ { = 2\left[ { - \sqrt 3  - \dfrac{\pi }{3} + \pi  + \pi  - \dfrac{\pi }{3}} \right]} $

$ { = 2\left[ { - \sqrt 3  - \dfrac{{2\pi }}{3} + 2\pi } \right] = 2\left[ {\dfrac{{6\pi  - 2\pi }}{3} - \sqrt 3 } \right]} $

$ { = 2\left[ {\dfrac{{4\pi }}{3} - \sqrt 3 } \right] = \left( {\dfrac{{8\pi }}{3} - 2\sqrt 3 } \right){\text{ sq}}{\text{. unit}}{\text{. }}} $

Q.9. Using integration, find the area of the region bounded by the triangle whose vertices are

1. (–1, 0), (1, 3) and (3, 2)

Ans:

We need line eq: for line AB, BC and AC

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points $( - 1,0)$ and $(1,3)$

eq: is $\dfrac{{y - 0}}{{3 - 0}} = \dfrac{{x - ( - 1)}}{{1 - ( - 1)}}$ eq: of line by 2-point form $\dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{x - {x_1}}}{{{x_2} - {x_1}}}$

$ \Rightarrow \dfrac{y}{3} = \dfrac{{x + 1}}{2}$

$ \Rightarrow 2{\text{y}} = 3{\text{x}} + 3$

$ \Rightarrow y = \dfrac{{3x}}{2} + \dfrac{3}{2}$

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so, for line BC points $(1,3)$ and $(3,2)$

so, for line AC points $( - 1,0)$ and $(3,2)$

eq: is $\dfrac{{y - 0}}{{2 - 0}} = \dfrac{{x - ( - 1)}}{{3 - ( - 1)}}$

$ \Rightarrow \dfrac{y}{2} = \dfrac{{x + 1}}{4}$

$ \Rightarrow y = \dfrac{x}{2} + \dfrac{1}{2}$

Required area is shaded area ⇒ [Area under line AB from x

 = -1 to x = 1 + Area under BC from x = 1 to x 

= 3 - Area under AC from x = -1 to x = 3]

$ { \Rightarrow \int_{ - 1}^1 {\left( {\dfrac{{3x}}{2} + \dfrac{3}{2}} \right)} dx + \int_1^3 {\left( {\dfrac{{ - x}}{2} + \dfrac{7}{2}} \right)} dx - \int_{ - 1}^3 {\left( {\dfrac{x}{2} + \dfrac{1}{2}} \right)} dx} \\ $

$ {\left. {\left. {\left. {\left. {\left. {\left. { \Rightarrow \dfrac{3}{2} \times \dfrac{{{x^2}}}{2}} \right]_{ - 1}^1 + \dfrac{{3x}}{2}} \right]_{ - 1}^1 + \dfrac{{ - 1}}{2} \times \dfrac{{{x^2}}}{2}} \right]_1^3 + \dfrac{{7x}}{2}} \right]_1^3 - \dfrac{{{x^2}}}{{2 \times 2}}} \right]_{ - 1}^3 - \dfrac{{1x}}{2}} \right]_{ - 1}^3} $ 

 $ { \ Rightarrow \left( {\left( {\dfrac{3}{2} \times \dfrac{1}{2}} \right) - \left( {\dfrac{3}{2} \times \dfrac{1}{2}} \right)} \right) + \left( {\left( {\dfrac{{3 \times 1}}{2}} \right) - \left( {\dfrac{{3 \times ( - 1)}}{2}} \right)} \right)} \\ $

$ { + \left( {\left( {\dfrac{{ - 1}}{2} \times \dfrac{9}{2}} \right) - \left( {\dfrac{{ - 1}}{2} \times \dfrac{{{1^2}}}{2}} \right)} \right) + \left( {\dfrac{{21}}{2} - \dfrac{7}{2}} \right)} \\ $

$ { - \left( {\dfrac{9}{4} - \dfrac{1}{4}} \right) - \left( {\dfrac{3}{2} - \left( {\dfrac{{ - 1}}{2}} \right)} \right)} $

$ { \Rightarrow \left( {\dfrac{3}{4} - \dfrac{3}{4}} \right) + \left( {\dfrac{3}{2} - \left( {\dfrac{{ - 3}}{2}} \right)} \right) + \left( {\left( {\dfrac{{ - 9}}{4}} \right)} \right.} \\ $

$ {\left. { - \left( {\dfrac{{ - 1}}{4}} \right)} \right) + \left( {\dfrac{{14}}{2}} \right) - \left( {\dfrac{8}{4}} \right) - \left( {\dfrac{3}{2} + \dfrac{1}{2}} \right)} $ 

$ { = (0) + \left( {\dfrac{6}{2}} \right) + \left( {\dfrac{{ - 8}}{4}} \right) + \left( {\dfrac{{14}}{2}} \right) - \left( {\dfrac{8}{4}} \right) - \left( {\dfrac{4}{2}} \right)} $

$ { = 0 + 3 - 2 + 7 - 2 - 2 = 4{\text{ sq}}{\text{. units}}{\text{. }}} $

2. (–2, 2) (0, 5) and (3, 2)

Ans: We have to find the area of the triangle whose vertices are A (– 2,2), B (0, 5), C (3, 2)

$ {{\text{ The equation of }}AB{\text{ , }}} \\ $

$ {{\text{y}} = {{\text{y}}_1} = \left( {\dfrac{{{{\text{y}}_2} - {{\text{y}}_1}}}{{{{\text{x}}_2} - {{\text{x}}_1}}}} \right)\left( {{\text{x}} - {{\text{x}}_1}} \right)} \\ $

$ {y - 2 = \left( {\dfrac{{5 - 2}}{{0 + 2}}} \right)(x + 2)} \\ $

$ {y - 2 = \dfrac{3}{2}(x + 2)} \\ $

$ {Y = \dfrac{3}{2}x + 5 \ldots {\text{ }}...{\text{ (i) }}} $

The equation of BC,

$y-5=(\frac{2-5}{3-0})(x-0)$

$ { = \dfrac{{ - 3}}{3}(x - 0)} \\ $

$ {y - 5 =  - x} \\ $

$ {Y = 5 - x \ldots {\text{ (ii) }}} \\ $

$ {{\text{ The equation of }}AC{\text{ , }}} \\ $

$ {{\text{y}} - 2 = \left( {\dfrac{{2 - 2}}{{3 + 2}}} \right)({\text{x}} + 2)} \\ $

$ {y - 2 = 0} \\ $ 

$ {{\text{y}} = 2 \ldots {\text{ (iii) }}} $ 

Now the required area $(A) = $

((Area between line A B and $x$ -axis) - (Area between line A C and $x$ - axis) from $x =  - 1$ to $x = 0)$

$ + [$ (Area between line ${\text{BC}}$ and ${\text{x}}$ -axis) - (Area between line ${\text{AC}}$ and ${\text{x}}$ -axis) from ${\text{x}} = 0$ to $x = 3$ ]

 Say, Area $A = {A_1} + {A_2}$

 $A_{1}=\int_{-2}^{0}[(\frac{3x}{2}+5)]dx$

$ { = \int_{ - 2}^0 {\left[ {\dfrac{{3x}}{2} + 5 - 2} \right]} dx} \\ $

$ { = \int_{ - 2}^0 {\left( {\dfrac{{3x}}{2} + 3} \right)} dx} \\ $

$ { = 3\left( {\dfrac{{{x^2}}}{4} + x} \right)_{ - 2}^0} \\ $

$ { = 3\left[ {(0) - \left( {\dfrac{4}{4} - 2} \right)} \right]} \\ $

$ { = 3} $ 

$ {{\text{ And, }}{A_2} = \int_0^3 {\left( {{y_2} - {y_3}} \right)} dx} \\  $ 

$ { = \int_0^3 {\left[ {(5 - x) - 2} \right]} dx} \\ $

$ { = \int_0^3 {\left[ {5 - x - 2} \right]} dx} \\ $

$ { = \int_0^3 {\left( {3 - x} \right)} dx} \\ $

$ { = \left( {3x - \dfrac{{{x^2}}}{2}} \right)_0^3} \\ $ 

$ { = \left[ {9 - \dfrac{9}{2}} \right] = \dfrac{9}{2}} $ 

So, the enclosed area of the triangle is 

\[3 + \dfrac{9}{2} = \dfrac{{15}}{2}\] sq. Units. 

Q.10. Using integration, find the area bounded by the lines.

(I) \[{\mathbf{x}}{\text{ }} + {\text{ }}{\mathbf{2y}}{\text{ }} = {\text{ }}{\mathbf{2}},{\text{ }}{\mathbf{y}}{\text{ }}--{\text{ }}{\mathbf{x}}{\text{ }} = {\text{ }}{\mathbf{1}}{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{2x}}{\text{ }} + {\text{ }}{\mathbf{y}}{\text{ }}--{\text{ }}{\mathbf{7}}{\text{ }} = {\text{ }}{\mathbf{0}}\]

Ans:

Given, x + 2y = 2 ...(I) 

 y – x = 1 ...(ii)

 2x + y = 7 ...(iii) 

 On plotting these lines, we have

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 Area of required region

$ = \int_{ - 1}^3 {\dfrac{{7 - y}}{2}} dy - \int_{ - 1}^1 {(2 - 2y)} dy - \int_1^3 {(y - 1)} dy$

$ = \dfrac{1}{2}\left[ {7y - \dfrac{{{y^2}}}{2}} \right]_{ - 1}^3 - \left[ {2y - {y^2}} \right]_{ - 1}^1 - \left[ {\dfrac{{{y^2}}}{2} - y} \right]_1^3$

$ = \dfrac{1}{2}\left( {21 - \dfrac{9}{2} + 7 + \dfrac{1}{2}} \right) - (2 - 1 + 2 + 1) - \left( {\dfrac{9}{2} - 3 - \dfrac{1}{2} + 1} \right)$

$ = 12 - 4 - 2 = 6$ sq. units

(ii) \[{\mathbf{y}}{\text{ }} = {\text{ }}{\mathbf{4x}}{\text{ }} + {\text{ }}{\mathbf{5}},{\text{ }}{\mathbf{y}}{\text{ }} = {\text{ }}{\mathbf{5}}{\text{ }}--{\text{ }}{\mathbf{x}}{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{4y}}{\text{ }}--{\text{ }}{\mathbf{x}}{\text{ }} = {\text{ }}{\mathbf{5}}.\]

Ans:

We have lines

$ {y = 4x + 5} \\ $ 

$ {y = 5 - x} \\ $

$ {4y = x + 5} $

 Solving(i)and(ii),we get point of intersection$(0,5)$Solving(ii)and(iii),we get point of intersection(3,2)

Solving(i)and(iii),we get point of intersection$(-1,1) 

These lines are plotted on coordinate plane as shown in the following figure.

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From the figure, area of the shaded region

$ {A = \int_{ - 1}^0 {(4x + 5)} dx + \int_0^3 {(5 - x)} dx - \int_{ - 1}^3 {\dfrac{{x + 5}}{4}} dx} $

$ { = \left[ {\dfrac{{4{x^2}}}{2} + 5x} \right]_{ - 1}^0 + \left[ {5x - \dfrac{{{x^2}}}{2}} \right]_0^3 - \dfrac{1}{4}\left[ {\dfrac{{{x^2}}}{2} + 5x} \right]_{ - 1}^3} \\ $

$ { = [0 - 2 + 5] + \left[ {15 - \dfrac{9}{2} - 0} \right] - \dfrac{1}{4}\left[ {\dfrac{9}{2} + 15 - \dfrac{1}{2} + 5} \right]} \\ $

$ { = 3 + \dfrac{{21}}{2} + \dfrac{1}{4} \cdot 24 = \dfrac{{15}}{2}{\text{ sq}}{\text{. units }}} $

Q. 11. Find the area of the region \[\{ \left( {{\mathbf{x}},{\text{ }}{\mathbf{y}}} \right):{\text{ }}{\mathbf{x}}{\text{ }}{\mathbf{2}}{\text{ }} + {\text{ }}{\mathbf{y}}{\text{ }}{\mathbf{2}} \leqslant {\mathbf{1}} \leqslant {\mathbf{x}}{\text{ }} + {\text{ }}{\mathbf{y}}\} .\]

Ans:

Given curves are 

${(x, y): {x_2} + {y_2}, ≤ 1 ≤ x + y}$.

(image will be uploaded soon)       

$ {{\text{ Hence, the required area }}} \\ $

$ { = \int_0^1 {(C - L)} dx} \\ $

$ { = \int_0^1 {\left( {\sqrt {1 - {x^2}}  - (1 - x)} \right)} dx} \\ $

$ { = \left( {\dfrac{{x\sqrt {1 - {x^2}} }}{2} + \dfrac{1}{2}{{\sin }^{ - 1}}\left( {\dfrac{x}{2}} \right) - x + \dfrac{{{x^2}}}{2}} \right)_0^1} \\ $

$ { = \left( {\dfrac{1}{2}{{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right) - 1 + \dfrac{1}{2}} \right)} \\ $ 

$ { = \left( {\dfrac{1}{2}{{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right) - \dfrac{1}{2}} \right)} \\ $

$ { = \dfrac{1}{2}\left( {\dfrac{\pi }{6} - 1} \right){\text{ sq}}{\text{.u}}{\text{. }}} $

Q. 12. Find the area of the region bounded by \[{\mathbf{y}}{\text{ }} = {\text{ }}\left| {{\mathbf{x}}{\text{ }}--{\text{ }}{\mathbf{1}}} \right|{\text{ }}{\mathbf{and}}{\text{ }}{\mathbf{y}}{\text{ }} = {\text{ }}{\mathbf{1}}.\]

Ans:

Area between two Curve 

If we have two functions .Area between two curves are

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Required Area = ½ × Base × Height

                          =1/2 ×2×1

                               =1

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Q.13. Find the area enclosed by the curve \[{\mathbf{y}}{\text{ }} = {\text{ }}{\mathbf{sin}}{\text{ }}{\mathbf{x}}\]

between x = 0 and x =\[\dfrac{{3\pi }}{2}\] and x-axis.

Ans:

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Area required = Area of OAB + Area of BC

$  {{\text{ Area OAB }}} \\ $

$  { = \int_0^\pi  y dx} \\ $

$  {y \to \sin x} \\ $

$  { = \int_0^\pi  {\sin } xdx} \\ $

$  { = [ - \cos x]_0^\pi } \\ $

$  { =  - [\cos \pi  - \cos 0]} \\ $

$  { =  - [ - 1 - 1]} \\ $

$  { =  - [ - 2]} \\ $

$  { = 2} $

Area BC

$ = \int_\pi ^{2\pi } y dx$

$ = \int_\pi ^{2\pi } {\sin } xdx$

$ = [ - \cos x]_\pi ^{\dfrac{{3\pi }}{2}}$

$ =  - [\cos \dfrac{{3\pi }}{2} - \cos \pi ]$

$ =  - [0 - ( - 1)]$

$ =  - 1$

Since area cannot be negative,

Area ${\text{BC}} = 1$

$  {{\text{ Hence, }}}\\ $

$  {{\text{ Area Required }}},{ = {\text{ Area }}OAB + {\text{ Area }}BC} \\ $

$  { = 2 + 1} \\ $

$  { = 3} $ 

Q. 14. Find the area bounded by semi-circle \[{\mathbf{y}} = {\mathbf{25}} - {{\mathbf{x}}^{\mathbf{2}}}\]and x-axis.

Ans:

Let the Area bounded By Semi-circle

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$  {\therefore A = \quad 2xy}\\ $

$  {\qquad {\text{ }}\therefore \dfrac{{dA}}{{dx}}},{ = 2x \cdot \dfrac{1}{{2\sqrt {25 - {x^2}} }} \cdot ( - 2x) + \sqrt {25 - {x^2}} },{ \cdot 2} \\ $

$  { = 2x\sqrt {25 - {x^2}} }\\ $

$  { = \dfrac{{2\left( {25 - {x^2}} \right) - 2{x^2}}}{{\sqrt {25 - {x^2}} }}}\\ $

$  { = \dfrac{{dA}}{{dx}} = },{0,{\text{ when }}x = y = \dfrac{{5\sqrt 2 }}{2}} $

$  {{\text{ By the First Derivative Test, the inscribed rectangle of maximum area has vertices }}} \\ $

$  {\left( { \pm \dfrac{{5\sqrt 2 }}{2},0} \right),{\text{ and }}\left( { \pm \dfrac{{5\sqrt 2 }}{2},\dfrac{{5\sqrt 2 }}{2}} \right)} \\ $

$  {\therefore {\text{ The length }} = 5\sqrt 2 ,\quad {\text{ the width }} = \dfrac{{5\sqrt 2 }}{2}} $

${\text{ The length }} = {\mathbf{5}}\sqrt {\mathbf{2}} ,\quad {\text{ the width }} = \dfrac{{{\mathbf{5}}\sqrt {\mathbf{2}} }}{{\mathbf{2}}}$

Q. 15. Find area of region given by \[\{ \left( {{\mathbf{x}},{\text{ }}{\mathbf{y}}} \right):{\text{ }}{{\mathbf{x}}^{\mathbf{2}}}\; \leqslant {\mathbf{y}} \leqslant \left| {\mathbf{x}} \right|\} .\]

Ans:

The required area is bounded between two curves $y = x^2$ and $y = |x|$.  

Both of these curves are symmetric about the y-axis and shaded region in the fig. shows the region whose area is required.

 Therefore, the required area

(image will be uploaded soon)       

 $A = 2 \times $ Area of the region ${R_1}$

Now, to find the point of intersection of the curves $y = |x|$ and $y = {x^2}$, we solve them simultaneously.

Clearly, the region ${R_1}$ is in the first quadrant, where $x > 0$,

$\therefore |x| = x$ or $\quad y = x$  ...(i)

 $y = {x^2}$

and

Solving these two equations, we get

$x = {x^2}$

or either $x = 0$ or $x = 1$

The limits are, when $x = 0,y = 0$ and when $x = 1,y = 1$. So, the points of intersection of the curves are ${\text{O}}(0,0)$ and ${\text{A}}(1,1)$.

Now, required Area $ = 2 \times $ area of line region ${R_1}$

(Tex translation failed) of the line $y = x$

$ - \left( y \right.$ of the parabola $\left. {\left. {y = {x^2}} \right)} \right]dx$

$ = 2\int_0^1 {\left( {x - {x^2}} \right)} dx = 2\left[ {\dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3}} \right]_0^1$

$ = 2\left[ {\dfrac{1}{2} - \dfrac{1}{3}} \right] = \dfrac{1}{3}$ Sq. unit

Q.16. Find area of smaller region bounded by ellipse \[\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{4} = 1\]

 and straight line \[{\mathbf{2x}}{\text{ }} + {\text{ }}{\mathbf{3y}}{\text{ }} = {\text{ }}{\mathbf{6}}.\]

Ans:

(image will be uploaded soon)       

 $ {{\text{ Area of shaded region }}} \\ $

 $ { = \int_0^3 {\left\{ {\dfrac{2}{3}\sqrt {9 - {x^2}}  - \dfrac{2}{3}(3 - x)} \right\}} dx} \\ $

$ { = \dfrac{2}{3}\left[ {\dfrac{x}{2}\sqrt {9 - {x^2}}  + \dfrac{9}{2}{{\sin }^{ - 1}}\dfrac{x}{3} + \dfrac{{{{(3 - x)}^2}}}{2}} \right]_0^3} \\ $

$ { = \dfrac{2}{3}\left[ {\left( {0 + \dfrac{9}{2} \cdot \dfrac{\pi }{2} + 0} \right) - \left( {0 + 0 + \dfrac{9}{2}} \right)} \right]} \\ $

$ { = \dfrac{2}{3}\left( {9\dfrac{\pi }{4} - \dfrac{9}{2}} \right)} \\ $

$ { = 3\left( {\dfrac{\pi }{2} - 1} \right){\text{sq}} \cdot {\text{ unit}}{\text{. }}} $

Q.17. Find the area of region bounded by the curve \[{{\mathbf{x}}^2}{\text{ }} = {\text{ }}{\mathbf{4y}}\]and line \[{\mathbf{x}}{\text{ }} = {\text{ }}{\mathbf{4y}}{\text{ }}--{\text{ }}{\mathbf{2}}.\]

Ans:

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Points of intersection are (2, 1) and (-1, 1/4)

Required Area of region (ΔOABO)

$  { = \int_{ - 1}^2 {\dfrac{{x + 2}}{4}} dx - \int_{ - 1}^2 {\dfrac{{{x^2}}}{4}} dx} \\ $

$  { = \dfrac{1}{4}\left[ {\dfrac{{{{(x + 2)}^2}}}{2} - \dfrac{{{x^3}}}{3}} \right]_{ - 1}^2} \\ $

$  { = \dfrac{1}{4}\left[ {\dfrac{{16}}{3} - \dfrac{5}{6}} \right] = \dfrac{{27}}{{24}}{\text{ sq}}{\text{.units }}} \\ $

$  { = \dfrac{9}{8}{\text{ squnits }}} $ 

Q. 18. Using integration find the area of region in first quadrant enclosed by x-axis, the line \[{\mathbf{x}} = \surd {\mathbf{3y}}\]and the circle \[{{\mathbf{x}}^{\mathbf{2}}} + {\text{ }}{{\mathbf{y}}^{\mathbf{2}}} = {\text{ }}{\mathbf{4}}.\]

Ans:

Given Equation of circle

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$x^{2}+y^{2}=4$

$  {{x^2} + {y^2} = {{(2)}^2}} $

$\therefore $ Radius $r = 2$

So, point $A$ is $(2,0)$

and point ${\text{B}}$ is $(0,2)$

 Let line $x = \sqrt 3 y$ intersect the circle at point $C$

Therefore, we have to find Area of AOC

 Finding point ${\text{C}}$

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We know that

$x = \sqrt 3 y$

Putting value of ${\text{x}}$ in equation of circle

${x^2} + {y^2} = 4$

$ {{{(\sqrt 3 y)}^2} + {y^2} = 4} \\ $

$ {3{y^2} + {y^2} = 4} \\ $

$ {4{y^2} = 4} \\ $

$ {{y^2} = 1} \\ $

$ {\therefore \quad y =  \pm 1} $

Now, finding value of $x$

$  {{\text{ When }}y = 1} \\ $ 

$  {x = \sqrt 3 y} \\ $

$  {x = \sqrt 3  \times 1} \\ $

$ {x = \sqrt 3 } $

$ {{\text{ When }}y =  - 1} \\ $

$ {x = \sqrt 3 y} \\ $

$ {x = \sqrt 3  \times  - 1} \\ $

$ {x =  - \sqrt 3 } $

$ {{\text{ Since point C is in}}{{\text{ }}^{{\text{st}}}}{\text{ quadrant }}} \\ $ 

$ {\therefore {\text{C is }}(\sqrt 3 ,1)} $

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Area OAC

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Area of $OAC = $ Area $OCX + $ Area XCA

 Area OCX

Area ${\text{OCX}} = \int_0^{\sqrt 3 } y dx$

$y \to $ Equation of line

Now,

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$  {x = \sqrt 3 y} \\ $

$  {y = \dfrac{x}{{\sqrt 3 }}} $ 

$  {{\text{ Therefore, }}} \\ $

$  {{\text{ Area OCX}} = \int_0^{\sqrt 3 } y dx} \\ $

$  { = \int_0^{\sqrt 3 } {\dfrac{x}{{\sqrt 3 }}} dx} \\ $

$  { = \dfrac{1}{{\sqrt 3 }}\int_0^{\sqrt 3 } x dx} \\ $

$  { = \dfrac{1}{{\sqrt 3 }}\left[ {\dfrac{{{x^2}}}{2}} \right]_0^{\sqrt 3 }} \\ $

$  { = \dfrac{1}{{2\sqrt 3 }}\left[ {{x^2}} \right]_0^{\sqrt 3 }} \\ $

$  { = \dfrac{1}{{2\sqrt 3 }}\left[ {{{(\sqrt 3 )}^2} - {{(0)}^2}} \right]} $ 

$\frac{1}{2\sqrt{3}}[3]$

$=\frac{\sqrt{3}}{2}$

Area XCA

Area $ \times CA = \int_{\sqrt 3 }^2 y dx$

$y \to $ Equation of circle

Now,

$ {{x^2} + {y^2} = 4} \\ $

$  {{y^2} = 4 - {x^2}} \\ $

$  {y =  \pm \sqrt {4 - {x^2}} } $ 

Since XCA is in ${1^{{\text{st }}}}$ Quadrant,

$ {{\text{ Value of }}y{\text{ will be positive }}} \\ $

$  {\therefore y = \sqrt {4 - {x^2}} }\\ $

$  {{\text{ Area  XCA}}},{ = \int_{\sqrt 3 }^2 {\sqrt {4 - {x^2}} } dx} \\ $ 

$  { = \int_{\sqrt 3 }^2 {\sqrt {{{(2)}^2} - {x^2}} } dx} $

It is of form

$\sqrt {{a^2} - {x^2}} dx = \dfrac{1}{2}x\sqrt {{a^2} - {x^2}}  + \dfrac{{{a^2}}}{2}{\sin ^{ - 1}}\dfrac{x}{a} + c$

Replacing a by 2, we get

$ = \left[ {\dfrac{1}{2}x\sqrt {{{(2)}^2} - {x^2}}  + \dfrac{{{{(2)}^2}}}{2}{{\sin }^{ - 1}}\dfrac{x}{2}} \right]_{\sqrt 3 }^2$

$ = \left[ {\dfrac{1}{2}x\sqrt {4 - {x^2}}  + 2{{\sin }^{ - 1}}\dfrac{x}{2}} \right]_{\sqrt 3 }^2$

$ = \dfrac{1}{2}(2)\sqrt {4 - {2^2}}  + 2{\sin ^{ - 1}}\dfrac{2}{2} - \dfrac{1}{2}(\sqrt 3 )\sqrt {4 - {{(\sqrt 3 )}^2}}  - 2{\sin ^{ - 1}}\dfrac{{\sqrt 3 }}{2}$

$ { = 0 + 2{{\sin }^{ - 1}}(1) - \dfrac{{\sqrt 3 }}{2}\sqrt {4 - 3}  - 2{{\sin }^{ - 1}}\dfrac{{\sqrt 3 }}{2}} \\ $

$ { = 2{{\sin }^{ - 1}}(1) - \dfrac{{\sqrt 3 }}{2} - 2{{\sin }^{ - 1}}\dfrac{{\sqrt 3 }}{2}} \\ $

$ { = \dfrac{{ - \sqrt 3 }}{2} + 2\left[ {{{\sin }^{ - 1}}(1) - {{\sin }^{ - 1}}\dfrac{{\sqrt 3 }}{2}} \right]} \\ $

$ { = \dfrac{{ - \sqrt 3 }}{2} + 2\left[ {\dfrac{\pi }{2} - \dfrac{\pi }{3}} \right]} \\ $

$ { = \dfrac{{ - \sqrt 3 }}{2} + 2\left[ {\dfrac{\pi }{6}} \right]} \\ $

$ { = \dfrac{{ - \sqrt 3 }}{2} + \dfrac{\pi }{3}} $

$ {{\text{ Therefore, }}} \\ $

$ {{\text{ Area of OAC}} = {\text{ Area OCX}} + {\text{ Area XCA }}} $

$ = \dfrac{{\sqrt 3 }}{2} - \dfrac{{\sqrt 3 }}{2} + \dfrac{\pi }{3}$

$ = \dfrac{\pi }{3}$ square units

$\therefore $ Required Area $ = \dfrac{\pi }{3}$ square units

Q.19. Find smaller of two areas bounded by the curve \[{\mathbf{y}}{\text{ }} = {\text{ }}\left| {\mathbf{x}} \right|{\text{ }}{\mathbf{and}}{\text{ }}{{\mathbf{x}}^{\mathbf{2}}} + {\text{ }}{{\mathbf{y}}^{\mathbf{2}}} = {\text{ }}{\mathbf{8}}\].

Ans:

Let the Parabola be:

(image will be uploaded soon)       

$  {{\text{ The required area }}} \\ $

$  { = \int_{ - 2}^2 {\sqrt {8 - {x^2}} } dx - \int_{ - 2}^0 {} xdx + \int_0^2 x dx} \\ $

$  { = 2\int_0^2 {\sqrt {8 - {x^2}} } dx - \left[ {\dfrac{{{x^2}}}{2}} \right]_{ - 2}^0 - \left[ {\dfrac{{{x^2}}}{2}} \right]_0^2} \\ $

$  { = 2\left[ {\dfrac{x}{2}\sqrt {8 - {x^2}}  + \dfrac{8}{2}{{\sin }^{ - 1}}\dfrac{x}{{2\sqrt 2 }}} \right]_0^2 - \dfrac{4}{2} - 2} \\ $

$  { = 2\left[ {2 + 4 \cdot \dfrac{\pi }{4}} \right] - 4 = 2\pi sq.unit} $ 

Q. 20. Find the area lying above x-axis and included between the \[{\mathbf{circle}}{\text{ }}{{\mathbf{x}}^{\mathbf{2}}} + {\text{ }}{{\mathbf{y}}^{\mathbf{2}}} = {\text{ }}{\mathbf{8x}}\]and the parabola \[{{\mathbf{y}}^{\mathbf{2}}} = {\text{ }}{\mathbf{4x}}.\]

Ans:

We have, given equations

${x^2} + {y^2} = 8x \ldots .$ (i)

and ${y^2} = 4x\quad  \ldots  \ldots  \ldots .$ (ii)

Equation (1) can be written as

${(x - 4)^2} + {y^2} = {(4)^2}$

So, equation (I) represents a circle with center $(4,0)$ and radius 4. 

Again, clearly equation (ii) represents parabola with vertex $(0,0)$ and axis as $x$ -axis. 

The curve (i) and (ii) are shown in figure and the required region is shaded. On solving equation (i) and (ii) we have points of intersection $0(0,0)$ and $A(4,4),C(4, - 4)$

 Now, we have to find the area of region bounded

(image will be uploaded soon)       

by $(i)$ and $(ii){\text{ }}$ above $x$ -axis. So required region is OBAO.

 Now, area of OBAO is

$A{\text{ }} = \int_0^4 {\left( {\sqrt {8x - {x^2}}  - \sqrt {4x} } \right)} dx = \int_0^4 {\left( {\sqrt {{{(4)}^2} - {{(x - 4)}^2}}  - 2\sqrt x } \right)} dx$

$ = \left[ {\dfrac{{(x - 4)}}{2}\sqrt {{{(4)}^2} - {{(x - 4)}^2}}  + \dfrac{{16}}{2}{{\sin }^{ - 1}}\dfrac{{(x - 4)}}{4} - 2 \times \dfrac{{2{x^{3/2}}}}{3}} \right]_0^4$

$  { = \left[ {8{{\sin }^{ - 1}}0 - \dfrac{4}{3}{{(4)}^{\dfrac{3}{2}}}} \right] - \left[ {8{{\sin }^{ - 1}}( - 1) - 0} \right]} \\ $

$  { = \left( {8 \times 0 - \dfrac{4}{3} \times 8} \right) - \left( {8 \times  - \dfrac{\pi }{2}} \right)} \\ $

$  { =  - \dfrac{{32}}{3} + 4\pi  = \left( {4\pi  - \dfrac{{32}}{3}} \right){\text{ sq}}{\text{.units }}} $

Q. 21. Using integration, find the area enclosed by the curve \[{\mathbf{y}}{\text{ }} = {\text{ }}{\mathbf{cos}}{\text{ }}{\mathbf{x}},{\text{ }}{\mathbf{y}}{\text{ }} = {\text{ }}{\mathbf{sin}}{\text{ }}{\mathbf{x}}\]

and x-axis in the interval (0, π/2).

Ans: The area bounded by the curves are in the chart below:

(image will be uploaded soon)       

 The previous version of my response had the area corresponding to ${\text{C}}$. However, after reading the question again, I guess the required area $ = {\text{A}} + {\text{B}}$

$AreaA = \int_0^{\dfrac{\pi }{4}} {(\cos x - \sin x)} .dx = [\sin x + \cos x]_0^{\dfrac{\pi }{4}}$

$A = \left[ {\sin \dfrac{\pi }{4} + \cos \dfrac{\pi }{4} - \sin 0 - \cos 0} \right] = \left[ {\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} - 0 - 1} \right] = [\sqrt 2  - 1]$

$\quad  = \int_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}} {(\sin x - \cos x)}  \cdot dx = {[ - \cos x - \sin x]_{{{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}}}}$

$B = \left[ { - \cos \dfrac{\pi }{2} - \sin \dfrac{\pi }{2} + \cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4}} \right] = \left[ { - 0 - 1 + \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }} + 0 + 1} \right] = [\sqrt 2 $

${\text{ Required Area }} = A + B = [\sqrt 2  - 1] + [\sqrt 2  - 1] = 2[\sqrt 2  - 1]$

Area C: (retained for academic exercise, if someone interested)

The required area $ = [y = \sin x]_0^{\dfrac{\pi }{4}} + [y = \cos x]_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}$

Due to the symmetry of $\sin x$ and $\cos x$ functions between $\left( {0,\dfrac{\pi }{4}} \right)$ and $\left( {\dfrac{\pi }{4},\dfrac{\pi }{2}} \right)$ :

$ = 2*\int_0^{\dfrac{\pi }{4}} {\sin } x.dx = 2*[ - \cos x]_0^{\dfrac{\pi }{4}}$

$ = 2*\left[ { - \cos \dfrac{\pi }{4} + \cos 0} \right] = 2*\left[ {1 - \dfrac{1}{{\sqrt 2 }}} \right] = 2 - \sqrt 2 $

Q. 22. Sketch the graph \[{\mathbf{y}}{\text{ }} = {\text{ }}\left| {{\mathbf{x}}{\text{ }}--{\text{ }}{\mathbf{5}}} \right|\]. Evaluate \[\int\limits_0^6 {|x - 5|} dx\].

Ans:

We have,

$y = |x - 5|$ intersect $x = 0$ and $x = 1$ at $(0,5)$ and $(1,4)$

Now,

$y = |x - 5|$

$ =  - (x - 5)$ For all $x \in (0,1)$

Integration represents the area enclosed by the graph from $x = 0$ to $x = 1$

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$  {A = \int_0^6 | y|dx} \\ $

$  { = \int_0^6 | x - 5|dx} \\ $

$ { = \int_0^6  -  (x - 5)dx} \\ $

$ { =  - \int_0^6 {(x - 5)} dx} \\$ 

$ { =  - \left[ {\dfrac{{{x^2}}}{2} - 5x} \right]_0^6} \\ $

$ { =  - \left[ {\left( {\dfrac{{36}}{2} - 30} \right) - (0 - 0)} \right]} \\ $

$  { = 12{\text{sq}} \cdot {\text{ units }}} $

Q.23. Find area enclosed between the curves\[,{\text{ }}{\mathbf{y}}{\text{ }} = {\text{ }}{\mathbf{4x}}{\text{ }}{\mathbf{and}}{\text{ }}{{\mathbf{x}}^{\mathbf{2}}} = {\text{ }}{\mathbf{6y}}\]

Ans: Here, ${x^2} = 6y$ is a parabola

And,

$y = 4x$ is a line

which intersects the parabola at points $A$ and $B$

We need to find Area of shaded region

First, we find Points A and B

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Finding points $A$ and $B$

Points $A{\text{ }}B$ are the intersection of curve and line

Putting in equation of curve, we get

$  {{x^2} = 6y} \\ $

$  {{{(\dfrac{y}{4})}^2} = 4y} \\ $

$  {\dfrac{{{y^{^2}}}}{{16}} = 4y} \\ $

$  {{y^2} - 64y = 0} \\ $

$  {{y^2} = 64y} \\ $

$  {y = 64} \\ $

$  {} $

y=64, y=0

 For y = 64 

$  {x = \dfrac{y}{4}} \\ $

$  {x = \dfrac{{64}}{4}} \\$ 

$  {x = 16} \\ $

$  {{\text{ So, point is }}\left( {16,64} \right)} $

For $y = 0$

$x = \dfrac{y}{4}$

$x = 0$

So, point is $(0,0)$ 

Point  is in 1stQuadrant (16,64)

Q.24.Using integration, find the area of the following region: 

$  {\left\{ {(x,y):|x - 1| \leqslant y \leqslant \sqrt {5 - {x^2}} } \right\}} $

Ans:

We have provided

$\left\{ {(x,y):|x - 1| \leqslant y \leqslant \sqrt {\left. {\left( {5 - {x^2}} \right)} \right\}} } \right.$

Equation of curve is $y = \sqrt {\left( {5 - {x^2}} \right)} $ or ${y^2} + {x^2} = 5$, which is a circle with center at $(0,0)$ 

and radius $5/2$.

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Equation of line is $y = |x - 1|$ Consider, $y = x - 1$ and $y = \sqrt {5 - {x^2}} $

Eliminating $y$, we get $x - 1 = \sqrt {5 - {x^2}} $

$ \Rightarrow \quad {x^2} + 1 - 2x = 5 - {x^2}$

$ \Rightarrow \quad 2{x^2} - 2x - 4 = 0$

$ \Rightarrow \quad {x^2} - x - 2 = 0$

$ \Rightarrow \quad (x - 2)(x + 1) = 0$

$ \Rightarrow \quad x = 2, - 1$

The Required Area is

 $ { = \int_{ - 1}^2 {\sqrt {5 - {x^2}} } dx - \int_{ - 1}^1 {( - x + 1)} dx - \int_1^2 {(x - 1)} dx} \\ $

 $ { = \left[ {\dfrac{x}{2}\sqrt {5 - {x^2}}  + \dfrac{5}{2}{{\sin }^{ - 1}}\dfrac{x}{{\sqrt 5 }}} \right]_{ - 1}^2 - \left[ { - \dfrac{{{x^2}}}{2} + x} \right]_{ - 1}^1 - \left[ {\dfrac{{{x^2}}}{2} - x} \right]_1^2} $

$ = \left( {1 + \dfrac{5}{2}{{\sin }^{ - 1}}\dfrac{2}{{\sqrt 5 }}} \right) + 1 - \dfrac{5}{2}{\sin ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 5 }}} \right) - \left( {\dfrac{{ - 1}}{2} + 1 + \dfrac{1}{2} + 1} \right) - \left( {2 - 2 - \dfrac{1}{2} + 1} \right)$

$ { = \dfrac{5}{2}\left( {{{\sin }^{ - 1}}\dfrac{2}{{\sqrt 5 }} + {{\sin }^{ - 1}}\dfrac{1}{{\sqrt 5 }}} \right) + 2 - 2 - \dfrac{1}{2}} \\ $

$ { = \dfrac{5}{2}{{\sin }^{ - 1}}\left[ {\dfrac{2}{{\sqrt 5 }}\sqrt {1 - \dfrac{1}{5}}  + \dfrac{1}{{\sqrt 5 }}\sqrt {1 - \dfrac{4}{5}} } \right] - \dfrac{1}{2}} $

$ { = \dfrac{5}{2}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{4}{5} + \dfrac{1}{5}} \right)} \right] - \dfrac{1}{2}} \\ $

$ { = \dfrac{5}{2}{{\sin }^{ - 1}}(1) - \dfrac{1}{2}} \\ $

$ { = \left( {\dfrac{{5\pi }}{4} - \dfrac{1}{2}} \right){\text{ sq}}{\text{. units }}} $

Important Formulas from Class 12 Maths Chapter 8 Applications of Integrals

Benefits of Important Questions for Class 12 Maths Chapter 8 By Vedantu

• Comprehensive Coverage: These questions encompass all key topics, ensuring a thorough understanding of areas under curves and between lines and curves.
  Exam-Oriented Preparation: Curated by experts, the questions align with the latest CBSE syllabus and exam patterns, aiding in effective exam preparation.

• Enhanced Problem-Solving Skills: Regular practice with these questions improves analytical abilities, enabling students to tackle various problem types confidently.

• Accessible Learning: Available as a free PDF download, these resources provide convenient access to quality study material anytime, anywhere.

• Time Management: Practicing these questions helps students develop efficient time management skills, ensuring they can complete exam sections within the allotted time.

• Self-Assessment: Attempting these questions allows students to assess their understanding and identify areas needing further revision, promoting a targeted and effective study approach.

Tips to Study Class 12 Maths Chapter 8: Applications of Integrals

• Comprehensive Coverage: These questions encompass all key topics, ensuring a thorough understanding of areas under curves and between lines and curves.

• Exam-Oriented Preparation: Curated by experts, the questions align with the latest CBSE syllabus and exam patterns, aiding in effective exam preparation.

• Enhanced Problem-Solving Skills: Regular practice with these questions improves analytical abilities, enabling students to tackle various problem types confidently.

• Accessible Learning: Available as a free PDF download, these resources provide convenient access to quality study material anytime, anywhere. 

• Time Management: Practicing these questions helps students develop efficient time management skills, ensuring they can complete exam sections within the allotted time.

• Self-Assessment: Attempting these questions allows students to assess their understanding and identify areas needing further revision, promoting a targeted and effective study approach.

Extra Practice Questions for Class 12 Maths Chapter 8 - Application of Integrals

1. Determine the region's area bounded by x =√k and x = k.

2. Determine the region's area bounded x2 = 4y, y = 2, y = 4, and the y – axis in the first quadrant.

3. Find the area bounded by the curve y = 4 – x2 and the lines y = 0 and y = 3.

4. Determine the area including between the parabolas y2 = 4ax and x2 = 4by.

5. What area is bounded by the parabola y2 = 4x and x2 = 4y?

6. Find the area bounded by the parabola y2 = 4ax. Also, find its latus rectum and the x-axis.

7. Determine the area of the curve y = sin x between 0 and π.

8. Calculate the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x = a/√2.

9. Find the areas of the region: {(x,y): x2 + y2 ≤ 1 ≤ x + 4}

10. Find the area bounded by the curves y = cos x, y= 0, and x = |1|.

Conclusion

Mastering Chapter 8, "Applications of Integrals," is essential for CBSE Class 12 Mathematics students, as it provides the tools to calculate areas under curves and between lines and curves—skills that are vital for advanced studies in mathematics and related fields. To aid in this endeavor, a curated set of important questions is available for free PDF download, aligning with the latest CBSE syllabus and exam patterns. Engaging with these resources will enhance your problem-solving abilities and deepen your understanding of integral applications, thereby boosting your confidence and performance in examinations.

Related Study Materials for Class 12 Maths Chapter 8 Application of Integrals

CBSE Class 12 Maths Chapter-wise Important Questions

CBSE Class 12 Maths Chapter-wise Important Questions and Answers cover topics from all 13 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.

Additional Study Materials for Class 12 Maths