MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 2

Area ofthe region bounded by

$y=|x|$ and

$\mathrm{y}=2$ is

0%
$4$ sq units
0%
$2$ sq. units
0%
$1$ sq. units
0%
$\displaystyle \frac{1}{2}$ sq. units

Explanation

$y=|x|$ and

$\mathrm{y}=2$
area of

$\triangle^{le}$ OAB

$=1/2\times(OC)\times(AB)$

$=1/2(2)(4)$

$=4 \:sq\:units$

The area of a region bounded by

$\mathrm{X}$ -axis and the curves defined by

$y=\tan x , 0\displaystyle \leq x\leq\frac{\pi}{4}$ and

$y=\displaystyle \cot x,\frac{\pi}{4}\leq x\leq\frac{\pi}{2}$ is:

0%
$log3$ sq. unlts
0%
$log5$ sq. unlts
0%
$log 1$ sq. unit
0%
$log 2$ sq. units

The area bounded by

$y=\cos x,\ y=x+1$ and

$y=0$ in the second quadrant is

0%
$\displaystyle \frac{3}{2}$ sq. units
0%
2 sq. units
0%
1 sq. unit
0%
$\displaystyle \frac{1}{2}$ sq,. units

Explanation

$y=\cos x,\ y=x+1$ and

$y=0$
Area of shaded region is required is (Area of curve OAB- area of

$\triangle^{le}$ OCB)

$\overset{0}{\underset{-\pi /2}{\int}}\cos x\:dx-\overset{0}{\underset{-1}{\int}}(x+1)dx$

$\sin x|_{-\pi /2}^{0}-\left ( \frac{x^2}{2}+x \right )|_{-1}^{0}$

$1-1/2=1/2 sq\:units$

The area bounded by tangent, normal and x-axis at

$\mathrm{P}(2,4)$ to the curve

$y=x^{2}$

0%
$34$
0%
$32$
0%
$36$
0%
$24$

Explanation

Slope of the tangent at

$(2,4)$

$=\dfrac{dy}{dx}|_{x=2}$

$=4$ .

Hence slope of normal at

$(2,4)$ is

$\dfrac{-1}{4}$ .

Therefore equation of tangent will be

$y-4=4(x-2)$

$y-4=4x-8$

$4x-y=4$

Hence it cuts the x axis at

$(1,0)$ .

Equation of normal will be

$y-4=\dfrac{-1}{4}(x-2)$

$4y-16=-x+2$

$x+4y=18$ .

Hence it cuts the x axis at

$(18,0)$ .

Now the normal and tangent meet each other at

$(2,4)$ .

Since normal is perpendicular to the tangent it is a right angled triangle.
The coordinates are

$(1,0),(18,0),(2,4)$ .

Hence
Length of hypotenuse= 17.
Length of base =

$\sqrt{17}$
Length of height =

$\sqrt{272}=4\sqrt{17}$ .

Hence

$Area=\dfrac{1}{2}(b\times h)$

$\dfrac{4\sqrt{17}\times \sqrt{17}}{2}$

$=2\times17$

$=34$ sq units.

Area of the region bounded by

$y=|x|$ and

$y=1-|x|$ is

0%
$\displaystyle \frac{1}{3}$ sq. units
0%
1 sq. units
0%
$\displaystyle \frac{1}{2}$ sq. unit
0%
2 sq. units

The area in square units bounded by the curves

$y=x^{3},\ y=x^{2}$ and the ordinates

${x}=1, {x}=2$ is

0%
$\displaystyle \frac{17}{12}$
0%
$\displaystyle \frac{12}{13}$
0%
$\displaystyle \frac{2}{7}$
0%
$\displaystyle \frac{7}{2}$

Explanation

The given curves are

$y=x^3$ and

$y=x^2$

And the limits of

$x$ are

$x=1, x=2$

Then, the required area is

$A=\displaystyle \int_{1}^{2}(x^3-x^2)dx$

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

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

$=\left ( \dfrac{16}{4}-\dfrac{1}{4} \right )-\left ( \dfrac{8}{3}-\dfrac{1}{3} \right )$

$=\dfrac{15}{4}-\dfrac{7}{3}=\dfrac{17}{12}$ sq units

The area, in square units of the region bounded by the parabolas

$y^{2}=4x$ and

$x^{2}=4y$ is

0%
$\dfrac{16}{3}$
0%
$\dfrac{32}{3}$
0%
$\dfrac{8}{3}$
0%
$\dfrac{4}{3}$

Explanation

The given parabolas are

$y^{2}=4x$ and

$x^{2}=4y$
The point of intersection of

$x^2=4y$ and

$y^2=4x$ will be

$x=4$ and

$x=0$

Then, the required area will be

$\overset{4}{\underset{0}{\displaystyle \int}}\left (\sqrt{4x} -\dfrac{x^2}{4} \right )dx$

$=\left[\sqrt{4}\times\dfrac{2}{3}x^{3/2}-\dfrac{x^3}{12}\right]_{0}^{4}$

$=\dfrac{4\times 8}{3}-\dfrac{4^3}{12}$

$=\dfrac{32}{3}-\dfrac{16}{3}$

$=\dfrac{16}{3}\:sq\:units$

The area bounded by the two curves

$y=\sqrt{x}$ and

$x=\sqrt{y}$ is:

0%
$\displaystyle \frac{1}{3}$ sq, units
0%
$\displaystyle \frac{2}{3}$ sq. units
0%
$\displaystyle \frac{1}{5}$ sq. units
0%
$\displaystyle \frac{1}{7}$ sq. units

Explanation

$y=\sqrt{x}$ and

$x=\sqrt{y}$
Area

$=\overset{1}{\underset{0}{\int}}(\sqrt{x}-x^2)dx$

$\quad\quad=\frac{2}{3}(1)-1/3\\\quad\quad=1/3\:sq\:units$

The area of the region bounded by

$x^{2}=8y,\ x=4$ and the

$\mathrm{x}$ -axis is

0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{8}{3}$
0%
$\displaystyle \frac{10}{3}$

Explanation

$y=\dfrac{x^2}{8};x=4$

Area

$=\overset{4}{\underset{0}{\int}}\dfrac{x^2}{8}\:dx$

$=\dfrac{x^3}{24}|_{0}^{4}$

$=\dfrac{4^3}{24}\\=\dfrac{64}{24}\\=\dfrac83\:sq\:units$

Area of the figure bounded by Y-axis,

$y=Sin^{-1}x,\ y=Cos^{-1}x$ and the first point of intersection from the origin is

0%
$2\sqrt{2}$
0%
$2\sqrt{2}+1$
0%
$\sqrt{2}-1$
0%
$\sqrt{2}+1$

Explanation

The curve

$y=\sin ^{ -1 }{ x }$ and

$y=\cos ^{ -1 }{ x }$ intesect at

$\displaystyle \left( \frac { 1 }{ \sqrt { 2 } } ,\frac { \pi }{ 4 } \right)$
Thus area

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

$\displaystyle ={ \left[ x\cos ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }-\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { -1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } +{ \left[ x\sin ^{ -1 }{ x } \right] }_{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }+\int _{ 0 }^{ \frac { 1 }{ \sqrt { 2 } } }{ x.\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } dx } \\ =\sqrt { 2 } -1$

The area bounded by the parabola

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

$y=x-6$ is

0%
$\displaystyle \frac{125}{3}$ sq. units
0%
$\displaystyle \frac{125}{6}$ sq. units
0%
$\displaystyle \frac{125}{4}$ sq. units
0%
$\displaystyle \frac{115}{3}$ sq. units

Explanation

$x=y^{2}$ and the line $y=x-6$
area in $QI$ + area in QIV
$|\overset {3 }{ \underset { -2}{ \int } } [(y+6)-y^2]dy|$
$=|\left | \dfrac{y^2}{2} +6y-y^3/3\right |_{-2}^{3}$
$=\left |\dfrac{9}{2} -\dfrac{4}{2}+6(5)-\left [ \dfrac{27}{3} +8/3\right ] \right |$
$\left | \dfrac{5}{2}+30-\left [ \dfrac{35}{3} \right ] \right | = \dfrac{15+180-70}{6} = \dfrac{125}{6}sq\:units.$

The area of the region bounded by

$y=x,\ y=x^{3}$ is:

0%
$\displaystyle \frac{1}{4}$ sq. units
0%
$\displaystyle \frac{1}{12}$ sq. units
0%
$\displaystyle \frac{1}{3}$ sq. units
0%
$\displaystyle \frac{1}{2}$ sq. units

Explanation

The curve intersect at

$x=-1,y=-1$ ;

$x=0,y=0$ and

$x=1,y=1$
Therefore area

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

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

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

The area bounded by the curve

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

$\mathrm{x}=4$ is:

0%
$\displaystyle \frac{32}{3}$ sq. units
0%
$\displaystyle \frac{16}{3}$ sq. units
0%
$\displaystyle \frac{8}{3}$ sq. units
0%
$\displaystyle \frac{4}{3}$ sq. units

Explanation

Symmetric about x-axis
So, area is

$2\overset{4}{\underset{0}{\int}}y\:dx$

$2\overset{4}{\underset{0}{\int}}\sqrt{x}\:dx$

$2\times\dfrac{2}{3}(2^3-0)=\dfrac{32}{3}sq\:units$

The area between the curve

$y=x^{2}$ and

$y=x+2$ is:

0%
$\displaystyle \frac{9}{2}$ sq. units
0%
$\displaystyle \frac{3}{2}$ sq. units
0%
9 sq. units
0%
6 sq. units

Explanation

point of intersection

$x^2-x-2=0$
2,-1

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

$3/2+2(3)-\left [ \frac{8}{3}+1/3 \right ]$

$3/2+6-3$

$6-3/2=9/2\:sq\:units$

The area of the region between the curve

$y=4x^{2}$ and the line

$y=6x-2$ is:

0%
$\displaystyle \frac{1}{9}$ sq. units
0%
$\displaystyle \frac{1}{12}$ sq. units
0%
$\displaystyle \frac{3}{2}$ sq. units
0%
$\displaystyle \frac{1}{5}$ sq. units

Explanation

The curve

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

$y=6x-2$ intersect at

$\displaystyle x=\frac { 1 }{ 2 } ,y=1$ and

$y=1,y=4$
Therefore area

$\displaystyle =\int _{ \frac { 1 }{ 2 } }^{ 1 }{ \left( 6x-2-4{ x }^{ 2 } \right) dx }$

$\displaystyle ={ \left[ \frac { 6{ x }^{ 2 } }{ 2 } -2x-\frac { 4{ x }^{ 3 } }{ 3 } \right] }_{ \frac { 1 }{ 2 } }^{ 1 }=\frac { 1 }{ 12 }$

The area bounded by the parabola

$y^{2}=4x$ and the line

$y=2x-4$ :

0%
9 sq. units
0%
5 sq. units
0%
4 sq. units
0%
2 sq. units

Explanation

$y^{2}=4x$ and the line

$y=2x-4$
area in

$QI$ +area in QIV

$QI=$

$\overset{4}{\underset{0}{\int}}2\sqrt{x}\:dx- area of \triangle ^{le}$

$=$

$\overset{4}{\underset{0}{\int}}2\sqrt{x}\:dx-1/2(2)(4)=\dfrac{32}{3}-4$

$QIV=$

$|\overset{1}{\underset{0}{\int}}-\sqrt{x}\:dx|+1/2\times1\times1$

$+\dfrac{4}{3}+1$

$QI$ +QIV

$= 12-3=9\:sq\:units$

The area bounded by the line

$\mathrm{x}=1$ and the curve

$\sqrt{\dfrac{y}{x}}+\sqrt{\dfrac{x}{y}}=4$ is

0%
$2\sqrt{3}$
0%
$\sqrt{3}$
0%
$3\sqrt{2}$
0%
$4\sqrt{3}$

Explanation

Let

$\sqrt{\dfrac{y}{x}}=t$ .
Then

$t+\dfrac{1}{t}=4$

$t^{2}-4t+1=0$

$t=\dfrac{4\pm\sqrt{12}}{2}$

$=2\pm\sqrt{3}$
Hence

$\dfrac{y}{x}=(2\pm\sqrt{3})^{2}$

$y=x(7-4\sqrt{3})$ and

$y=x(7+4\sqrt{3})$
Hence the required area is the area of the triangle formed by the lines

$y=x(7-4\sqrt{3})$ and

$y=x(7+4\sqrt{3})$ and

$x=1$ .
Hence we get the area
=(Area of the right angled triangle formed by

$y=(7+4\sqrt{3})$ , x-axis and x=1 )-(Area of the right angled triangle formed by

$y=(7-4\sqrt{3})$ , x-axis and x=1 )

$=\dfrac{1}{2}[(7+4\sqrt{3})\times 1]-\dfrac{1}{2}[(7-4\sqrt{3})\times1)]$

$=\dfrac{1}{2}[7+4\sqrt{3}-7+4\sqrt{3}]$

$=4\sqrt{3}$ units.

Area of the region enclosed by

$y^{2}=8x$ and

${y}=2{x}$ is

0%
$\dfrac{4}{3}$
0%
$\dfrac{3}{4}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{2}$

Explanation

Given curves are $y^2=8x$ and $y=2x$
Let's find out their intersection point

$\Rightarrow 4x^{2}=8x$

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

$\Rightarrow x(x-2)=0$

$\Rightarrow x=0,2$

The required area is
$A=\displaystyle \overset{2}{\underset{0}{\int}}(\sqrt{8x}-2x)dx$
$=\displaystyle \overset{2}{\underset{0}{\int}}\sqrt{8x}\:dx-\overset{2}{\underset{0}{\int}}2x\:dx$
$=\sqrt{8}\times\dfrac{2}{3}\times [x^{3/2}]_{0}^{2}-[x^2]_{0}^{2}$
$=\dfrac{2\sqrt{8}}{3}\sqrt{8}-4$
$=\dfrac{16}{3}-4=\dfrac{4}{3}\:sq\:units$

The area between the curves

$y=\sqrt{x}$ and

$y=x^{3}$ is

0%
$\displaystyle \frac{1}{12}$ sq. units
0%
$\displaystyle \frac{5}{12}$ sq. units
0%
$\displaystyle \frac{3}{5}$ sq. units
0%
$\displaystyle \frac{4}{5}$ sq. units

Explanation

$y=\sqrt{x}$ and

$y=x^{3}$

The point of intersection of these curves will be

$\sqrt{x}=x^{3}\Rightarrow x=0,1$

Then, the area is given by

$A=\displaystyle \overset {1}{ \underset { 0}{ \int } } (\sqrt{x}-x^3)dx$

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

$=\dfrac{2}{3}(1)-\dfrac{1}{4}$

$=\dfrac{8-3}{12}$

$=\dfrac{5}{12}sq\:units.$

The area bounded by the parabola

$x^{2}=4ay,\ \mathrm{x}$ -axis and the straight line

$\mathrm{y}=2\mathrm{a}$ is:

0%
$16\sqrt{2}a^{2}$ sq. units
0%
$\displaystyle \frac{16\sqrt{2}}{3}a^{2}$ sq. units
0%
$\displaystyle \frac{32\sqrt{2}}{3}a^{2}$ sq. units
0%
$\displaystyle \frac{32\sqrt{2}}{5}a^{2}$ sq. units

Explanation

So, area of the shaded region
$-\overset{2\sqrt{2}a}{\underset{0}{\int}}x^2/4a\:dx+\overset{2\sqrt{2}a}{\underset{0}{\int}}(2a)\:dx$
$2a(2\sqrt{2a})-\dfrac{}{4a}\dfrac{x^3}{3}|_{0}^{2\sqrt{2a}}$
$=4\sqrt{2}a^2-\dfrac{(2^{3/2}a)^3}{4a\times 3}$
$4\sqrt{2}a^2-\dfrac{2\times2\sqrt{2}a^2}{3}$
$=\dfrac{16\sqrt{2}a^2}{3}sq\:units.$

The area bounded by

$\mathrm{y}=\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\ \mathrm{y}=\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x}$ between any two successive intersections is:

0%
$2$
0%
$\sqrt{2}$
0%
$2\sqrt{2}$
0%
4

Explanation

$\sqrt{2}\overset {5\pi/4}{ \underset { \pi/4}{ \int } } \sin (x-\pi /4)dx$

$=-\sqrt{2}\cos (x-\pi /4)|_{\pi /4}^{5\pi /4}$

$=-\sqrt{2}[\cos \pi-\cos 0]$

$=2\sqrt{2}sq\:units.$

The area bounded by the curves

$y=\sin x,y=$ cosx and the

$\mathrm{y}$ -axis and the first point of intersection is:

0%
$\sqrt{2}$ sq,.units
0%
$\sqrt{2}-1$ sq. units
0%
$2+\sqrt{2}$ sq. units
0%
0 sq, units

Explanation

$\overset {\pi/4}{ \underset { 0}{ \int } } (\cos x-\sin x)\:dx$

$\overset {\pi/4}{ \underset { 0}{ \int } } \cos x\:dx-\overset {\pi/4}{ \underset { 0}{ \int } } \sin x\:dx$

$\sin x|_{0}^{\pi /4}+\cos x|_{0}^{\pi /4}$

$=1/\sqrt{2}+1/\sqrt{2}-1$

$=\sqrt{2}-1 sq\:units.$

Assertion(A): The area bounded by

$y^{2}=4x$ and

$x^{2}=4y$ is

$\displaystyle \frac{16}{3}$ sq. units.

Reason(R): The area bounded by

$y^{2}=4ax$ and

$x^{2}=4ay$ is

$\displaystyle \frac{16a^2}{3}$ sq. units

0%
Both A and R are true and R is the correct explanation of A.
0%
Both A and R are true but R is not the correct explanation of A.
0%
A is true but R is false.
0%
A is false but R is true.

Explanation

Given

$y^2=4ax$ and

$x^2=4ay$

Since both curves intersect,

$\therefore x=\dfrac{y^2}{4a}$

$\Rightarrow (\dfrac{y^2}{4a})^2=4ay$

$\Rightarrow y^4=64a^3y$

$\Rightarrow y[y^3-(4a)^3]=0$

$\Rightarrow y=0,4a.$

When

$y=0,x=0$ and when

$y=4a,x=4a,$

Area

$A=\int_{0}^{4a}\left(\sqrt(4ax)-\dfrac{x^2}{4a}\right)dx$

on solving we will get

$A=\dfrac{16}{3a^2}$

Now if we put

$a=1$ ,

we will get

$A=\dfrac{16}{3}$

I : The area bounded by

$x=2\cos\theta,\ y=3\sin\theta$ is

$36\pi$ sq. units.
II: The area bounded by

$x=2\cos\theta,\ y=2\sin\theta$ is

$4\pi$ sq.units.
Which of the above statement is correct?

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I nor II.

Explanation

$I$ :    $\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{3} \right )^2=1$
           $Area=\pi \times 2 \times3$
             $=6\pi sq\:units.$
$I$       $\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{2} \right )^2=1$
              $\Rightarrow x^2+y^2=4$
              $\pi (2)^2$
              $=4\pi\:sq\:units.$

The area of the triangle formed by the positive X-axis and the normal and tangent to the circle

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

$(1,\sqrt{3})$ in sq. units is:

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

Explanation

the tangent of

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

$(1,\sqrt{3}) \:is\: \sqrt{3}y+x=4$ and that of normal is

$y=\sqrt{3}x$
So, the area of

$\triangle ^{le}=1/2\times4\times\sqrt{3}=2\sqrt{3}\:sq\;units$

The area between the curves

$y=\tan x$ ,

$y=\cot x$ and

$\mathrm{x}$ -axis

$(0\displaystyle \leq x\leq\frac{\pi}{2})$ is

0%
$\log 2$
0%
$2 \log 2$
0%
$\displaystyle \frac{1}{2}$ $\log2$
0%
$1$

Explanation

$y=\tan x$ ,

$y=\cot x$ and

$\mathrm{x}$ -axis

$(0\displaystyle \leq x\leq\frac{\pi}{2})$

$\overset {\pi/4}{ \underset { 0}{ \int } } \tan x\:dx+\overset {\pi/2}{ \underset { \pi/4}{ \int } } (\cot x)dx$

$-\log |\cos x||_{0}^{\pi /4}+\log|\sin x||_{\pi /4}^{\pi /2}$

$-\log(1/\sqrt{2})+\log(1)-\log(1/\sqrt{2})$

$\log 2$

I: The area bounded by the line

$\mathrm{y}=\mathrm{x}$ and the curve

$y=x^{3}$ is

$1/_{2}$ sq. units.
II: The area bounded by the curves

$y=x^{3}$ and

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

$x=1$ ,

$x=2$ is

$\frac{7}{12}$ sq. units.
Which of the above statement is correct?

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I nor II.

Explanation

$SI$ :

$\overset {1}{ \underset { 0 }{ \int } } (x-x^3)dx=\left (\dfrac{x^2 }{2}-\dfrac{x^4}{4} \right )\bigg|_{0}^{1}$

$=2(\dfrac12-\dfrac14)=2(1/4)sq\:units$

$=\dfrac12sq\:units.$

$SII:$

$\overset {2}{ \underset { 1 }{ \int } } (x^3-x^2)dx$

$=\dfrac{x^4}{4}-\dfrac{x^3}{3}\bigg|_{1}^{2}$

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

$=\dfrac{15}{4}-\dfrac{7}{3}=\dfrac{45-28}{12}=\dfrac{17}{12}sq\:units.$

only

$I$ is true

Assertion(A): The area bounded by

$y^{2}=4x$ and

$y=x$ is

$\displaystyle \frac{8}{3}$ sq. units.

Reason(R): The area bounded by

$y^{2}=4ax$ and

$y=mx$ is

$\displaystyle \frac{8a^{2}}{3m^{3}}$ sq. units.

0%
Both A and R are true and R is the correct explanation of A.
0%
Both A and R are true but R is not the correct explanation of A.
0%
A is true but R is false.
0%
A is false but R is true.

Explanation

$\overset {4a/m^2 }{ \underset { 0 }{ \int } } (\sqrt{4x}-x)dx$

$\frac{2}{3}\times2a^{1/2}x^{3/2}-\frac{x^2}{2}|_{0}^{4a/m^2}$

$=8a^2/3m^3$ (R is true)
A is true by inspection
So, A,R are true and R explains A.

Area bounded by

$\displaystyle \mathrm{f}(\mathrm{x})=\max.(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})$

$\displaystyle \forall 0\leq x\leq\frac{\pi}{2}$ and the co-ordinate axis is equal to:

0%
$\displaystyle \frac{1}{\sqrt{2}}$ sq.units
0%
$\sqrt{2}$ sq.units
0%
$2$ sq.units
0%
$1$ sq. unit

Explanation

$\displaystyle \mathrm{f}(\mathrm{x})=\max.(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})$

$\displaystyle \forall 0\leq x\leq\frac{\pi}{2}$

$\overset {\pi/4}{ \underset { 0}{ \int } } \cos x\:dx+\overset {\pi/2}{ \underset { \pi/4}{ \int } } \sin x\:dx$

$\sin x|_{0}^{\pi /4}-cos x|_{0}^{\pi /4}$

$=(1/\sqrt{2})-[0-1/\sqrt{2}]$

$=\sqrt{2}sq\:units.$

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 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 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

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

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=

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

$=1\:sq\:units.$

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

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

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

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 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 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

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

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

The area bounded by the curve

$x^2=4y$ and straight line

$x=4y-2$ is

0%
$\frac{3}{8}$
0%
$\frac{5}{8}$
0%
$\frac{7}{8}$
0%
$\frac{9}{8}$

The area bounded by the curve

$f(x) = ce^x(c > 0)$ , the x-axis and the two ordinates x = p and x = q is proportional to

0%
$f(p) . f(q)$
0%
$|f(p)-f(q)|$
0%
$f(p) + f(q)$
0%
$\sqrt{f(p)f(q)}$

Explanation

Area =

$\int _{ p }^{ q }{ { ce }^{ x } }$

$=c({ e }^{ q }-{ e }^{ p })$

$= f(q)-f(p) = |f(p) - f(q)|$

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 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 =

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

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

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

$\frac { 16 }{ 3 }$

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

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

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

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

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

$\frac { 49 }{ 12 }$

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

Semicircles are drawn outside by taking every side of regular hexagon as a diameter. The perimeter of hexagon is 60 cm. Find the area of complete figure formed as such.(

$\pi$ = 3.14) (

$\sqrt3$ = 1.73)

0%
495 c $m^2$
0%
259.5c $m^2$
0%
235.5c $m^2$
0%
695.5c $m^2$

Explanation

The area of a hexagon=

$\quad \dfrac { 3\sqrt { 3 } }{ 2 } { side }^{ 2 }.\quad$

There are 6 semicircles=3 circles.

radius of each circle=r=

$\quad \dfrac { 1 }{ 2 } \times$ side of the hexagon.

ar.complete figure=ar.hexagon+3(ar.circle).

Here ,

perimeter of the regular hexagon=60cm.

So one side=

$\quad \dfrac { 60 }{ 6 } cm=10cm.\\ \therefore \quad ar.hexagon\\ =\dfrac { 3\sqrt { 3 } }{ 2 } { side }^{ 2 }\\ =\dfrac { 3\sqrt { 3 } }{ 2 } { 10 }^{ 2 }{ cm }^{ 2 }\\ =\dfrac { 3\times 1.73\times 100 }{ 2 } { cm }^{ 2 }\\ =259.5{ cm }^{ 2 }.\quad$

Again radius of each circle=r=

$\quad \dfrac { 1 }{ 2 } \times$ side of the hexagon=

$\dfrac{10}{2}cm=5cm.$

$\quad ar.(3circles)=3\times \pi \times { r }^{ 2 }\\ =3\times 3.14\times 5\times 5{ cm }^{ 2 }\\ =235.5{ cm }^{ 2 }.\quad$

So ar.complete figure=ar.hexagon+3(ar.circle).

$=(259.5+235.5)$

$\quad { cm }^{ 2 }\quad$

=495

$\quad { cm }^{ 2 }$

Ans- Option A

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

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

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

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

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

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

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

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 2 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 2 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

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