MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 3

The area enclosed between the curves

$\displaystyle y={ x }^{ 3 }$ and

$\displaystyle y=\sqrt { x }$ is, (in square units):

0%
$\displaystyle \frac { 5 }{ 3 }$
0%
$\displaystyle \frac { 5 }{ 4 }$
0%
$\displaystyle \frac { 5 }{ 12 }$
0%
$\displaystyle \frac { 12 }{ 5 }$

Explanation

Required area

$\displaystyle \int_0^1 (\sqrt x -x^3)dx=\left( \dfrac{2x^{3/2}}{3}-\dfrac{x^4}{4}\right)_0^1=\dfrac{2}{3}-\dfrac{1}{4}-0=\dfrac{5}{12}$

The area of the portion of the circle

$x^{2}+y^{2}=1$ , which lies inside the parabola

${y}^{2}=1-x$ is

0%
$\displaystyle \frac{\pi}{2}-\frac{2}{3}$
0%
$\displaystyle \frac{\pi}{2}+\frac{2}{3}$
0%
$\displaystyle \frac{\pi}{2}+\frac{4}{3}$
0%
$\displaystyle \frac{\pi}{2}-\frac{4}{3}$

Explanation

The given curve intersect at

$(0,1)$ and

$(0,-1)$
Therefore, the required area

$\displaystyle =\left[\frac { 1 }{ 2 } \pi { r }^{ 2 }\right]_{0}^{1}+2\int _{ 0 }^{ 1 }{ \left( 1-{ y }^{ 2 } \right) dy }$

$\displaystyle =\frac { 1 }{ 2 } \pi { \left( 1 \right) }^{ 2 }+2{ { \left[ y-\frac { { y }^{ 3 } }{ 3 } \right] } }_{ 0 }^{ 1 }=\frac { \pi }{ 2 } +\frac { 4 }{ 3 }$

Area of the region bounded by

$y=e^{x},y=e^{-x},x=0$ and

$x=1$ in sq. units is:

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

Explanation

$y=e^{x},y=e^{-x}$

Then, area bounded by these two curves s given by

$A=\displaystyle \overset{1}{\underset{0}{\int}}(e^{x}-e^{-x})dx=\overset{1}{\underset{0}{\int}}e^{x}dx-\overset{1}{\underset{0}{\int}}e^{-x}dx$

$=(e^1-1)+1(e^{-1}-1)$

$=e+e^{-1}-2$

$=\left(\sqrt{e}-\dfrac{1}{\sqrt{e}}\right)^2sq\:units.$

The area of the region bounded by the curves

$\mathrm{y}=\sqrt{x}$ and

$y=\sqrt{4-3x}$ and

$\mathrm{y}=0$ is:

0%
4/9
0%
16/9
0%
8/9
0%
9/2

Explanation

$\mathrm{y}=\sqrt{x}$ ; $y=\sqrt{4-3x}$
$\overset{1}{\underset{0}{\int}}\sqrt{x}dx+\overset{4/3}{\underset{1}{\int}}\sqrt{4-3x}dx$
$\dfrac{2}{3}1+\dfrac{2}{9}(4-3x)^{3/2}|_{1}^{4/3}$
$\dfrac{2}{3}-\dfrac{2}{9}(0-1)$
$=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}sq\:units$

The area of the region between the curve

$y=x^{3}$ and the lines

$y=-x$ and

$y=1$ is:

0%
5 sq. units
0%
$\displaystyle \frac{4}{5}$ sq. units
0%
$\displaystyle \frac{5}{4}$ sq. units
0%
$\displaystyle \frac{3}{5}$ sq. units

Explanation

$|\overset{1}{\underset{0}{\int}}-y \:dy|+|\overset{1}{\underset{0}{\int}}y^{1/3}\:dy|$
$=\left |\dfrac{-y^2}{2}|_{0}^{1} \right |+\left |\dfrac{3}{4} y^{4/3}|_{0}^{1} \right |$
$=\dfrac{2}{4}+\dfrac{3}{4}=\dfrac{5}{4}sq\:units$

For which of the following values of

$\mathrm{m}$ is the area of the region bounded by the curve

$y=x-x^{2}$ and the line

$y=mx$ equal to

$\displaystyle \frac{9}{2}$ :

0%
$-4$
0%
$2$
0%
$1$
0%
$-2$

Explanation

$\overset {1-m}{\underset { 0 }{ \int } }(x-x^2-mx)dx$
$\overset {1-m}{\underset { 0 }{ \int } }[x(1-m)-x^2]dx$
$\dfrac{(1-m)x^2}{2}-\dfrac{x^3}{3}|_{0}^{1-m}$
$\dfrac{(1-m)^3}{2}-\dfrac{(1-m)^3}{3}$ $\rightarrow \dfrac{(1-m)^3}{6}=9/2$
$(1-m)=3$
$m=1-3 =-2$

The area of the region bounded by

$y=|x-1|$ and

$\mathrm{y}=1$ in sq. units is:

0%
1
0%
1/2
0%
2
0%
3

Explanation

$y=|x-1|$
area of the

$\triangle ^{le}=1/2 \:base\times\:height$

$=1/2(2)(1)=1sq\:unit.$

The area bounded by the parabolas

${y}=4{x}^{2},\ y=\displaystyle \dfrac{x^{2}}{9}$ and the line

${y}=2$ is

0%
$\displaystyle \frac{20\sqrt{2}}{3}$
0%
$\displaystyle \frac{10\sqrt{2}}{3}$
0%
$\displaystyle \frac{40\sqrt{2}}{3}$
0%
$\displaystyle \frac{5\sqrt{2}}{3}$

Explanation

The area bounded by the curves ${y}=4{x}^{2},\ y=\displaystyle \dfrac{x^{2}}{9}$ and the line $y=2$ is
$A=2\times \displaystyle \overset{2}{\underset{0}{\int}}\left (3\sqrt{y}-\dfrac{\sqrt{y}}{2} \right )dy$
$=2\left [\left[\dfrac{2}{3} 3 y^{3/2}\right]_{0}^{2} -\left[\dfrac{2}{3}\dfrac{y^{3/2}}{2} \right ]_{0}^{2} \right ]$
$=2\left [ 2(\sqrt{8})-\dfrac{\sqrt{8}}{3} \right ]$
$=2\left [ \sqrt{8}\left ( \dfrac{5}{3} \right ) \right ]=\dfrac{20\sqrt{2}}{3}sq\:units.$

The area between the parabolas

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

$y^{2}=-4a(x-a)$ in sq. units is

0%
$\displaystyle \frac{4a^{2}}{3}$
0%
$\displaystyle \frac{8a^{2}}{3}$
0%
$\displaystyle \frac{12a^{2}}{3}$
0%
$\displaystyle \frac{16a^{2}}{3}$

Explanation

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

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

$4\overset{0}{\underset{-a}{\int}}\sqrt{4a(x+a)}dx=4\times 2\sqrt{a}\times \dfrac{2}{3}(x+a)^{3/2}|_{-a}^{0}$

$=\dfrac{16}{3}\sqrt{a}(a^{3/2})$

$=\dfrac{16a^2}{3}sq\:units.$

The area bounded by

$y=|x-1|,\ y=0$ and

$|x|=2$ is

0%
$4$
0%
$5$
0%
$3$
0%
$2$

Explanation

${\textbf{Step -1: Sketching the rough figure in the Coordinate plane according to the given equation.}}$

${\text{Firstly from the first curve}}$

${\text{Given function is }} = \left| {\left. {{\text{x}} - {\text{1}}} \right|} \right.$

$\Rightarrow {\text{y}} = {\text{x}} - {\text{1 if x > 1}}$

$= - {\text{x}} + 1{\text{ if x < 1}}$

${\text{For 2nd function }}\left| {\left. {\text{x}} \right| = 2} \right.$

$\Rightarrow {\text{x = + 2 or x}} = - 2$

${\textbf{Step -2: Finding the required coordinates}}$

${\text{So the point will be for A }} = \left( {1,0} \right)$

${\text{As EF is x}} = 2{\text{ line so E }} = \left( {2,0} \right)$

${\text{F}} = \left( {2,1} \right)$

${\text{D}} = \left( { - 2,0} \right)$

${\text{Put }} - 2{\text{ in }}\left( { - {\text{x}} + 1} \right){\text{ so y}} = 3$

${\text{C}} = \left( {-2,3} \right)$

${\textbf{Step -3: Finding the total Area.}}$

${\text{Total area}} = \Delta CDA + \Delta AEF$

${\text{Area of those }}\Delta {\text{ will be }}\dfrac{1}{2} \times {\text{base}} \times {\text{height}}$

$= \ \dfrac{1}{2} \times {\text{DA}} \times {\text{CD + }}\dfrac{1}{2} \times {\text{AF}} \times {\text{EF}}$

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

$= \dfrac{1}{2} \times 3 \times 3 + \dfrac{1}{2} \times 1 \times 1$

$= \dfrac{9}{2} + \dfrac{1}{2}$

$= \dfrac{{10}}{2}$

$= 5{\text{ square unit}}$

${\textbf{ Hence, the area bounced by given curves is 5 square units. Thus, option B is the}}$

${\textbf{correct answer.}}$

The area lying in the first quadrant between the curves

$x^{2}+y^{2}=\pi^{2}$ and

$y=\sin x$ and y- axis is

0%
$\displaystyle \frac{\pi^{3}-8}{4}$ sq. units
0%
$\displaystyle \frac{\pi^{3}+8}{4}$ sq. units
0%
$4(\pi^{3}-8)$ sq. units
0%
$\displaystyle \frac{\pi-8}{4}$ sq. units

Explanation

Area of the circle with radius

$\pi$ is

${\pi}^{3}$

Then, area of the arc of circle in the

$first quadrant is$

$\dfrac{\pi^3 }{4}sq\:units$ $ .......

$(i)$
Also, the area of the curve

$y=\sin x$ is

$\overset {\pi }{ \underset { 0 }{ \int } } \sin x\:dx=2\:sq\:units$ .......

$(ii)$
So, the required area is obtained by taking out

$(ii)$ from

$(i)$ i.e.

$(i)-(ii)$

Thus, the reuqired area

$=\dfrac{\pi^3 }{4}-2 = \dfrac{\pi ^3-8}{4}sq\:units.$

The area bounded by

$y=x^{2}, y=[x+1],\ x\leq 1$ and the

$\mathrm{y}$ -axis is

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{2}{3}$
0%
1
0%
$\displaystyle \frac{7}{3}$

Explanation

Required area

$\displaystyle =\int _{ 0 }^{ 1 }{ \sqrt { y } } dy=\frac { 2 }{ 3 }$

The area of the region of the plane bounded by

$max(|x|, |y|) \leq 1$ and

$xy\leq \dfrac {1}{2}$ is

0%
$\dfrac{1}{2} + ln \ 2$ sq. units
0%
$3 + ln\ 2$ sq. units
0%
$\dfrac{31}{4}$ sq. units
0%
$1 + 2\ ln \ 2$ sq. units

The area of the region bounded by the curves

$y=xe^{x}, y=xe^{-x}$ and the line

$\left| x \right| =1,y=0$ is:

0%
$4$
0%
$3$
0%
$2$
0%
$1$

Explanation

Since

$\left| x \right| =1\quad \therefore x=\pm 1$

$y=x{ e }^{ \left| x \right| }=\begin{cases} { x }e^{ -x }\quad -1<x<0 \\ { xe }^{ x }\quad 0\le x<1 \end{cases}$
Therefore

$=\left| \int _{ -1 }^{ 0 }{ { xe }^{ -x } } dx \right| +\left| \int _{ 0 }^{ 1 }{ { xe }^{ x } } dx \right| \\ =\left| \left[ { -xe }^{ -x }-{ e }^{ -x } \right] _{ -1 }^{ 0 } \right| +\left| \left[ { xe }^{ x }-{ e }^{ x } \right] _{ 0 }^{ 1 } \right| =2$

The area enclosed by the curves

$\mathrm{x}^{2}=\mathrm{y},\ \mathrm{y}=\mathrm{x}+2$ and

$\mathrm{x}$ -axis is:

0%
$\dfrac{5}{6}$
0%
$\dfrac{5}{4}$
0%
$\dfrac{5}{2}$
0%
$\dfrac{5}{3}$

Explanation

$\mathrm{x}^{2}=\mathrm{y},\ \mathrm{y}=\mathrm{x}+2$

$\left | \overset{-1}{\underset{-2}{\int}}(x+2) dx \right |+\left |\overset{0}{\underset{-1}{\int}}-x^2\:dx \right |$

$=\dfrac{x^2}{2}|_{-2}^{-1}+2(1)$

$=\dfrac{1}{2}-2+2+1/3=\dfrac{5}{6}sq\:units.$

Area of the region bounded by

$y=x-[x],\ y=[x]$ and

$\mathrm{x}$ -axis in

$[$ 0,2

$]$ is:

0%
$\displaystyle \frac{5}{2}$
0%
$\displaystyle \frac{3}{2}$
0%
1
0%
2

Explanation

$y=x-[x],\ y=[x]$
The parallelogram Area=

$2(1/2\times1\times1)$

$=1\:sq\:units.$

Area bounded by the curve

$\mathrm{y}=\mathrm{x}+\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x}$ and its inverse function between the ordinates

$\mathrm{x}=0$ and

$\mathrm{x}=2\pi$ is

0%
8 $\pi$ sqp. units
0%
4 $\pi$ sq. units
0%
8 sq. units
0%
3 $\pi$ sq. units

Explanation

Inverse function is the mirror image with respect to

$y=x$
Then area bounded by

$x+\sin x$ and its inverse function is

$4\int _{ 0 }^{ \pi }{ \left( x+\sin { x } -x \right) dx } =8$

The ratio in which the area bounded by the curves

$y^2=4x$ and

$x^2=4y$ is divided by the line

$x = 1$ is

0%
64 : 49
0%
15 : 34
0%
15 : 49
0%
None o fthese

Explanation

Area =

$\int _{ 0 }^{ 4 }{ \sqrt { 4x } } -\frac { { x }^{ 2 } }{ 4 }$
=

$2\times \frac { 2 }{ 3 } \times 8-\frac { (4)^{ 3 } }{ 12 }$
=

$\frac { 32 }{ 3 } -\frac { 4 }{ 3 } \times 4$
=

$\frac { 16 }{ 3 }$

${ A }_{ 1 }=\int _{ 0 }^{ 1 }{ \sqrt { 4x } -\frac { x^{ 2 } }{ 4 } }$
=

$2\times \frac { 2 }{ 3 } \times -\frac { 1 }{ 12 }$
=

$\frac { 4 }{ 3 } \times \frac { 1 }{ 12 } =\frac { 15 }{ 12 }$

${ A }_{ 2 }=\frac { 16 }{ 3 } -\frac { 15 }{ 12 }$
=

$\frac { 64-15 }{ 12 }$
=

$\frac { 49 }{ 12 }$

$\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 15 }{ 49 }$

The area of the region enclosed by the curves

$\mathrm{y}=\mathrm{x},\ \mathrm{x}=\mathrm{e},\ \mathrm{y}=1/\mathrm{x}$ and the positive

$\mathrm{x}$ -axis is:

0%
1/2 square units
0%
1 square units
0%
3/2 square units
0%
5/2 square units

Find the area bounded on the right by the line x+2=y, on the left by the parabola

$y=x^{2}$ and above the x-axis

0%
$9/2$ sq.units
0%
$2$ sq.units
0%
$3$ sq.units
0%
$5/3$ sq.units

Explanation

$x+2=y$ and

$y=x^2$ intersect at

$x=2$ and

$x=-1$

Therefore,

$A=\int _{ -1 }^{ 2 }{ \left( x+2-{ x }^{ 2 } \right) dx }$

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

$=\dfrac{1}{2} + 2 - \dfrac{1}{3} - 0$

$=\dfrac { 9 }{ 2 }$

The area of the portion of the circle

${ x }^{ 2 }+{ y }^{ 2 }=1$ , which lies inside the parabola

${ y }^{ 2 }=1-x$ , is

0%
$\displaystyle \frac { \pi }{ 2 } -\frac { 2 }{ 3 }$
0%
$\displaystyle \frac { \pi }{ 2 } +\frac { 2 }{ 3 }$
0%
$\displaystyle \frac { \pi }{ 2 } -\frac { 4 }{ 3 }$
0%
$\displaystyle \frac { \pi }{ 2 } +\frac { 4 }{ 3 }$

Explanation

The required area

$\displaystyle =2\int _{ 0 }^{ 1 }{ \sqrt { 1-x } dx } +2.\left( \frac { 1 }{ 4 } .\pi .{ 1 }^{ 2 } \right)$

$\displaystyle =-2{ \left[ \frac { 2{ \left( 1-x \right) }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ 1 }+\frac { \pi }{ 2 } =\frac { \pi }{ 2 } -\frac { 4 }{ 3 }$

The area bounded by the parabolas

${ y }^{ 2 }=4a\left( x+a \right)$ and

${ y }^{ 2 }=-4a\left( x-a \right)$ is:

0%
$\displaystyle \frac { 16 }{ 3 } { a }^{ 2 }$
0%
$\displaystyle \frac { 8 }{ 3 } { a }^{ 2 }$
0%
$\displaystyle \frac { 4 }{ 3 } { a }^{ 2 }$
0%
none of these

Explanation

The required area

$\displaystyle =4\int _{ 0 }^{ a }{ \sqrt { 4a\left( a-x \right) } dx }$

$\displaystyle =4\times 2\sqrt { a } .{ \left[ \frac { -2{ \left( a-x \right) }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ a }=\frac { 16 }{ 3 } { a }^{ 2 }$

If the area enclosed by the parabolas

$\displaystyle\ y= a-x^{2}$ and

$\displaystyle\ y=x^{2}$ is

$18\sqrt{2}$ sq.units. Find the value of 'a'

0%
1
0%
2
0%
5
0%
9

Explanation

The intersection point is

$(\sqrt{\dfrac{a}{2}},\dfrac{a}{2}),(-\sqrt{\dfrac{a}{2}},\dfrac{a}{2})$

$\int y_{1}-y_{2}.dx$

$\int a-2x^{2}.dx$

$=[ax-\dfrac{2x^{3}}{3}]_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}}$

$=2[ax-\dfrac{2x^{3}}{3}]_{0} ^{\sqrt{\frac{a}{2}}}$

$2[\dfrac{a\sqrt{a}}{\sqrt{2}}-\dfrac{2a\sqrt{a}}{6\sqrt{2}}]$

$=\sqrt{2}[a\sqrt{a}-\dfrac{a\sqrt{a}}{3}]$

$=\dfrac{2\sqrt{2}a\sqrt{a}}{3}=18\sqrt{2}$

$2a\sqrt{a}=54$

$a\sqrt{a}=27=3^{3}$
Squaring both sides, we get

$a^{3}=3^{6}$

$a=3^{2}$
Hence

$a=9$ .

The area of the region bounded by

$y=\mid x-1\mid$ and

$y=1$ is

0%
$1$
0%
$2$
0%
$\dfrac{1}{2}$
0%
None of these

Explanation

$y=\left|x-1\right|$

for

$x\gt1, y=x-1$

for

$x\lt1, y=-x+1$

Here area can be calculated as the area of the triangle.

Base of triangle,

$b=2$

Height of triangle,

$h=1$

Area of triangle,

$A=\dfrac{1}{2}\times b\times h$

$A=\dfrac{1}{2}\times 2\times 1$

Area,

$A=1cm^2$

The area bounded by the curve

$\displaystyle y=2x-x^{2}$ and the straight line

$\displaystyle y+x=0$ is given by:

0%
$\displaystyle \frac{9}{2}$
0%
$\displaystyle \frac{43}{6}$
0%
$\displaystyle \frac{35}{6}$
0%
None of these

Explanation

From figure required area

$=\int _{ 0 }^{ 2 }{ \left( 2x-{ x }^{ 2 } \right) dx } -\int _{ 2 }^{ 3 }{ \left( { 2x-{ x }^{ 2 }-\left( -x \right) } \right) dx }$

$\displaystyle =\int _{ 0 }^{ 2 }{ \left( 2x-{ x }^{ 2 } \right) dx } -\int _{ 2 }^{ 3 }{ \left( 3x-{ x }^{ 2 } \right) } dx={ \left[ \frac { 2{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }-{ \left[ \frac { 3{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 2 }^{ 3 }=\frac { 9 }{ 2 }$

The area bounded by the circle

${ x }^{ 2 }+{ y }^{ 2 }=8$ , the parabola

${ x }^{ 2 }=2y$ and the line

$y=x$ in

$y\ge 0$ is

0%
$\displaystyle \frac { 2 }{ 3 } +2\pi$
0%
$\displaystyle \frac { 2 }{ 3 } -2\pi$
0%
$\displaystyle \frac { 2 }{ 3 } +\pi$
0%
$\displaystyle \frac { 2 }{ 3 } -\pi$

Explanation

The required area

$\displaystyle =\int _{ -2 }^{ 2 }{ \sqrt { 8-{ x }^{ 2 } } dx } -\int _{ -2 }^{ 0 }{ \frac { 1 }{ 2 } { x }^{ 2 }dx } -\int _{ 0 }^{ 2 }{ xdx }$

$\displaystyle =2\int _{ 0 }^{ 2 }{ \sqrt { 8-{ x }^{ 2 } } dx } -{ \left[ \frac { { x }^{ 3 } }{ 6 } \right] }_{ -2 }^{ 0 }-{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 2 }$

$\displaystyle =2{ \left[ \frac { x }{ 2 } \sqrt { 8-{ x }^{ 2 } } +\frac { 8 }{ 2 } \sin ^{ -1 }{ \frac { x }{ 2\sqrt { 2 } } } \right] }_{ 0 }^{ 2 }-\frac { 4 }{ 3 } -2$

$\displaystyle =2\left[ 2+4.\frac { \pi }{ 4 } \right] -\frac { 10 }{ 3 } =\frac { 2 }{ 3 } +2\pi$

The area bounded by the curve

$y=\sqrt{x}$ , the line

$2y+3=x$ and the

$x$ -axis in the first quadrant is

0%
$9$
0%
$\displaystyle \frac{27}{4}$
0%
$36$
0%
$18$

Explanation

Given curves are

$y=\sqrt { x }$ ...(1)
and

$2y-x+3=0$ ...(2)
Solving (1) and (2), we get

$2\sqrt { x } -{ \left( \sqrt { x } \right) }^{ 2 }+3=0\\ \Rightarrow { \left( \sqrt { x } \right) }^{ 2 }-2\sqrt { x } -3=0\\ \Rightarrow \left( \sqrt { x } -3 \right) \left( \sqrt { x } +1 \right) =0\\ \Rightarrow \sqrt { x } =3$

$\because \sqrt { x } =-1$ is not possible

$\therefore y=3$
Hence required area

$\displaystyle =\int _{ 0 }^{ 3 }{ \left( { x }_{ 2 }-{ x }_{ 1 } \right) } dy=\int _{ 0 }^{ 3 }{ \left( { \left( 2y+3 \right) }-y^{ 2 } \right) } dy$

$\displaystyle =\left[ { y }^{ 2 }+3y-\frac { { y }^{ 3 } }{ 3 } \right] _{ 0 }^{ 3 }=9+9-9=9$

The area of the region bounded by the curves

$\displaystyle y=\sqrt{x}$ and

$\displaystyle y=\sqrt{4-3x}$ and

$\displaystyle y=0$ is

0%
$\dfrac{4}{9}$
0%
$\dfrac{16}{9}$
0%
$\dfrac{8}{9}$
0%
None of these

Explanation

From figure required area

$\displaystyle =\int _{ 0 }^{ 1 }{ \sqrt { x } dx } +\int _{ 1 }^{ \frac { 4 }{ 3 } }{ \sqrt { 4-3x } dx }$

$\displaystyle ={ \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 1 }+{ \left[ -\frac { 1 }{ 3 } \frac { { \left( 4-3x \right) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 1 }^{ \frac { 4 }{ 3 } }=\frac { 8 }{ 9 }$

Area of the region bounded by the curves

$y=x-1$ and

$y=3-|x|$ is

0%
$3$
0%
$4$
0%
$6$
0%
$2$

Explanation

Consider the curves

$y = |x-1|$ &

$y = 3-|x|$

when

$x< 0$

$3-(-x) = (-x+1)$

$\Rightarrow 3+x = -x+1$

$\Rightarrow 2x = -2$

$x = -1$

Put

$x = -1$ in

$y = |x-1|$ we have

$y = 2$

When

$x>0$

$3-x = x- 1$

$\Rightarrow 3+1 = 2x$

$x = 2$

put

$x = 2$ in

$y = |x-1|$ we have

$y = 1$

Thus pt. of intersection is

$(-1,2),(2,1)$

$\therefore$ required area

$\displaystyle = \int_{-1}^{0}(3+x+x-1)dx+\int_{0}^{1}(3-x+x-1)dx$

$\displaystyle +\int_{1}^{2}3-x-x+1dx$

$\displaystyle \Rightarrow \int_{-1}^{0}(2x+2)dx+\int_{0}^{1}2dx+\int_{1}^{2}(4-2x)dx$

$\displaystyle \Rightarrow \left ( \frac{2x^{2}}{2} \right )^{1}+(2x)_{-1}^{0}+(2x)_{0}^{1}+(4x)_{1}^{2}-\left ( \frac{2x^{2}}{2} \right )_{1}^{2}$

$\Rightarrow -1+2(-(-1))+2+4(2-1)-(2^{2}-1^{2})$

$\Rightarrow -1+2+2+4-3 = \boxed{4}$

The area bounded by the curve

$\displaystyle y=\sin x$ and

$\displaystyle y=\cos x,\forall 0\leq x\leq \pi /2$ is

0%
$\displaystyle 2\left( \sqrt { 2 } -1 \right)$
0%
$\displaystyle 2\sqrt{2} \left ( \sqrt{3} -1\right )$
0%
$\displaystyle 2\left ( \sqrt{2} +1\right )$
0%
$\displaystyle \sqrt{3}-1$

Area bounded by the curves

$\displaystyle y=xe^{x}$ and

$\displaystyle y=xe^{-x}$ and the line

$\displaystyle \left| x \right| =1$ is

0%
$\displaystyle 1$
0%
$\displaystyle \dfrac 4 e$
0%
$e$
0%
$\displaystyle -1$

The area common to the curves

$\displaystyle y^{2}=x$ and

$x^{2}=y$ is

0%
$1$
0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{1}{3}$
0%
None of these

Explanation

The given curves intersect at

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

$y=0$ or

$y=1$

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

$x=0$ or

$x=1$

$\displaystyle =\int _{ x=0 }^{ x=1 }{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) } dx=\int _{ x=0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 2 } \right) } dx$

$\displaystyle =\left[ \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ \dfrac { 3 }{ 2 } } -\dfrac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }=\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 }$

The area of the region bounded by the parabola

$\left ( y-2 \right )^{2}=x-1$ , the tangent to the parabola at the point

$(2, 3)$ and the

$x$ -axis is:

0%
$6$
0%
$9$
0%
$12$
0%
$3$

Explanation

$\left ( y-2 \right )^{2}=x-1$
Differentiating w.r.t.

$x$ , we have

$2\left ( y-2 \right ){y}'=1$

$\Rightarrow$

$\displaystyle {y}'=\frac{1}{2\left ( y-2 \right )}$

$(2, 3),$

${y}'=\dfrac{1}{2}$
The equation of the tengent to the parabola at

$(2, 3)$ is

$\displaystyle y-3=\frac{1}{2}\left ( x-2 \right )$

$\Rightarrow$

$x-2y+4=0$

$\therefore$ The area of the bounded region

$=\displaystyle \int_{0}^{3}\left [ \left ( y-2 \right )^{2}+1-\left ( 2y-4 \right ) \right ]dy$

$=\displaystyle \int_{0}^{3}\left ( y^{2}-6y+9 \right )dy=\int_{0}^{3}\left ( y-3 \right )^{2}dy$

(Let

$3-y=t$ )

$\Rightarrow dy=-dt$

When

$y=0,t=3$ and

$y=3, t=0$

$\therefore \displaystyle \int_{0}^{3}\left ( 3-y \right )^{2}dy=\int_{0}^{3}t^{2}dt=\left [ \frac{t^{3}}{3} \right ]_{0}^{3}=\frac{3^{3}}{3}=9$

The area bounded by

$y= x^{2}$ and

$y= 1-x^{2}$ is

0%
$\dfrac{\sqrt{8}}{3}$
0%
$\dfrac{16}{3}$
0%
$\dfrac{32}{3}$
0%
$\dfrac{17}{3}$

Explanation

Required are

$\displaystyle =2\left[ \int _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }{ \left( 1+{ x }^{ 2 } \right) } dx-\int _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }{ { x }^{ 2 } } dx \right]$

$\displaystyle ={ 2\left[ x+\dfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }-{ 2\left[ \dfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }=\dfrac { \sqrt { 8 } }{ 3 }$

The area bounded by

$\displaystyle \begin{vmatrix}y\end{vmatrix}=1-x^{2}$ is

0%
$8/3$
0%
$4/3$
0%
$16/3$
0%
None of these

Explanation

Clearly given curve is symmetrical about both the axis,
Hence required area

$=\displaystyle 4.\int_0^1(1-x^2)dx =4. \left[x-\frac{x^3}{3}\right]_0^1=8/3$

The area common to the circle

$x^2+y^2=16a^2$ and the parabola

$y^2=6ax$ is

0%
$\displaystyle \frac { 4{ a }^{ 2 } }{ 3 } \left( 4\pi -\sqrt { 3 } \right)$
0%
$\displaystyle \frac { 4{ a }^{ 2 } }{ 3 } \left( 8\pi -3 \right)$
0%
$\displaystyle \frac { 4{ a }^{ 2 } }{ 3 } \left( 4\pi +\sqrt { 3 } \right)$
0%
None of these

Explanation

Solving the two, we get

$\displaystyle { x }^{ 2 }+6a-16{ a }^{ 2 }=0$

$\displaystyle \left( x+8a \right) \left( x-2a \right) =0$

$\therefore \quad x=2a$

$\therefore \quad y=\pm 2\sqrt { 3a }$

The required common area

$=2[APOA]$

$=2\int { ydx } +2\int { ydx }$

$\displaystyle =2\int _{ 0 }^{ 2a }{ \sqrt { 6a } \sqrt { x } dx } +2\int _{ 2a }^{ 4a }{ \sqrt { { \left( 4a \right) }^{ 2 }-{ x }^{ 2 } } dx }$

$\displaystyle =2.\sqrt { 6a } \frac { 2 }{ 3 } \left[ { x }^{ 3/2 } \right] _{ 0 }^{ 2a }+2.\left[ \frac { x }{ 2 } \sqrt { { \left( 4a \right) }^{ 2 }-{ x }^{ 2 } } +\frac { 1 }{ 2 } { \left( 4a \right) }^{ 2 }\sin ^{ -1 }{ \frac { x }{ 4a } } \right] _{ 2a }^{ 4a }$

$\displaystyle =2.\sqrt { 6a } \frac { 2 }{ 3 } \left( 2a \right) \sqrt { 2a } +2\left[ \left( 0-a2\sqrt { 3a } \right) +8{ a }^{ 2 }\left( \sin ^{ -1 }{ 1 } -\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right]$

$\displaystyle =2.2\sqrt { 3 } .\frac { 4 }{ 3 } { a }^{ 2 }+2\left[ -2\sqrt { 3 } { a }^{ 2 }\left( \frac { \pi }{ 2 } -\frac { \pi }{ 6 } \right) \right]$

$\displaystyle =\frac { 16 }{ 3 } \sqrt { 3 } { a }^{ 2 }-4\sqrt { 3 } { a }^{ 2 }+16{ a }^{ 2 }\frac { \pi }{ 3 }$

$\displaystyle =\frac { 4\sqrt { 3 } { a }^{ 2 } }{ 3 } +\frac { 16\pi { a }^{ 2 } }{ 3 } =\frac { 4{ a }^{ 2 } }{ 3 } \left( 4\pi +\sqrt { 3 } \right)$

Let T be the triangle with vertices

$\displaystyle (0,0),(0,c^{2})$ and

$\displaystyle (c,c^{2})$ and let R be the region between

$y = cx$ and

$\displaystyle y=x^{2}$

0%
Area $\displaystyle (R)=\frac{c^{3}}{6}$
0%
Area of $\displaystyle R=\frac{c^{3}}{3}$
0%
$\displaystyle \lim_{c\rightarrow 0^{+}}\frac{Area(T)}{Area(R)}=3$
0%
$\displaystyle \lim_{c\rightarrow 0^{+}}\frac{Area(T)}{Area(R)}=\frac{3}{2}$

Explanation

Area(T)

$\displaystyle ==\frac { c.{ c }^{ 2 } }{ 2 } =\frac { { c }^{ 3 } }{ 2 }$
Area(R)

$\displaystyle =\frac { { c }^{ 3 } }{ 2 } -\int _{ 0 }^{ c }{ { x }^{ 2 }dx } =\frac { { c }^{ 3 } }{ 2 } -\frac { { c }^{ 3 } }{ 3 } =\frac { { c }^{ 3 } }{ 6 }$

$\displaystyle \therefore \lim _{ c\rightarrow { 0 }^{ + } }{ \frac { Area(T) }{ Area(R) } } =\lim _{ c\rightarrow { 0 }^{ + } }{ \frac { \frac { { c }^{ 3 } }{ 2 } }{ \frac { { c }^{ 3 } }{ 6 } } } =3$

The area of the closed figure bounded by

$y = x, y = -x$ & the tangent to the curve

$\displaystyle y=\sqrt{x^{2}-5}$ at the point (3, 2) is

0%
$5$
0%
$\displaystyle 2\sqrt{5}$
0%
$10$
0%
$\displaystyle \frac{5}{2}$

The area of the region(s) enclosed by the curves

$\displaystyle y=x^{2}$ and

$\displaystyle y=\sqrt{\left | x \right |}$ is:

0%
1/3
0%
2/3
0%
1/6
0%
1

Explanation

The curve

$y={ x }^{ 2 }$ and

$y=\sqrt { x }$ intersect at

$O(0,0)$ and

$P(1,1)$
Therefore required area

$\displaystyle =\int _{ 0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 2 } \right) dx } ={ \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 3 }$

Let 'a' be a positive constant number. Consider two curves

$\displaystyle C_{1}:y=e^{x},C_{2}:y=e^{a-x}$ . Let S be the area of the part surrounded by

$\displaystyle C_{1}$ ,

$\displaystyle C_{2}$ and the y-axis, then

0%
$\displaystyle \lim_{a\rightarrow \infty }S=1$
0%
$\displaystyle \lim_{a\rightarrow 0}\frac{S}{a^{2}}=\frac{1}{4}$
0%
Range of S is $\displaystyle \left ( 0, \infty \right )$
0%
S(a) is neither odd nor even

Explanation

$\displaystyle C_{1}:y=e^{x},C_{2}:y=e^{a-x}$

Point of intersection

$e^x={ e }^{ \left( a-x \right) }$

$x=\dfrac{a}{2},y={ e }^{ \dfrac { a }{ 2 } }$

$S=\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { C }_{ 2 } } dx-\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { C }_{ 1 } } dx$

$S=\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { e }^{ a-x } } dx-\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { e }^{ x } } dx={ \left[ { e }^{ a-x } \right] }_{ 0 }^{ \dfrac { a }{ 2 } }-{ \left[ { e }^{ x } \right] }_{ 0 }^{ \dfrac { a }{ 2 } }$

$S=-({ e }^{ \dfrac { a }{ 2 } }-{ e }^{ a })-({ e }^{ \dfrac { a }{ 2 } }-1)=({ e }^{ a }{ -2e }^{ \dfrac { a }{ 2 } }+1)={ \left( ({ e }^{ \dfrac { a }{ 2 } }-1) \right) }^{ 2 }$

$\lim _{ a\rightarrow 0 } \dfrac { s }{ a^{ 2 } } =\left[ \dfrac { { ({ e }^{ \dfrac { a }{ 2 } }-1) }^{ 2 } }{ \dfrac { a }{ 2 } } \right] *\dfrac { { a }^{ 2 } }{ 4{ a }^{ 2 } } =\dfrac { 1 }{ 4 }$

Area enclosed by the curves

$\displaystyle y=\ln x;y=\ln\left | x \right |;y=\left | \ln x \right |$ and

$\displaystyle y=\left | \ln\left | x \right | \right |$ is equal to

0%
$2$
0%
$4$
0%
$8$
0%
Cannot be determined

Explanation

Required area =

$2 \displaystyle \int_{0}^{1} |log |x||dx$

$=2[(x|log|x||)]_{0}^{1}- \displaystyle\int_{0}^{1}\left(-\dfrac{1}{2}\right)xdx$

$=2[(1-0)+(x)_{0}^{1}]$

$= 4 sq.units$

1060203_261282_ans_e0efae8d4052462f87e87106bc8a3bda.JPG

The area bounded by the parabola

$y=x^2+1$ and the straight line

$x+y=3$ is given by

0%
$\displaystyle\frac{45}{7}$
0%
$\displaystyle\frac{25}{4}$
0%
$\displaystyle\frac{\pi}{18}$
0%
$\displaystyle\frac{9}{2}$

Explanation

The two meet at

$3-x={ x }^{ 2 }+1\Rightarrow { x }^{ 2 }+x-2=0\Rightarrow \left( x+2 \right) \left( x-1 \right) =0\Rightarrow x=-2,1$

$\therefore$ Required area

$\displaystyle =\int _{ -2 }^{ 1 }{ \left[ \left( 3-x \right) -\left( { x }^{ 2 }+1 \right) \right] dx }$

$\displaystyle =\int _{ -2 }^{ 1 }{ \left( 2-x-{ x }^{ 2 } \right) dx } ={ \left[ 2x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ -2 }^{ 1 }$

$\displaystyle =\left( 2-\frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right) -\left( -4-\frac { 4 }{ 2 } +\frac { 8 }{ 3 } \right) =\frac { 9 }{ 2 }$

The area of the figure bounded by the lines

$x= 0,\: x= \dfrac{\pi}{2},\: f\left ( x \right )= \sin x$ and

$g\left ( x \right )= \cos x$ is

0%
$2\left ( \sqrt{2}-1 \right )$
0%
$\sqrt{3}-1$
0%
$2\left ( \sqrt{3}-1 \right )$
0%
$2\left ( \sqrt{2}+1 \right )$

Explanation

For

$x=0\ to\ \dfrac{\pi}{4}\ \cos x>\sin x\ and\ for\ x=\dfrac{\pi}{4}\ to\ \dfrac{\pi}{2} \sin x>\cos x$

So, Required area

$=\displaystyle \int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \left( \cos { x } -\sin { x } \right) dx } +\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \left( -\cos { x } +\sin { x } \right) dx }$

$\displaystyle ={ \left[ \sin { x } +\cos { x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }+{ \left[ -\cos { x } -\sin { x } \right] }_{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }=2\left( \sqrt { 2 } -1 \right)$

The area of the figure bounded by the curves

$\displaystyle y=\ln x$ &

$\displaystyle y=\left ( \ln x \right )^{2}$ is

0%
$e+1$
0%
$e-1$
0%
$3-e$
0%
$1$

Explanation

Given curves,

$\displaystyle y=\ln x$ &

$\displaystyle y=\left ( \ln x \right )^{2}$

Points of intersection of given curves are,

$(1,0)$ and

$(e,1)$
Hence required area is,

$A = \displaystyle \int_{1}^{e}\left ( \ln x-(\ln x)^{2} \right )dx$
On solving it by parts we get

$A\displaystyle =[x\ln x-x]_{1}^{e}-\left[x(\ln x)^{2}-\int_{1}^{e} 2\ln x dx\right]$

$A\displaystyle =[x\ln x-x]_{1}^{e}-\left[x(\ln x)^{2}- 2x\ln x +2x\right]_{1}^{e}$

$A=\left[x\ln x-x-x(\ln x)^{2}+2x\ln x-2x\right]_{1}^{e}$

$A=\displaystyle \left|3x\left ( \ln x-1 \right )\right|_{1}^{e}-|x(\ln x)^{2}|_{1}^{e}=3-e$

The area bounded by

$\displaystyle y={ xe }^{ |X| }$ and

$\displaystyle |x|=1$ is -

0%
4
0%
6
0%
1
0%
2

Explanation

$I=\int_{-1} ^{1} ydx$

$=\int_{-1} ^{1} xe^{|x|}dx$

$=\int_{-1} ^{0} xe^{-x}dx+\int_{0} ^{1} xe^{x}dx$

$=[-xe^{-x}+\int e^{-x}dx]_{-1} ^{0}+[xe^{x}-\int e^{x}dx]_{0} ^{1}$

$=[-xe^{-x}-e^{-x}]_{-1}^{0}+[xe^{x}-e^{x}]_{0} ^{1}$

$=[-1-(e-e)]+[e-e-(-1)]$

$=\left| -1 \right| +\left| 1 \right|$

we take modulus because area can not be negative and this function is symmetry about y axis.we have to put modulus otherwise area will be zero

so,

ans is

$2$

$y = (x)$ is the solution of equation

$ydx + dy = e^x y^ 2 dy, (0) = 1$ and area bounded by the curve

$y= (x)$ ,

$y = e^x$ and

$x = 1$ is A, then

0%
curve $y = (x)$ is passing through $(2,e)$
0%
curve $y = (x)$ is passing through $( 1, \displaystyle \frac{1}{e})$
0%
$A = e - \displaystyle \frac{2}{\sqrt e} + 3$
0%
$A = e + \displaystyle \frac{2}{\sqrt e} - 3$

Tangent is drawn from

$\displaystyle \left( 1,0 \right)$ to

$\displaystyle y={ e }^{ x }$ , then the area bounded between the coordinate axes and the tangent is equal to -

0%
$\displaystyle \frac { e }{ 2 }$
0%
$\displaystyle e$
0%
$\displaystyle \frac { { e }^{ 2 } }{ 2 }$
0%
$\displaystyle { e }^{ 2 }$

Explanation

Let

$P(t,e^t)$ be any point on

$y=e^x$

Slope of tangent at any point P on

$y=e^{x}$ is

$y'(t)=e^t$ ........

$(1)$

Since the tangent passes through

$(1,0)$

Therefore, slope of tangent is given by two point form i.e.

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{e^t}{t-1}$

From

$(1)$ :

$\dfrac{e^t}{t-1}=e^t$

$\Rightarrow t=2$

Therefore, equation of tangent is

$y-0=e^2(x-1)$

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

It is in intercept form

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

Therefore, area with co-ordinate axis is

$A=\dfrac{|a||b|}{2}=\dfrac{e^2}{2}$

Area bounded by

$\displaystyle y=2\sqrt { x }$ and

$x=3\sqrt { y }$ is equal to (in sq. units)

0%
12
0%
8
0%
10
0%
6

Explanation

First we have to find the intersection point.
Hence,

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

$\Rightarrow 18\sqrt{x}=x^{2}$

$\Rightarrow \sqrt{x}(18-x\sqrt{x})=0$

$\Rightarrow x=0$ and

$x\sqrt{x}=18$

$\Rightarrow x^{3}=18^{2}$
Now the area bounded by the curve is

$I=\displaystyle \int_{0} ^{18^{\frac{2}{3}}} 2\sqrt{x}-\dfrac{x^{2}}{9} dx$

$=\left [\dfrac{4x^{\frac{3}{2}}}{3}-\dfrac{x^{3}}{27}\right]_{0} ^{18^{\frac{2}{3}}}$

$=\dfrac{4\times 18}{3}-\dfrac{18^{2}}{27}$

$=24-12$

$=12$ sq units.

The area bounded by the parabola

$\displaystyle { x }^{ 2 }=8y$ and line

$\displaystyle x-2y+8=0$ is-

0%
36
0%
72
0%
18
0%
9

Explanation

Solving parabola

$\displaystyle { x }^{ 2 }=8y$
and line

$\displaystyle x-2y+8=0$ ,
we get

$\displaystyle { x }^{ 2 }=4\left( x+8 \right)$

$\displaystyle \Rightarrow \quad x=8,-4$

$\displaystyle \therefore$ Area of shaded region

$\displaystyle =\frac { 1 }{ 2 } \left[ 8+2 \right] \times 12-\int _{ -4 }^{ 8 }{ \frac { { x }^{ 2 } }{ 8 } dx }$

$\displaystyle =60-\frac { 1 }{ 8 } .\frac { { x }^{ 3 } }{ 3 } { | }_{ -4 }^{ 8 }=60-24=36$

The area bounded by the curves

$x^2+y^2\le 8$ and

$y^2\ge 4x$ lying in the first quadrant is not equal to

0%
$\displaystyle 32\left( \frac { \pi }{ 8 } -\frac1 3\right)$
0%
$\displaystyle \frac { 32 }{ 3 } \left( \frac { 3\pi }{ 8 } -1 \right)$
0%
$\displaystyle 4\pi -\frac { 32 }{ 3 }$
0%
$\displaystyle \frac { 1 }{ 3 } \left( 12\pi -32 \right)$

Explanation

Given

$x^2+y^2\le 8x$

Let

$x^2+y^2=8x$ ...(1)

$y^2\ge 4x, y^2=4x$ ...(2)

From (1) and (2), we have

$x^2=4x\Rightarrow x=0,x=4$

$\therefore$ Set of points

$($ or points of intersection

$)$ are

$(0,0),(4,4)$ .

Required area

$\displaystyle =\int _{ 0 }^{ 4 }{ \sqrt { { 4 }^{ 2 }-{ \left( x-4 \right) }^{ 2 } } } -\int _{ 0 }^{ 4 }{ 2\sqrt { x } dx } \quad \quad \quad \left( \because { x }^{ 2 }+{ y }^{ 2 }=8x \right)$

$\displaystyle \Rightarrow { \left( x-4 \right) }^{ 2 }+{ y }^{ 2 }={ 4 }^{ 2 }\Rightarrow { y }^{ 2 }={ 4 }^{ 2 }-{ \left( x-4 \right) }^{ 2 }$

$\displaystyle ={ \left[ \frac { \left( x-4 \right) }{ 2 } \sqrt { { 4 }^{ 2 }-{ \left( x-4 \right) }^{ 2 } } +\frac { 16 }{ 2 } \sin ^{ -1 }{ \left( \frac { x-4 }{ 4 } \right) } \right] }_{ 0 }^{ 4 }-2\times \frac { 2 }{ 3 } { \left[ { x }^{ 3/2 } \right] }_{ 0 }^{ 4 }$

$\displaystyle =8\times \frac { \pi }{ 2 } -\frac { 4 }{ 3 } \times { \left( 4 \right) }^{ 3/2 }=\left( 4\pi -\frac { 32 }{ 3 } \right) =\frac { 32 }{ 3 } \left( \frac { 3\pi }{ 8 } -1 \right)$ sq. units.

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 3 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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