MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 4

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 bounded between the parabolas

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

$x^2 = 9y$ , and the straight line

$y = 2$ is:

0%
$10\sqrt2$
0%
$20\sqrt2$
0%
$\dfrac{10\sqrt2}{3}$
0%
$\dfrac{20\sqrt2}{3}$

Explanation

The given two parabolas are

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

$x^2=9y$

We have to find the area bounded between the parabolas and the straight line

$y=2$

The required area is equal to the shaded region in the drawn figure.

The area of the shaded region is twice the area shaded in first quadrant.

$\Rightarrow$ Required area

$=2\displaystyle\int_0^2\left(3\sqrt{y}-\dfrac{\sqrt{y}}{2}\right)dy$

$=2\displaystyle\int_0^2\left(\dfrac{6\sqrt{y}-\sqrt{y}}{2}\right)dy$

$=2\displaystyle\int_0^2\left(\dfrac{5\sqrt{y}}{2}\right)dy$

$=5\left(\dfrac{y^{\dfrac{3}{2}}}{3/2}\right)^{2}_{0}$

$=\dfrac{10}{3}\left(2^{\dfrac{3}{2}}-0\right)$

$=\dfrac{20\sqrt{2}}{3}$

1446627_760476_ans_eae73dea63a640b8b1f68c18875f0895.png

The parabola

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

${ x }^{ 2 }=4y$ divide the square region bounded by the lines

$x=4, y=4$ and the coordinate axes. If

${ S }_{ 1 }, { S }_{ 2 }$ and

${ S }_{ 3 }$ are respectively the areas of these parts numbered from top-to-bottom, then

${ S }_{ 1 } : { S }_{ 2 } : { S }_{ 3 }$ is

0%
$1 : 1 : 1$
0%
$2 : 1 : 2$
0%
$1 : 2 : 3$
0%
$1 : 2 : 1$

Explanation

${ S }_{ 1 }={ S }_{ 3 }=\displaystyle\int _{ 0 }^{ 4 }{ \dfrac { { x }^{ 2 } }{ 4 } dx }$

$={ \left[ \dfrac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }=\left[ \dfrac { 64 }{ 12 } -0 \right] =\dfrac { 16 }{ 3 }$
Now,

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

$={ \left[ 2{ x }^{ { 3 }/{ 2 } }\cdot \dfrac { 2 }{ 3 } -\dfrac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }=\dfrac { 16 }{ 3 }$

$\therefore { S }_{ 1 } : { S }_{ 2 } : { S }_{ 3 }=\dfrac { 16 }{ 3 } : \dfrac { 16 }{ 3 } : \dfrac { 16 }{ 3 } = 1 : 1 : 1$

The area enclosed between the parabolas

$y^{2} = 16x$ and

$x^{2} = 16y$ is

0%
$\dfrac {64}{3} sq. units$
0%
$\dfrac {256}{3} sq. units$
0%
$\dfrac {16}{3} sq. units$
0%
None of these

Explanation

Both parabolas cut each other at

$(0,0)$ and

$(16,16)$

Area enclosed by these parabolas

$=\displaystyle\int_0^{16}4\sqrt x dx-\int_0^{16} \dfrac{x^2}{16}dx$

$=\Bigl[\dfrac{2\times 4\times x^{3/2}}{3}\Bigr]_0^{16}-\Big[\dfrac{x^3}{16\times 3}\Bigr]_0^{16}$

$=\dfrac{2\times 4^4}{3}-\dfrac{4^4}{3}=\dfrac{256}{3}$ sq. units

The area of the region described by

$\begin{Bmatrix} (x,,y)/x^2 +y^2 \leq 1 and\ y^2\leq1-x\end{Bmatrix}$ is

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

Area of the region bounded by the curves

$y={ 2 }^{ x },y=2x-{ x }^{ 2 },x=0$ and

$x=2$ is given by

0%
$\cfrac { 3 }{ \log { 2 } } -\cfrac { 4 }{ 3 }$
0%
$\cfrac { 3 }{ \log { 2 } } +\cfrac { 4 }{ 3 }$
0%
$3\log { 2 } -\cfrac { 4 }{ 3 }$
0%
$3\log { 2 } +\cfrac { 4 }{ 3 }$

Explanation

Let the required area be

$A$ sq. units. Then,

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

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

$={ \left[ \cfrac { { 2 }^{ x } }{ \log { 2 } } -{ x }^{ 2 }+\cfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }$

$=\cfrac { 4 }{ \log { 2 } } -4+\cfrac { 8 }{ 3 } \cfrac { 1 }{ \log { 2 } }$

$=\cfrac { 3 }{ \log { 2 } } -\cfrac { 4 }{ 3 } sq.units$

The area enclosed by the curves

$|y + x|\leq 1, |y - x|\leq 1$ and

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

0%
$\left (2 + \dfrac {\pi}{2}\right )$ sq. units
0%
$\left (2 - \dfrac {\pi}{2}\right )$ sq. units
0%
$\left (3 + \dfrac {\pi}{2}\right )$ sq. units
0%
$\left (3 - \dfrac {\pi}{2}\right )$ sq.units

Explanation

The curves

$y + x\leq 1$

$and$

$|y - x|\leq 1$

$form a square$

$ABCD$

$and$

$2x^{2} + 2y^{2} = 1$

$is a circle of radius$

$\dfrac {1}{\sqrt {2}}$

$and centre$

$(0, 0)$

$. Therefore, required area$

$=$

$Area of square$

$-$

$Area of circle$

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

Area bounded by

$f(x)=max.\left( \sin { x } ,\cos { x } \right)$

$\quad \forall 0\le x\le \cfrac { \pi }{ 2 }$ and the co-ordinate axis is equal to

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

Explanation

$f(x)=max(\sin x,\cos x)$

$0<x<\pi/4, \cos x>\sin x$

$\pi/4<x<\pi/2, \cos x>\sin x$

$f(x)=\cos x, 0<x<\pi/4\\ =\sin x,\pi/4<x<\pi /2$

Area

$=|\int _{ 0 }^{ \pi /4 }{ \cos { x } } dx|+|\int _{ \pi /4 }^{ \pi /2 }{ \sin { x } } dx|$

$|\sin \pi/4-\sin 0|+|\cos\pi/2+\cos \pi/4|$

$=\sqrt 2$ units

$^2$

1141982_698655_ans_6ff327a8ab22444e8a69a2d8301a402c.PNG

What is the area of the rectangle , whose length is

$5\sqrt 3\ cm$ and breadth is

$5\ cm$ .

0%
$43.3\ cm^2$
0%
$\sqrt {75}\ cm^2$
0%
$25\ cm^2$
0%
None of these

Explanation

Area of rectangle having length

$l$ and breadth

$b=l\times b$

$l=5\sqrt3 cm$ and

$b=5 cm$ , then Area

$=5\sqrt3\times 5cm^2\\ =25\sqrt3=25\times1.73=43.25cm^2$

The area bounded by the circles

${ x }^{ 2 }+{ y }^{ 2 }=1, { x }^{ 2 }+{ y }^{ 2 }=4$ in the first Quadrant is

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

Explanation

$x^2+y^2=1$ and

$x^2+y^2=4$ are concentric circles

They form Ring in between them.

Area is given by

$\pi(4-1)=3\pi$

In first Quadrant it is divided by

$4$

So, Area is given by

$\dfrac{3\pi}{4}$

Find the area of a shaded portion.

0%
$270cm^2$
0%
$360cm^2$
0%
$180cm^2$
0%
$150cm^2$

Explanation

Area of shaded region

$=ar(PQRS)-ar(\triangle ORS)$

$=20\times 18-\cfrac{1}{2}\times20\times 18\\=360-180=180cm^2$

1174914_879024_ans_3b80c740f0554aaab35b2f7954ec53d5.png

If the area bounded by the curve

$y=a{ x }^{ 2 }\quad$ and

$x=a{ y }^{ 2 },\left( a>0 \right)$ is

$3sq.units$ , then the value of

$a$ is

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

Explanation

The parabolas intersect at

$(0,0)$ and

$(\dfrac{1}{a},\dfrac{1}{a})$

Area bounded

$=\displaystyle \int_{0}^{\frac{1}{a}}(\sqrt{\dfrac{x}{a}}-ax^2)dx$

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

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

$=\left [ \dfrac{2}{3a^2} - \dfrac{1}{3a^2} \right ]$

$= \dfrac{1}{3a^2}$

It is given that area

$= 3$ sq.units

Therefore,

$\dfrac{1}{3a^2}=3$

$\implies a = \dfrac{1}{3}$

Hence, answer is option (B).

If the area of the region bounded by the curves,

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

$y=0$ and

$x=t(t > 1)$ is

$1$ sq. unit, then t is equal to?

0%
$\displaystyle\frac{4}{3}$
0%
$e^{{2}/{3}}$
0%
$\displaystyle\frac{3}{2}$
0%
$e^{{3}/{2}}$

Explanation

Given,

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

Area bounded by the curves is the region ABCD.

Therefore, area

$=\displaystyle \int_{0}^{1}x^2dx + \int_{1}^{t}\dfrac{1}{x}dx$

$=\left [ \dfrac{x^3}{3} \right ]_{0}^{1} +\left [ \ln (x) \right ]_{1}^{t}$

$=\dfrac{1}{3}+\ln(t)$

It is given that area enclosed is

$1$

$\Rightarrow \dfrac{1}{3}+\ln(t)=1$

$\Rightarrow \ln(t)=\dfrac{2}{3}$

$\Rightarrow t = e^{\frac{2}{3}}$

Hence, answer is option (B)

The area of the closed figure bounded by the following curves.
y = 7x -

$2x^2$ , x + y = 7/2= 8 sq m

0%
True
0%
False

Find the area of the closed figure bounded by the following curves

$y =$

$\sqrt{x}, y \, = \, \sqrt{4 - 3x}$ , y = 0.

0%
$\dfrac 89$
0%
$\dfrac 79$
0%
$\dfrac 59$
0%
$\dfrac 13$

Explanation

Given curve,

$y = \sqrt{x} , y = \sqrt{4-3x}, y = 0$

$y > 0, x \ge 0 , y - 3x \ge 0$ (basic square root conditions)

$\Rightarrow x \le 4/3$ __(1)

Now,

$y = \sqrt{4 - 3x} \Rightarrow y^2 = 4-3x$

$x = \dfrac{4 - y^2}{3} = \dfrac{(2-y)(2 + y)}{3}$

It represents a parabola cutting y-axis at

$(0,2), (0,-2)$ & x-axis at

$(4/3, 0)$

Ref. image

Point of intersection

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

$\Rightarrow 4-3x = x$

$x = 1 \Rightarrow y = \sqrt{1} = 1$

$(1, 1)$ is req. point

Area bounded by curves (shaded)

$= \displaystyle \int_0^1 \sqrt{x} dx + \int_1^{4/3} \sqrt{4 - 3x} dx$

$= \dfrac{2}{3} \left[x^{3/2}\right]_0^1 + \left[\dfrac{2}{3} \times \dfrac{(4 - 3x)}{(-3)}^{3/2} \right]^{4/3}_1$

$= \dfrac{2}{3} (1 - 0) + \left(\dfrac{-2}{9} \right) (0-1)$

$= \dfrac{2}{3} + \dfrac{2}{9} = \dfrac{8}{9}$

$\therefore$ option A is correct.

1451409_890351_ans_63ed3975669d4d1cb2571e9f0753e47d.png

The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:

0%
1:2
0%
2:3
0%
2:5
0%
3:4
0%
3:5

Explanation

Let side of square

$=a$

& radii of circle

$=r$

$(b=a/2)$

${ r }^{ 2 }=\dfrac { { a }^{ 2 } }{ 4 } +{ a }^{ 2 }=\dfrac { { 5a }^{ 2 } }{ 4 }$

Area of square

$={ a }^{ 2 }=\dfrac { { 4r }^{ 2 } }{ 5 }$

Semicircle

${ a }^{ 2 }={ 2r }^{ 2 }$

$\dfrac { Ar.\quad of\quad sq }{ Ar.\quad of\quad circle } =\dfrac { { a }^{ 2 } }{ { \pi r }^{ 2 } } =\dfrac { { 2a }^{ 2 } }{ { \pi a }^{ 2 } } =\dfrac { 2 }{ \pi }$

Ar of

${ \left( square \right) }_{ circle }={ a }^{ 2 }={ 2r }^{ 2 }$

Ratio

$\rightarrow \dfrac { { 4r }^{ 2 } }{ 5\times { 2 }r^{ 2 } } =\dfrac { 2 }{ 5 }$

1225634_922042_ans_46a3b85350294fb6afe6148900a5cdf6.jpg

The area enclosed by the curves

$f(x) = \vert sin x - cos x \vert + \vert cos x + sin x \vert \ \text {and} \ g(x) = 2\vert cos x + sin x \vert , 0 \leq x \leq \pi$

0%
$2(2 - \sqrt2)$
0%
$4(2 + \sqrt2)$
0%
$4(2 - \sqrt2)$
0%
$2(2 + \sqrt2)$

The area of figure bounded by the curve

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

$y=-x$ is

0%
$\frac { 9 }{ 2 }$
0%
$9$
0%
$\frac { 7 }{ 2 }$
0%
$7$

$\int { [g(x)-f(x)]dx=5 }$ , then the area between two curves for 0 < x < 2, is

0%
5
0%
10
0%
15
0%
20

The area bounded by the curve

$y=x^2$ and

$y \, = \, \dfrac{2}{1 \, + \, x^2}$ is

$\lambda\ sq.\ unit$ , then the value of

$[\lambda ]$ is

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

Explanation

Required area =

$\int_{-1}^{1}\left(\dfrac{2}{1+x^2} \, - \, x^2\right)dx$

$2\int_{0}^{1}\left(\dfrac{2}{1+x^2} \, - \, x^2\right)dx$

$2\left(2tan^{-1}x \, - \, \dfrac{x^3}{3}\right)^1_0 \, = \, 2\left(\dfrac{\pi}{2} \, - \, \dfrac{1}{3}\right) \, = \, \pi \, - \, \dfrac{2}{3}$

$\therefore \left[\pi \, - , \dfrac{2}{3}\right] \,= \, 2$
1065584_993334_ans_82f45e6add23411182f84b496e3f95af.png

Area of region bounded by

$x=0$ ,

$y=0$

$x=2$ ,

$y=2$ ,

$y\le {e}^{x}$ &

$y\ge lnx$ is

0%
$6-4ln2$
0%
$4 ln2-2$
0%
$2ln2-4$
0%
$6-2ln2$

Explanation

$Area=\int _{ 0 }^{ 2 }{ \left( x-\ln { x } \right) dx } +\int _{ 0 }^{ 2 }{ \left( x-\ln { y } \right) dy } \\ Area={ \left[ \cfrac { { x }^{ 2 } }{ 2 } -\left( x\ln { x-x } \right) \right] }_{ 0 }^{ 2 }+{ \left[ \cfrac { { y }^{ 2 } }{ 2 } -\left( y\ln { y-y } \right) \right] }_{ 0 }^{ 2 }\\ Area=4-4\ln { 2 } +2\\ Area=6-4\ln { 2 }$

1137403_994284_ans_cd71ffcf45054004a099d042e19b8750.png

The area bounded by the curves

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

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

0%
$1$
0%
$2$
0%
$2\sqrt{ 2 }$
0%
$4$

Explanation

We can see that it is a square of side

$\sqrt{2}$ units

Hence

$Area=a^{2}$

$Area=(\sqrt{2})^{2} =2$ units

993279_1006974_ans_02bb54f69e084b69aa109ceedbf3b93d.png

Area common to the curves

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

${ x }^{ 2 }$ +

${ y }^{ 2 }$

$= 4ax$ is equal to

0%
$(9\sqrt { 3 } +4\pi )\cfrac { { a }^{ 2 } }{ 3 }$
0%
$(9\sqrt { 3 } +4\pi ){ a }^{ 2 }$
0%
$(9\sqrt { 3 } -4\pi )\cfrac { { a }^{ 2 } }{ 3 }$
0%
none of the above

Explanation

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

$\Rightarrow$

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

$\int_{0}^{2a}{\sqrt{ax}}dx+\int_{3a}^{4a}{\sqrt{(4ax-x^{2})}dx}$

$\Rightarrow$

$\left(\dfrac{4\pi}{6}-3^{{3}/{2}}\right)a^{2}+\dfrac{2}{3}\sqrt{a}(3^{{3}/{2}}a^{{3}/{2}})$

$\Rightarrow$

$a^{2}\left(\dfrac{4\pi}{6}+2\sqrt{3}-\dfrac{3\sqrt{3}}{6}\right)$

Due to symmetry

$=\dfrac{a^{2}}{3}[4\pi+9\sqrt{3}]$

993370_1018355_ans_4dcf847cb22d44c0b483e4c7daf5283a.png

Find the area bounded by the curves

$x = a\cos t, y = b\sin t$ in the first quadrant

0%
$\dfrac{\pi ab}{4}$
0%
$\dfrac{\pi a^2b}{4}$
0%
None of these
0%
$\dfrac{\pi ab^2}{4}$

Explanation

$x=a\cos t$ ,

$y=b\sin t$

$\rightarrow$

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

It's a hyperbola whose area in any quadrant is given by

$\dfrac{\pi ab}{4}$

996417_1032146_ans_c5f0c9620c08468da530aabdbb5e3dc6.jpeg

A point P moves in xy-plane In such a way that [|x|] + [|y|] = 1 were [.] denotes the greatest integer function. Area of the region representing all possible positions of the point 'P' is equal to

0%
$4$ sq. units
0%
$16$ sq. units
0%
$2 \sqrt{2}$ sq. units
0%
$8$ sq. units

Explanation

Graph of the given function is in the figure,

From figure , graph is a square with side=

$2$ units.

Therefore , Area=

$2\times2=4$ sq. units

1035836_1028258_ans_d6338c0da0aa421a9cb8233f98377e85.jpg

The area common to the circles

$r=a\sqrt{2}$ and

$r=2a\cos{\theta}$ is:

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

Explanation

The equations of the circles are

$r=a\sqrt{2}$ ................

$\left(1\right)$
and

$r=2a\cos{\theta}$ .................

$\left(2\right)$
eqn

$\left(1\right)$ represents a circle with centre at

$\left(0,0\right)$ and radius

$a\sqrt{2}$
eqn

$\left(2\right)$ represents a circle symmetrical about

$OX$ with centre at

$\left(a,0\right)$ and radius

$a$
The circles are shown in fig.At their point of intersection

$P$ ,eliminating

$r$ from

$\left(1\right)$ and

$\left(2\right)$ ,

$a\sqrt{2}=2a\cos{\theta}$

$\Rightarrow \cos{\theta}=\dfrac{1}{\sqrt{2}}$
or

$\theta=\dfrac{\pi}{4}$

$\therefore$ Required Area

$=2\times$ area

$OAPQ$ (By symmetry)

$=2$ (area

$OAP+$ area

$OPQ$ )

$=2\left[\dfrac{1}{2}\int_{0}^{\frac{\pi}{4}}{{r}^{2}d\theta}+\dfrac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{r}^{2}d\theta\right]$ for

$\left(1\right)$ and

$\left(2\right)$ respectively.

$=\int_{0}^{\frac{\pi}{4}}{\left(a\sqrt{2}\right)}^{2}d\theta+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{{\left(2a\cos{\theta}\right)}^{2}}d\theta$

$=2{a}^{2}\left|\theta\right|_{0}^{\frac{\pi}{4}}+4{a}^{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\dfrac{1+\cos{\theta}}{2}}d\theta$

$=2{a}^{2}\left(\dfrac{\pi}{4}-0\right)+2{a}^{2}\left|\theta+\dfrac{\sin{2\theta}}{2}\right|_{\frac{\pi}{4}}^{\frac{\pi}{2}}$

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

${a}^{2}\left(\pi-1\right)$
949765_1035248_ans_4ad9f68bbd104fa2bdb479db244f8e2f.png

The area under the curve

$y= 2\sqrt x$ bounded by the lines

$x=0$ and

$x= 1$ is

0%
$\frac{4}{3}$
0%
$\frac{2}{3}$
0%
$1$
0%
$\frac{8}{3}$

Explanation

$A=(0,0)$ ,

$B=(1,2)$ ,

$C=(1,0)$

$=2\int_{0}^{1}{\sqrt{x}}dx$

$[\dfrac{2x^{\dfrac{3}{2}}}{\dfrac{3}{2}}]_{0}^{1}$

$=\dfrac{4}{3}$

997859_1034197_ans_ad523056e7784a058df8dc09628239ee.png

Area of the region, bounded by the parabolas

$3x^{2}=16y$ and

$4y^{2}=9x$ , is

0%
$4$
0%
$6$
0%
$8$
0%
$16$

Explanation

Solving

$3x^2=16y,4y^2=9x$

$4 \bigg(\large\frac{3x^2}{16}\bigg)^2=9x$

$\large\frac{36x^4}{256}=9x$

$x^4=64x$

$x(x^3−43)=0=>x=0,4$

Required area= Area

$4y^2=9x$ between

$x=0,4$ -area

$3x^2=16y$ between

$x=0,4$

$\qquad= \large\int \limits_0^4 \large\frac{3x^2}{16})dx=\large\frac{3}{2}.\frac{2}{3}x^{3/2}-\large\frac{x^3}{16}\bigg]_0^4$

$\qquad= x^{\large\frac{3}{2}} -\large\frac{x^3}{16}\bigg]_0^4$

$\qquad= 4 \sqrt 4 -4$

$=8−4=4squnits$

The area common to the cardioids

$r=a\left(1+\cos{\theta}\right)$ and

$r=a\left(1-\cos{\theta}\right)$ is:

0%
$\left(\dfrac{3\pi}{2}+4\right){a}^{2}$
0%
$\left(\dfrac{3\pi}{2}-4\right){a}^{2}$
0%
$\left(\dfrac{\pi}{2}+4\right){a}^{2}$
0%
$\left(\dfrac{\pi}{2}-4\right){a}^{2}$

Explanation

The cardioid

$r=a\left(1+\cos{\theta}\right)$ is

$ABCO{B}^{\prime}A$ and the cardioid

$r=a\left(1-\cos{\theta}\right)$ is

$O{C}^{\prime}B{A}^{\prime}{B}^{\prime}O$
Both the cardioids are symmetrical about the initial line

$OX$ and intersect at

$B$ and

${B}^{\prime}$

$\therefore$ Required Area

$=2$ Area

$O{C}^{\prime}BCO$

$=2$ [area

$O{C}^{\prime}BO+$ area

$OBCO$ ]

$=2\left[\left(\int_{0}^{\frac{\pi}{2}}{\dfrac{1}{2}{r}^{2}}d\theta\right)_{r=a\left(1-\cos{\theta}\right)}+\int_{\frac{\pi}{2}}^{\pi}\left({\left(1+\cos{\theta}\right)}^{2}d\theta\right)_{r=a\left(1+\cos{\theta}\right)}\right]$

$={a}^{2}\left[\int_{0}^{\frac{\pi}{2}}{\left(1-2\cos{\theta}+{\cos}^{2}\theta\right)d\theta}+\int_{\frac{\pi}{2}}^{\pi}{\left(1+2\cos{\theta}+{\cos}^{2}\theta\right)}d\theta\right]$

$={a}^{2}\left[\int_{0}^{\pi}{\left(1+{\cos}^{2}\theta\right)d\theta}-2\int_{0}^{\frac{\pi}{2}}{\cos{\theta}d\theta}+2\int_{\frac{\pi}{2}}^{\pi}{\cos{\theta}d\theta}\right]$

$={a}^{2}\left[\int_{0}^{\pi}{\left(1+\dfrac{1+\cos{2\theta}}{2}\right)d\theta}-2\left|\sin{\theta}\right|_{0}^{\frac{\pi}{2}}+2\left|\sin{\theta}\right|_{\frac{\pi}{2}}^{\pi}\right]$

$={a}^{2}\left[\left|\dfrac{3}{2}\theta+\dfrac{\sin{2\theta}}{4}\right|_{0}^{\pi}-2\left(1-0\right)+2\left(0-1\right)\right]$

$=\left(\dfrac{3\pi}{2}-4\right){a}^{2}$
949791_1035304_ans_8465b28a69f54583b4def60fa7d032a6.png

Area bounded by

$y = x^2$ and

$y = \dfrac{2}{1 + x^2}$ is:

0%
$\pi - \dfrac{1}{3}$
0%
$\pi - \dfrac{2}{3}$
0%
$2 \pi - \dfrac{1}{3}$
0%
none of these

Explanation

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

$x^{2}=\dfrac{1}{1+x^{2}}$ , For intersection Points

$(x^{2})^{2}+x^{2}-2=0$

$x^{2}=1$

$\therefore x=\pm 1$

Area bounded

$=\displaystyle\int_{-1}^{1}\left(\dfrac{2}{1+x^{2}}-x^{2}\right)dx$

$=2\tan^{-1}(x)-\dfrac{x}{3}\left.\dfrac{}{}\right|_{-1}^{1}$

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

The area between the curves

$y= \tan x, y=2 \sin x$ and

$x$ -axis in

$-\dfrac{\pi}{3} \leq x \leq \dfrac{\pi}{3}$ is

0%
$2-\ln2$
0%
$3-ln2$
0%
$4-\ln2$
0%
None of these

Explanation

Due to symmetry

$2\int_{0}^{\dfrac{\pi}{3}}{2\sin x-\tan x}dx$

$=2[-2\cos x+\ln \cos x]_{0}^{\dfrac{\pi}{3}}$

$|2\ln|\cos x|1-4\cos x|_{0}^{\dfrac{\pi}{3}}$

$=2-2\ln 2$

997865_1034272_ans_4d1060c488554d27a455ef152306871a.png

The area of the region bounded by the curve

${a}^{4}{y}^{2}=\left(2a-x\right){x}^{5}$ is to that of the circle whose radius is

$a$ , is given by the ratio

0%
$4:5$
0%
$5:8$
0%
$2:3$
0%
$3:2$

Explanation

Given curve

${a}^{4}{y}^{2}=\left(2a-x\right){x}^{5}$
cut off

$x-$ axis,when

$y=0$

$0=\left(2a-x\right){x}^{5}$ is

${A}_{1}=\int_{0}^{2a}{\dfrac{\sqrt{\left(2a-x\right)}{x}^{\frac{5}{2}}}{{a}^{2}}dx}$
Put

$x=2a{\sin}^{2}{\theta}$

$\therefore dx=4a\sin{\theta}\cos{\theta}d\theta$

$\therefore {A}_{1}=\int_{0}^{{\frac{\pi}{2}}}{\dfrac{\sqrt{2a}\cos{\theta}{\left(2a\right)}^{\frac{5}{2}}{\sin}^{5}{\theta}\times 4a\sin{\theta}\cos{\theta}}{{a}^{2}}d\theta}$

$=32{a}^{2}\int_{0}^{\frac{\pi}{2}}{{\sin}^{6}{\theta}{\cos}^{2}{\theta}d\theta}$

$=32{a}^{2}\dfrac{\left(5.3.1\right)\left(1\right)}{8.6.4.2}.\dfrac{\pi}{2}$ (by Walli's formula)

$=\dfrac{5\pi{a}^{2}}{8}$
Area of circle,

${A}_{2}=\pi{a}^{2}$

$\therefore \dfrac{{A}_{1}}{{A}_{2}}=\dfrac{5}{8}$

$\Rightarrow {A}_{1}:{A}_{2}=5:8$

The area of the region described by

$A=((x,y): {x}^{2}+{y}^{2}\le1)$ and

$B=((x,y):{y}^{2}\le1-x)$

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

Explanation

Area

$ABCA$

$\dfrac{\pi}{2}r^{2}=\dfrac{\pi}{2}$

Due to symmetry

$ADCA=2AODA$

$=2\int_{0}^{1}{\sqrt{1-x}}$

$=\dfrac{-2[(1-x)^{{3}/{2}}]_{0}^{1}}{\dfrac{3}{2}}$

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

Hence the total area =

$\dfrac{\pi}{2}$ +

$\dfrac {4}{3}$

997936_1036157_ans_0c4e6a1c368e495dafb067bfaa5ec666.png

Area bounded by the curves

$y=\left[\dfrac{{x}^{2}}{64}+2\right], y=x-1$ and

$x=0$ above

$x-$ axis is (

$\left[.\right]$ denotes the greatest integer function.)

0%
$2$ .sq.unit
0%
$3$ .sq.unit
0%
$4$ .sq.unit
0%
none of these

Explanation

From the figure shown

$-8<x<8\Rightarrow y=2$

$\therefore$ Required Area

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

$=4$ .sq.unit.
959114_1040538_ans_446261271275473ca2e65a6daf4a63c4.png

Area of the region bounded by the curves

$y\left|y\right|\pm x\left|x\right|=1$ and

$y=\left|x\right|$ is:

0%
$\dfrac{\pi}{8}$ sq.unit
0%
$\dfrac{\pi}{4}$ sq.unit
0%
$\dfrac{\pi}{2}$ sq.unit
0%
$\pi$ sq.unit

Explanation

In First quadrant

${y}^{2}\pm{x}^{2}=1$
and in Second quadrant

${y}^{2}\pm{x}^{2}=1$
i.e.,

${y}^{2}+{x}^{2}=1$
and

${y}^{2}-{x}^{2}=1$
and in Third quadrant

$-{y}^{2}-{x}^{2}=1$
and in fourth quadrant

$-{y}^{2}\pm{x}^{2}=1$
i.e.,

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

$\left(\because -{x}^{2}-{y}^{2}\neq1\right)$

$\therefore$ Required Area

$=$ area of

$OABCO$

$=\dfrac{1}{4}$ (area of circle)

$=\dfrac{1}{4}\pi{r}^{2}$

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

$=\dfrac{\pi}{4}$ sq.units
957935_1037675_ans_3eb19cca9c8142608141ddf649cc613e.png

Find the area included between the parabolas

$y^{2} = x$ and

$x = 3 - 2y^{2}$ .

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

Explanation

$A=(1,1)$ ,

$B=(0,\sqrt{\dfrac{3}{2}})$ ,

$C=(0,-\sqrt{\dfrac{3}{2}})$ ,

$D=(1,-1)$

Due to symmetry we can just calculate the area in the first quadrant and multiply by 2

$=\int_{0}^{1}{3-2y^{2}}dy -\int_{0}^{1}y^{2}dy$

$=[3y-\dfrac{2y^{3}}{3}]-\dfrac{y^{3}}{3}=3-\dfrac{2}{3}-\dfrac{1}{3}$

Area of first quadrant

$=2$

$Total\ \text {area} = 4$

998810_1039915_ans_6ac02a76d67c49e9a46849aa249b62db.JPG

The area of the region bounded by

$1-{y}^{2}=\left|x\right|$ and

$\left|x\right|+\left|y\right|=1$ is

0%
$\dfrac{1}{3}$ .sq.unit
0%
$\dfrac{2}{3}$ .sq.unit
0%
$\dfrac{4}{3}$ .sq.unit
0%
$1$ .sq.unit

Explanation

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

$|x|+|y|=1$

$\begin{aligned} \text { Area } &=\int_{0}^{1} x d y \\ &=\int_{0}^{1}\left(1-y^{2}\right) d y \end{aligned}$

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

$=1-1 / 3$

$=\dfrac{2}{3}$ sq.unit (Area of one region)

$\begin{aligned} \text { Area } &=\dfrac{2}{3}-\dfrac{1}{2} \\ &=4\left[\dfrac{2}{3}-\dfrac{1}{2}\right] \\ &=\dfrac{4}{6}=\dfrac{2}{3} \text { squnits } \end{aligned}$

Correct option is (B).

1989637_1037716_ans_1f123858018b42d59f045d9a9247bdb7.jpg

The area of the figure bounded by the curves

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

$y=3-\left|x\right|$ is

0%
$1$ sq.unit
0%
$2$ sq.unit
0%
$3$ sq.unit
0%
$4$ sq.unit

Explanation

Since,

$y=\left|x-1\right|=\begin{cases}x-1;x\ge 1 \\ 1-x, x<1 \end{cases}$ ..................

$\left(1\right)$
and

$y=3-\left|x\right|=\begin{cases} 3-x; x\ge 0 \\ 3+x; x<0\end{cases}$ ..................

$\left(2\right)$
Solving eqns

$\left(1\right)$ and

$\left(2\right)$ , we get

$x=2$ and

$x=-1$

$\therefore$ Required Area

$=\left|\int_{-1}^{2}{\left(3\left|x\right|-\left|x-1\right|\right)}\right|$

$=\left|\int_{-1}^{0}{\left(2x+2\right)dx}+\int_{0}^{1}{2.dx}+\int_{1}^{2}{\left(4-2x\right)dx}\right|$

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

$=4$ .sq.unit.

The area between the curve

$y=2{x}^{4}-{x}^{2},$ the

$x-$ axis and the ordinates of two minima of the curve is:

0%
$\dfrac{7}{120}$ sq.unit
0%
$\dfrac{9}{120}$ sq.unit
0%
$\dfrac{11}{120}$ sq.unit
0%
$\dfrac{13}{120}$ sq.unit

Explanation

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

$\therefore \dfrac{dy}{dx}=8{x}^{3}-2x,$ for max. or min.

$\dfrac{dy}{dx}=0$

$x=-\dfrac{1}{2},0,\dfrac{1}{2}$
Then,

$\left(\dfrac{{d}^{2}y}{d{x}^{2}}\right)_{x=-\frac{1}{2}}>0, \left(\dfrac{{d}^{2}y}{d{x}^{2}}\right)_{x=0}<0$
and

$\left(\dfrac{{d}^{2}y}{d{x}^{2}}\right)_{x=\frac{1}{2}}>0$

$\therefore$ Required Area

$=\left|\int_{-\frac{1}{2}}^{\frac{1}{2}}{\left(2{x}^{4}-{x}^{2}\right)dx}\right|$

$=\left[\dfrac{2{x}^{5}}{5}-\dfrac{{x}^{3}}{3}\right]_{-\frac{1}{2}}^{\frac{1}{2}}$

$=\dfrac{7}{120}$ .sq.unit.

The area of a loop bounded by the curve

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

0%
$a$
0%
$2a^2$
0%
$0$
0%
$2a$

The area enclosed between the curves

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

${y}^{2}=x$ is equal to

0%
$\dfrac{1}{3}$ .sq.unit
0%
$2\int_{0}^{1}{\left(x-{x}^{2}\right)dx}$
0%
area of the region $\left\{\left(x,y\right):{x}^{2}\le y\le \left|x\right|\right\}$
0%
none of the above

Explanation

$\because$ Curves

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

${y}^{2}=x$ intersect at

$\left(0,0\right)$ and

$\left(1,1\right)$ .

$\therefore$ Required Area

$=\int_{0}^{1}{\left(\sqrt{x}-{x}^{2}\right)dx}$

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

$=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}$ .sq.unit.
Also, both curves

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

${y}^{2}=x$ are symmetrical about

$y=x$

$\therefore$ Required Area

$=2\int_{0}^{1}{\left(x-{x}^{2}\right)dx}$
Option

$\left(c\right):{x}^{2}\le y\le \left|x\right|$

$y={x}^{2},y=\left|x\right|$ point of intersection is

$\left(0,0\right)$ and

$\left(1,1\right)$ .

$\therefore$ Required Area

$=2\int_{0}^{1}{\left(x-{x}^{2}\right)dx}$

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

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

$=1-\dfrac{2}{3}=\dfrac{1}{3}$ .sq.unit
953231_1041115_ans_469fc3cdd058451da91c4b568f918145.png

The area bounded by the curves

$\left|x\right|+\left|y\right|\ge 1$ and

${x}^{2}+{y}^{2}\le 1$ is:

0%
$2$ sq.unit
0%
$\pi$ sq.unit
0%
$\left(\pi-2\right)$ sq.unit
0%
$\left(\pi+2\right)$ sq.unit

Explanation

From the figure shown

$\therefore$ Required Area

$=$ area of the circle

$-$ area of the square.

$=\pi{\left(1\right)}^{2}-{\left(\sqrt{2}\right)}^{2}$

$=\left(\pi-2\right)$ .sq.unit
953163_1041023_ans_78694511878c49feb834ab5118e707d1.png

The area bounded by the curves

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

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

0%
$1$ .sq.unit
0%
$2$ .sq.unit
0%
$2\sqrt{2}$ .sq.unit
0%
$4\sqrt{2}$ .sq.unit

Explanation

Required area

$=$ shaded area

$={\left(\sqrt{2}\right)}^{2}=2$ .sq.unit.
959195_1040649_ans_39572af002644707a2c912b74de78fed.png

Let

$f\left(x\right)={x}^{2}-3x+2$ be a function,for all

$x\in R$ . On the basis of given information, answer the given question

The number of solutions of

${x}^{2}+{y}^{2}=2$ is,

0%
$4$
0%
$6$
0%
$8$
0%
$5$

Let

$f\left(x\right)={x}^{2}-3x+2$ be a function, for all

$x\in R$ . On the basis of given information, answer the given question.

The area bounded by

$f\left(x\right),$ the

$x-$ axis and

$y-$ axis is,

0%
$\dfrac{1}{3}$ .sq.unit
0%
$\dfrac{2}{3}$ .sq.unit
0%
$\dfrac{3}{5}$ .sq.unit
0%
$\dfrac{5}{6}$ .sq.unit

The area of the figure bounded by two branches of the curve

${\left(y-x\right)}^{2}={x}^{3}$ and the straight line

$x=1$ is:

0%
$\dfrac{1}{3}$ sq.unit
0%
$\dfrac{4}{5}$ sq.unit
0%
$\dfrac{5}{4}$ sq.unit
0%
$3$ sq.unit

Explanation

Given curve is

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

$\Rightarrow y-x=\pm x\sqrt{x}$
or

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

$\therefore$ Required Area

$=\left|\int_{0}^{1}{\left[(x+x\sqrt{x}\right)-\left(x-x\sqrt{x})\right]dx}\right|$

$=\left|2\int_{0}^{1}{{x}^{\frac{3}{2}}dx}\right|$

$=2\left[\dfrac{{x}^{\frac{5}{2}}}{\frac{5}{2}}\right]_{0}^{1}$

$=\left|2.\dfrac{2}{5}\right|=\dfrac{4}{5}$ .sq.unit.
959187_1040622_ans_387d15bdc4ea477fafb6051494034dbc.png

Let

$f$ and

$g$ be continuous function on

$a\le x\le b$ and set

$p\left(x\right)=$ max

$\left\{f\left(x\right),g\left(x\right)\right\}$ and

$q\left(x\right)=$ min

$\left\{f\left(x\right),g\left(x\right)\right\}$ , the area bounded by the curves

$y=p\left(x\right),y=q\left(x\right)$ and the ordinates

$x=a$ and

$x=b$ is given by

0%
$\int_{a}^{b}{\left(f\left(x\right)-g\left(x\right)\right)dx}$
0%
$\int_{a}^{b}{\left(p\left(x\right)-q\left(x\right)\right)dx}$
0%
$\int_{a}^{b}{\left|p\left(x\right)-q\left(x\right)\right|dx}$
0%
$\int_{a}^{b}{\left|f\left(x\right)-g\left(x\right)\right|dx}$

Explanation

Here,curve

$ABCD$ is max

$\left\{f\left(x\right),g\left(x\right)\right\}=p\left(x\right)$ and curve

$EBCF$ is min

$\left\{f\left(x\right),g\left(x\right)\right\}=q\left(x\right)$
Thus, the area can be determine by either

$\left(c\right)$ or

$\left(d\right)$
953274_1041218_ans_a4f983dc24e7457f9aee19880671ee05.png

Area bounded by the curve

$y=\ln{x}, y=0$ and

$x=3$ is

0%
$\left(\ln{9}-2\right)$ .sq.unit
0%
$\left(\ln{27}-2\right)$ .sq.unit
0%
$\ln\left(\dfrac{27}{{e}^{2}}\right)$ .sq.unit
0%
$>3$ .sq.unit

Explanation

Required Area

$=\int_{1}^{3}{\ln{x}dx}$

$=\left[x\ln{x}-x\right]_{1}^{3}$

$=\left(3\ln{3}-2\right)$

$=\ln{27}-2$ .sq.unit(option

$\left(b\right)$ is correct)

$=\ln{27}-\ln{{e}^{2}}$

$=\ln{\left(\dfrac{27}{{e}^{2}}\right)}$ .sq.unit(option

$\left(c\right)$ is correct)
Also,

$\ln{\left(\dfrac{27}{{e}^{2}}\right)}>3$

$\Rightarrow \ln{\left(\dfrac{27}{{e}^{2}}\right)}>3\ln{e}$

$\Rightarrow \ln{27}-\ln{{e}^{2}}>3\ln{e}$

$\Rightarrow \ln{27}-2\ln{e}>3\ln{e}$

$\Rightarrow \ln{27}>3\ln{e}+2\ln{e}$

$\Rightarrow \ln{27}>5\ln{e}$

$\Rightarrow \ln{27}>\ln{{e}^{5}}$ which is false.

953460_1041488_ans_2715451c52af4b1fbdcb5ce12803351b.png

The area bounded by

$y=2-\left|2-x\right| , y=\dfrac{3}{\left|x\right|}$ is

0%
$\dfrac{5-4\ln{2}}{3}$ .sq.unit
0%
$\dfrac{2-\ln{3}}{2}$ .sq.unit
0%
$\dfrac{4-3\ln{3}}{2}$ .sq.unit
0%
none of these

Explanation

$y=2-\left|2-x\right|, y=\dfrac{3}{\left|x\right|}$

$y=\begin{cases}x; x\le 2 \\ 4-x; x\ge 2\end{cases}$ and

$y=\begin{cases} \dfrac{3}{x}; x>0 \\-\dfrac{3}{x}; x<0 \end{cases}$
Hence,Required Area

$PQRSP=$ area

$PQRP+$ area

$PRSP$

$=\left|\int_{\sqrt{3}}^{2}{\left(x-\dfrac{3}{x}\right)dx}\right|+\left|\int_{2}^{3}\left[\left(4-x\right)-\dfrac{3}{x}\right]dx\right|$

$=\left[\dfrac{{x}^{2}}{2}-3\ln{x}\right]_{\sqrt{3}}^{2}+\left[4x-\dfrac{{x}^{2}}{2}-3\ln{x}\right]_{2}^{3}$
On simplification,we get

$=\dfrac{4-3\ln{3}}{2}$ .sq.unit
959576_1040928_ans_5bfc89b21e084172a70db7ecc44aa06e.png

$f\left(x+y\right)=f\left(x\right)+f\left(y\right)-xy$ for all

$x,y\in R$ and

$\lim _{ h\rightarrow 0 }{ \frac { f(h) }{ h } } =3$ , then the area bounded by the curves

$y=f\left(x\right)$ and

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

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

Explanation

Put

$x=y=0$

$\Rightarrow f\left(0\right)=0$
Now,

${f}{'}\left(x\right)=\displaystyle \lim_{h\rightarrow 0}{\dfrac{f\left(x+h\right)-f\left(x\right)}{h}}$

$=\displaystyle \lim_{h\rightarrow 0}{\dfrac{f\left(x\right)+f\left(h\right)-hx-f\left(x\right)}{h}}$

$=\displaystyle \lim_{h\rightarrow 0}{\dfrac{f\left(h\right)}{h}}-x=3-x$

$\therefore f\left(x\right)=3x-\dfrac{{x}^{2}}{2}+c$ and

$f\left(0\right)=0$

$\Rightarrow c=0$

$\therefore f\left(x\right)=3x-\dfrac{{x}^{2}}{2}$
Required Area

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

$=\left(\dfrac{3{x}^{2}}{2}-\dfrac{3{x}^{3}}{6}\right)_{0}^{2}$

$=6-4=2$

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 4 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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