MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 3

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

$\displaystyle y=\sin x$ 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 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

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

$y = (x)$ 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%
$A = e + \displaystyle \frac{2}{\sqrt e} - 3$

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

$=2[APOA]$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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 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 of the smaller portion between curves

$x^2 + y^2 = 8$ and

$y^2 = 2x$ is

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

Explanation

Given curves

$x^2 + y^2 = 8$ ....(1)
Center is (0,0) and radius

$r=2\sqrt{2}$

$y^2 = 2x$ ....(2)
Solving (1) and (2), we get

$x^2+2x=8$

$\Rightarrow x=2,-4$
At

$x=2 \Rightarrow y=2,-2$
Hence, the point of intersection of the curves is

$(2,2)$ and

$(2,-2)$

Now, required area

$=2(area OPA)$

$=2[ar(OPB)+ar(PBA)]$

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

$=2\sqrt{2}[\dfrac{x^{3/2}}{3/2}]_{0}^2 +2 [\dfrac{x}{2}\sqrt{8 - x^2}+4\sin^{-1}{\dfrac{x}{2\sqrt{2}}}]_{2}^{2\sqrt{2}}$

$=\dfrac{4\sqrt{2}}{3} (2\sqrt{2})+2(0+{2}{\pi}-2-\pi)$

Required Area

$= \displaystyle \left( 2 \pi + \frac{4}{3} \right )$ sq.units

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 between the parabola

$y =x^2$ and the line

$y = x$ is

0%
$\dfrac{1}{6}$ sq. units
0%
$\dfrac{1}{3}$ sq. units
0%
$\dfrac{1}{2}$ sq. units
0%
None of these

Explanation

For points of intersection of the equations

$y=x^2$ and

$y=x$

$\therefore x^2-x=0 \Rightarrow x(x-1)=0$

$x=0$ or

$x=1\Rightarrow Y=0$ or

$y =1$
Hence, the coordinates of their points of intersection are O(0, 0) and P (1, 1).

$\therefore$ Required area (shaded region)

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

$= \displaystyle \left [ \left ( \frac{1}{2} - \frac{1}{3} \right ) - 0 \right ] = \frac{1}{6}$ sq. units
1099966_356923_ans_0a9a4f91656e4900b4e1a2b72cd60b3a.jpg

Area lying in the first quadrant and bounded by the circle

$x^2 + y^2 = 4$ and the lines

$x = 0$ and

$x = 2$ is

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

Explanation

The area bounded by the circle and the lines,

$x=0$ and

$x=2$ , in the first quadrant is represented as shaded region in the plot.

$\therefore$ Area

$OAB=\int_0^2y dx$

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

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

$=2\left (\dfrac {\pi}{2}\right )=\pi$ sq. units
Thus, the correct answer is A.

The area bounded by

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

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

0%
$\cfrac { 20 }{ 3 }$ sq. units
0%
$\cfrac { 16 }{ 3 }$ sq. units
0%
$\cfrac { 14 }{ 3 }$ sq. units
0%
$\cfrac { 10 }{ 3 }$ sq. units

Explanation

The intersection point of a curves

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

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

$O(0,0)$ and

$A(4,4)$

$\therefore$ Required area

$\displaystyle \int _{ 0 }^{ 4 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx }$

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

$={ \left[ \cfrac { 2{ x }^{ 3/2 } }{ 3/2 } -\cfrac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }\quad$

$=\cfrac { 4 }{ 3 } \left[ { 4 }^{ 3/2 }-0 \right] -\cfrac { 1 }{ 12 } \left[ { 4 }^{ 3 }-0 \right]$

$=\cfrac { 32 }{ 3 } -\cfrac { 16 }{ 3 } =\cfrac { 16 }{ 3 }$ sq. units

The area of the region bounded by the y-axis,

$y = \cos x$ and

$y = \sin x, 0\leq x \leq \dfrac {\pi}{2}$ is

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

Explanation

Refer image 1.

Finding point of Intersection

$B$

Solving

$y=\cos x$ and

$y=\sin x$

$\cos x=\sin x$

Refer image 2.

$x=\dfrac{\pi}{4}$ , both are equal

Also,

$y=\cos x=\cos \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$

So,

$B=\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)$

Refer image 3.

Refer image 4.

Area

$ABCO$

Area ABCO

$\displaystyle \int^{\pi/4}_0 ydx$

Here,

$y=\cos x$

Thus,

Area ABCO=

$\displaystyle \int^{\pi/4}_0 \cos x dx$

$=[\sin x]^{\pi/4}_0$

$=[\sin\dfrac{\pi}{4}-\sin 0]$

$=\dfrac{1}{\sqrt{2}}-0$

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

Refer image 5.

Area

$BCO$

Area BCO

$\displaystyle \int^{\pi/4}_0 ydx$

Here,

$y=\sin x$

Thus,

Area BCO =

$\displaystyle \int^{\pi/4}_0 \sin x dx$

$=-[\cos x]^{\pi/4}_0$

$=-[\cos\dfrac{\pi}{4}-\cos(0)]$

$=-[\dfrac{1}{\sqrt{2}}-1]$

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

Therefore

Area Required = Area ABCD - Area BCO

$=\dfrac{1}{\sqrt{2}}-[1-\dfrac{1}{\sqrt{2}}]$

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

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

$=\sqrt{2}-1$

$\therefore$ Option

$B$ is correct.

1521667_428566_ans_d0ca9efb62ec4cc7a2bba45b0ea202ff.png

Area bounded between the curve

$x^2=y$ and the line

$y=4x$ is

0%
$\displaystyle\frac{32}{3}$ sq unit
0%
$\displaystyle\frac{1}{3}$ sq unit
0%
$\displaystyle\frac{8}{3}$ sq unit
0%
$\displaystyle\frac{16}{3}$ sq unit

Explanation

Given curves are

$x^2=y$ and

$y=4x$
Intersection points are

$(0, 0)$ and

$(4, 16)$

$\therefore$ Required area

$=\int^4_0(4x-x^2)dx$

$=\left[\displaystyle\frac{4x^2}{2}-\frac{x^3}{3}\right]^4_0$

$=\displaystyle\left[32-\frac{64}{3}\right]$

$=\displaystyle\frac{32}{3}$ sq unit

The area enclosed between the curve

$\displaystyle y=1+{ x }^{ 2 }$ , the y-axis and the straight line

$\displaystyle y=5$ is given by

0%
$\displaystyle \frac { 14 }{ 3 }$ sq unit
0%
$\displaystyle \frac { 7 }{ 3 }$ sq unit
0%
$\displaystyle 5$ sq unit
0%
$\displaystyle \frac { 16 }{ 3 }$ sq unit

Explanation

Given curve is

$\displaystyle y=1+{ x }^{ 2 }$ and line

$\displaystyle y=5$

$\displaystyle \therefore$ Required area

$\displaystyle =\int _{ 1 }^{ 5 }{ x } dy=\int _{ 1 }^{ 5 }{ \sqrt { y-1 } dx }$

$\displaystyle ={ \left[ \frac { { \left( y-1 \right) }^{ { 3 }/{ 2 } } }{ { 3 }/{ 2 } } \right] }_{ 1 }^{ 5 }=\frac { 2 }{ 3 } \left[ { \left( 4 \right) }^{ { 3 }/{ 2 } }-0 \right]$

$\displaystyle =\frac { 16 }{ 3 }$ sq unit.

The area bounded by the parabolas

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

$y=2$ is

0%
$\dfrac{5\sqrt{2}}{3}$ sq units
0%
$\dfrac{10\sqrt{2}}{3}$ sq units
0%
$\dfrac{15\sqrt{2}}{3}$ sq units
0%
$\dfrac{20\sqrt{20}}{3}$ sq units

Explanation

$y=4x^2\;\;\;\dots(i)$

$y=\dfrac{x^2}{4}\;\;\;\dots(ii)$

$\therefore\;A=\int_0^2\begin{bmatrix}\dfrac{\sqrt{y}}{2}-3\sqrt{y}\end{bmatrix}dy$

$=\begin{pmatrix}\dfrac{1}{2}-3\end{pmatrix}\int_0^2\sqrt{y}dy$

$=\begin{pmatrix}\dfrac{-5}{2}\end{pmatrix}.\dfrac{2}{3}\begin{bmatrix}y^{{3}/{2}}\end{bmatrix}_0^2$

$=-\dfrac{5}{3}(2\sqrt{2}-0)=\begin{vmatrix}\dfrac{-10\sqrt{2}}{3}\end{vmatrix}$

$=\dfrac{10\sqrt{2}}{3}\Rightarrow\text{Area of bounded Area}=2A=\dfrac{20\sqrt{20}}{3}$

The area of the circle

$x^2+y^2=16$ exterior to the parabola

$y^2=6x$ is

0%
$\dfrac {4}{3}(4\pi -\sqrt 3)$
0%
$\dfrac {4}{3}(4\pi +\sqrt 3)$
0%
$\dfrac {4}{3}(8\pi -\sqrt 3)$
0%
$\dfrac {4}{3}(8\pi +\sqrt 3)$

Explanation

The given equations are

$x^2+y^2=16........ (1)$

$y^2=6x ....... (2)$

Area bounded by the circle and parabola

$=2[Area (OADO)+Area (ADBA)]$

$=2[\displaystyle \int_{0}^{2}\sqrt {16x}dx+\int_{2}^{4}\sqrt {16-x^2}dx]$

$=2\left [\sqrt 6\left \{\dfrac {x^{\dfrac {3}{2}}}{\dfrac {3}{2}}\right \}_{0}^{2}\right ]+2\left [\dfrac {x}{2}\sqrt {16-x^2}+\dfrac {16}{2}\sin^{-1}\dfrac {x}{4}\right ]_{2}^{4}$

$=2\sqrt 6\times \dfrac {2}{3}\left [x^{\dfrac {3}{2}}\right ]_{0}^{2}+2\left [8\cdot \dfrac {\pi}{2}-\sqrt {16-4}-8 \sin^{-1}\left (\dfrac {1}{2}\right )\right ]$

$=\dfrac {4\sqrt 6}{3}(2\sqrt 2)+2\left [4\pi-\sqrt {12}-8\dfrac {\pi}{6}\right ]$

$=\dfrac {16\sqrt 3}{3}+8\pi - 4\sqrt 3-\dfrac {8}{3}\pi$

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

$=\dfrac {4}{3}[\sqrt 3+4\pi]$

$=\dfrac {4}{3}[4\pi +\sqrt 3]$ sq. units

Area of circle

$=\pi(r)^2$

$=\pi(4)^2$

$=16 \pi$ sq. units

$\therefore$ Required area

$=16\pi -\dfrac {4}{3}[4\pi +\sqrt 3]$

$=\dfrac {4}{3}[4\times 3\pi -4\pi -\sqrt 3]$

$=\dfrac {4}{3}(8\pi -\sqrt 3)$ sq. units

Thus, the correct answer is C.

The area bounded by the curve

$y=x|x|$ ,

$x$ -axis and the ordinates

$x=-1$ and

$x=1$ is given by

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

Explanation

$y=x^2$ if

$x > 0$ and

$y=-x^2$ if

$x < 0$
Required area

$=\displaystyle \int_{-1}^{1}y dx$

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

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

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

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

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

Thus, the correct answer is C.

Area bounded by the curves

$y=x^3$ , the

$x$ -axis and the ordinates

$x=-2$ and

$x=1$ is

0%
$-9$
0%
$-\dfrac {15}{4}$
0%
$\dfrac {15}{4}$
0%
$\dfrac {17}{4}$

Explanation

Required area

$=-\displaystyle \int_{-2}^{0}x^3 dx+\int_{0}^{1}x^3 dx$

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

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

$=\left (\dfrac {1}{4}+4\right )=\dfrac {17}{4}$ sq. units
Thus, the correct answer is D.

The area of the region bounded by the curves

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

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

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

Explanation

Given curves are

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

$x={y}^{2}$ , which is the form of parabola.
The point of intersection,

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

$\Rightarrow x={ x }^{ 4 }\Rightarrow x\left( 1-{ x }^{ 3 } \right) =0$

$\Rightarrow x=0$ and

$1={ x }^{ 3 }$

$\Rightarrow x=0$ and

$x=1$
When

$x=0$ , then

$y=0$
When

$x=1$ , then

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

$\therefore$ The point of intersection is

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

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

$\therefore$ Area of shaded region

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

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

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

$=\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 }$ sq units

Area of the region bounded by

$y = |x|$ and

$y = |x| + 2$ , is

0%
$4\ sq. units$
0%
$3\ sq. units$
0%
$2\ sq. units$
0%
$1\ sq. units$

Explanation

The graph is shown in image

$1^{st}$ quadrant , intersection point of

$y=x$ and

$y=-x+2$ is

$(1,1)$

similarly in

$2^{nd}$ quadrant , intersection point of

$Y=X$ and

$y=-x+2$ is

$(-1,1)$

now

$A=2\Bigg[\int_{0}^{1}\bigg[-x+2\bigg]-\int_{0}^{1}\bigg[x\bigg]\Bigg]$

$A=2\times{1}$

$A=2 sq.unit$

The area included between the parabolas

$x^2=4y$ and

$y^2=4x$ is (in square units)

0%
$\dfrac{4}{3}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{16}{3}$
0%
$\dfrac{8}{3}$

Explanation

One intersection point of the two parabolas is the origin, while the other can be found out by substituting either equation into the other.

Thus we have

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

which yields

$x^4 = 64x$ i.e.

$x = 0,4$

Area included between the two curves can be found out by

$\displaystyle \int_0^4 \left (2\sqrt{x} - \dfrac{x^2}{4}\right) dx$

$=\left [ \dfrac{4x^{1.5}}{3} - \dfrac{x^3}{12}\right] _0^4$

$= \dfrac{32}{3} - \dfrac{64}{12}$

$= \dfrac{32}{6}$

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 included between the parabolas

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

$\displaystyle { x }^{ 2 }=4y$ is

0%
$\displaystyle \frac { 8 }{ 3 }$ sq unit
0%
$\displaystyle 8$ sq unit
0%
$\displaystyle \frac { 16 }{ 3 }$ sq unit
0%
$\displaystyle 12$ sq unit

Explanation

We know that, the area of region bounded by the parabolas

$\displaystyle { y }^{ 2 }=4ax$ and

$\displaystyle { x }^{ 2 }=4by$ is

$\displaystyle \frac { 16 }{ 3 } ab$ sq.unit.
Therefore,

$\displaystyle { y }^{ 2 }=4ax$ and

$\displaystyle { x }^{ 2 }=4y$ is

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

$\displaystyle \left( \because a=1,b=1 \right)$

The area of the region bounded by the graph of

$y = \sin x$ and

$y = \cos x$ between

$x = 0$ and

$x = \dfrac {\pi}{4}$ is

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

Explanation

Area bounded by the curves

$y=\sin x$ and

$y=\cos x$ is given by

$A=\displaystyle \int _{0} ^{\tfrac{\pi}{4}} (\cos x-\sin x) dx$

$=[\sin x+\cos x]_{0} ^{\tfrac{\pi}{4}}$

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

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

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

Area of the region satisfying

$x \le 2, y \le |x|,x-axis$ and

$x\ge 0$ is:

0%
$4$ sq unit
0%
$1$ sq unit
0%
$2$ sq unit
0%
None of these

Explanation

Required area = Area of shaded region OAB

$=^2_oydx=^2_0xdx=\frac{x^2}{2}^2_0$

$=$ 2 sq unit
Alternate Solution
Required area = Area of

$\Delta$ OAB

$=\frac{1}{2}\times 2\times 2$

$=$ 2 sq unit

The area of the region bounded by the curves

$y = x^{3}, y = \dfrac {1}{x}, x = 2$ is

0%
$4 - \log_{e}2$
0%
$\dfrac {1}{4} + \log_{e}2$
0%
$3 - \log_{e}2$
0%
$\dfrac {15}{4} - \log_{e}2$

Explanation

First of all we draw the graph,

$y = x^{3}, y = \dfrac {1}{x}, x = 2$

$\therefore$ Required area i.e.,

$(OMPNO)$

$=\displaystyle \int_{0}^{1} x^{3} dx + \int_{1}^{2} \dfrac {1}{x}dx$

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

The area (in square units) bounded by the curves

$y^{2} = 4x$ and

$x^{2} = 4y$ is

0%
$\dfrac {64}{3}$
0%
$\dfrac {16}{3}$
0%
$\dfrac {8}{3}$
0%
$\dfrac {2}{3}$

Explanation

$y^{2} = 4x$

$x^{2} = 4y$

solving two equations

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

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

$\dfrac {y^{4}}{4\times 4} = 4y$

$y^{4} = 64y$

$y(y^{2} - 64) = 0$

$y = 0, y^{3} = 64$

$y = 4$

$\int_{0}^{4} \left (\dfrac {y^{2}}{4} - 2\sqrt {y}\right )dy$

$= \left [\dfrac {y^{3}}{4\times 3} - \dfrac {y^{\dfrac {B}{2}}}{3}\right ]_{0}^{4}$

$= \left |\dfrac {-16}{3}\right | = \dfrac {16}{3}$

The line

$2y=3x+12$ cuts the parabola

$4y=3x^2$ . What is the area enclosed by the parabola and the line?

0%
$27$ square unit
0%
$36$ square unit
0%
$48$ square unit
0%
$54$ square unit

Explanation

The graph is shown in the image.

The intersection point of

$2y=3x+12$ and

$4y=3x^2$

$2\Big(\dfrac{3x^2}{4}\Big)=3x+12$

$x^2-2x-8=0$

$x=-2,4$

Hence, intersection point are

$(-2,3)$ and

$(4,12)$

Now,

$A=\int_{-2}^{4}\Bigg[\dfrac{3x+12}{2}\Bigg]-\int_{-2}^{4}\Bigg[\dfrac{3x^2}{4}\Bigg]$

$A=\Bigg[\dfrac{3x^2}{4}+6x \Bigg]\Bigg|_{-2}^4 -\Bigg[\dfrac{x^3}{4}\Bigg]\Bigg|_{-2}^4$

$A=9+36-18$

$A=27$

$sq.unit$

The area in the first quadrant between

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

$y = sin x$ is

0%
$\dfrac{\pi^3 - 8}{4}$
0%
$\dfrac{\pi^3}{4}$
0%
$\dfrac{\pi^3 - 16}{4}$
0%
$\dfrac{\pi^3 - 8}{2}$

Explanation

$x^2 + y^2 = \pi^2$ is a circle of radius

$\pi$ and centre at origin.

Required area

= Area of circle (1st quadrant)

$- \int_0^{\pi} sin x dx$

$= \dfrac{\pi. \pi^2}{4} - [-cos x]_{0}^{\pi} = \dfrac{\pi^3}{4} + (cos \pi - cos 0)$

$= \dfrac{\pi^3}{4} + (-1 - 1) = \dfrac{\pi^3}{4} - 2 = \dfrac{\pi^3 - 8}{4}$

Hence, option A is correct.

Consider the curves

$y = \sin x$ and

$y = \cos x$ .

What is the area of the region bounded by the above two curves and the lines

$x = 0$ and

$x = \dfrac {\pi}{4}$ ?

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

Explanation

The area enclosed by

$y=sinx, y=cosx,x=0,x=\frac { \pi }{ 4 }$ is given by

$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ (\cos x-\sin x)dx }$

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

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

$=\sqrt { 2 } -1$

The area bounded by the curves

$y = \cos x$ and

$y = \sin x$ between the ordinates

$x = 0$ and

$x = \dfrac {3\pi}{2}$ is

0%
$(4\sqrt {2} - 2)sq\ units$
0%
$(4\sqrt {2} + 2)sq\ units$
0%
$(4\sqrt {2} - 1)sq\ units$
0%
$(4\sqrt {2} + 1)sq\ units$

Explanation

Required area

$= \int_{0}^{\pi /4} (\cos x - \sin x) dx + \int_{\pi/4}^{5\pi/4}(\sin x - \cos x) dx + \int_{5\pi/4}^{3\pi/2} (\cos x - \sin x)dx$

$= [\sin x + \cos x]_{0}^{\pi/4} + [-\cos x - \sin x]_{\pi/4}^{5\pi/4} + [\sin x + \cos x]_{5\pi/4}^{3\pi/2}$

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

$=\dfrac{8}{\sqrt{2}} - 2$

$= (4\sqrt {2} - 2)sq\ units$

Hence, option A is correct.

Consider the curves

$y = \sin x$ and

$y = \cos x$ .

What is the area of the region bounded by the above two curves and the lines

$x = \dfrac {\pi}{4}$ and

$x = \dfrac {\pi}{2}$ ?

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

Explanation

The area enclosed by

$y=\sin x, y=\cos x, x=\dfrac { \pi }{ 2}, x=\dfrac { \pi }{ 4 }$ is given by

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

$=|-\sin x-\cos x|_{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }$

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

$=\sqrt { 2 } -1$

The area of the region bounded by the lines

$y = 2x + 1, y = 3x + 1$ and

$x = 4$ is

0%
$16\ sq.unit$
0%
$\dfrac {121}{3}\ sq.unit$
0%
$\dfrac {121}{6}\ sq.unit$
0%
$8\ sq.unit$

Explanation

$A$ (Shaded region)

$= A(\triangle ABD) - A(\triangle ABC) = \dfrac {1}{2} [4\times 12 - 4 \times 8] = \dfrac {1}{2} (48 - 32) = 8\ sq.units$ .
OR

$A (\triangle ACD) = \dfrac {1}{2}\begin{vmatrix} 0& 1 & 1\\ 4 & 9 & 1\\ 4 & 13 & 1\end{vmatrix}$

$= \dfrac {1}{2}\times 16 = 8$

The area bounded by the curves

$y=\cos x$ and

$y=\sin x$ between the ordinates

$x=0$ and

$x=\displaystyle\frac{3\pi}{2}$ is?

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

Explanation

We have,

$y=\cos x$ and

$y=\sin x$ between

$x=0$ and

$x=\displaystyle\frac{3\pi}{2}$

Required area A is given by

$A=\displaystyle\int^{3\pi /2}_0|\cos x-\sin x|dx$

$\Rightarrow A=\displaystyle\int^{\pi /4}_0|\cos x-\sin x|dx+\int^{5\pi/4}_{\pi /4}|\cos x -\sin x|dx+\displaystyle \int^{3\pi /2}_{5\pi/4}|\cos x - \sin x|dx$

$\Rightarrow \displaystyle A=\int^{\pi /4}_0(\cos x-\sin x)dx+\displaystyle \int^{5\pi/4}_{\pi /4}(\sin x -\cos x)dx+ \displaystyle \int^{3\pi /2}_{5\pi /4}(\cos x -\sin x)dx$

$\Rightarrow A=\displaystyle [\sin x+\cos x]^{\pi /4}_0+[-\cos x -\sin x]^{5\pi /4}_{\pi/4}+[\sin x+ \cos x]^{3\pi /2}_{\pi /4}$

$\Rightarrow A=(\sqrt{2}-1)+(\sqrt{2}+\sqrt{2})+(-1+\sqrt{2})$

$\Rightarrow A=4\sqrt{2}-2$ .

The line

$x=\dfrac{\pi}{4}$ divide the area of the region bounded by

$y=\sin x, y = \cos x$ and X-axis

$\left(0 \le x \le \frac{\pi}{2}\right)$ into two regions of areas

$A_1$ and

$A_2$ . Then,

$A_1:A_2$ equals

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

Explanation

Area

$A_1=\displaystyle \int_0^{\frac{\pi}{4}}\sin x\,dx$

$=-[\cos x]_0^{\frac{\pi}{4}}$

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

and Area

$A_2=\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos x\,dx$

$=\displaystyle [\sin x]^{\frac{\pi}{2}}_{\frac{\pi}{4}}=\left[1-\dfrac{1}{\sqrt{2}}\right]=\dfrac{\sqrt{2}-1}{\sqrt{2}}$

$\therefore A_1:A_2 = \dfrac{\sqrt{2}-1}{\sqrt{2}}: \dfrac{\sqrt{2}-1}{\sqrt{2}}=1:1$

Area bounded by the curves

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

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

0%
$\cfrac { 8 }{ 3 }$ sq. units
0%
$\cfrac { 3 }{ 8 }$ sq. units
0%
$\cfrac { 3 }{ 2 }$ sq. units
0%
None of these

Explanation

$\textbf{Step-1: Finding the bounded Area}$

$\text{On plotting these curves we can easily see area bounded by them}$

$\textbf{Step-2: And Area bounded by them is given as}$

$\Rightarrow 2\int\limits_{0}^{1}{[(2-{{x}^{2}})-{{x}^{2}}]}dx$

$\Rightarrow 2\int\limits_{0}^{1}{[2-2{{x}^{2}}]}dx$

$\Rightarrow 2\times \left| 2x-\dfrac{2{{x}^{3}}}{3} \right|_{0}^{1}$

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

$\Rightarrow \dfrac{8}{3}$

$\textbf{Therefore, area bounded by these curves is (A) }$ $\mathbf {\dfrac{8}{3}}$

The area bounded by the parabola

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

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

0%
$\cfrac { 16 }{ 3 } { a }^{ 2 }$
0%
$\cfrac { 8 }{ 3 } { a }^{ 2 }$
0%
$\cfrac { 4 }{ 3 } { a }^{ 2 }$
0%
None of these

Explanation

Required area

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

$=4\times2\sqrt { a } { \left[ \cfrac { -2{ (a-x) }^{ 3/2 } }{ 3 } \right] }_{ 0 }^{ 4 }=\cfrac { 16 }{ 3 } { a }^{ 2 }$

Consider an ellipse

$\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$ What is the area included between the ellipse and the greatest rectangle inscribed in the ellipse?

0%
$ab(\pi -1)$
0%
$2ab(\pi -1)$
0%
$ab(\pi -2)$
0%
None of the above

Explanation

As the rectangle lies inside a standard ellipse, the vertices of the rectangle can be assumed to be

$(h,k),(h,-k),(-h,k)$ and

$(-h,-k)$ without any loss of generality.

$\therefore \dfrac{h^2}{a^2}+\dfrac{k^2}{b^2}=1\implies k=b\sqrt{1-\dfrac{h^2}{a^2}}$

Clearly, the sides of the rectangle based on the coordinates chosen can be determined as

$2h$ and

$2k$ .

$\therefore$ area of rectangle

$=A(h)=2h\times 2k=4h b\sqrt{1-\dfrac{h^2}{a^2}}\implies 4b\sqrt{h^2-\dfrac{h^4}{a^2}}$

Now, let

$P(h)=h^2-\dfrac{h^4}{a^2}$ . Thus, the maximization of

$A(h)$ boils down just to the maximization of

$P(h)$ .

Differentiating

$P(h)$ wrt

$h$ , we have

$2h-\dfrac{4h^3}{a^2}=0\implies h=\dfrac{2h^3}{a^2}\implies 2h^2=a^2\implies h=\dfrac{a}{\sqrt{2}}\implies k=\dfrac{b}{\sqrt{2}}$

Hence,

$A_{max}=2ab$

$\therefore$ area included between this ellipse and rectangle

$=\pi ab-2ab=ab(\pi-2)$ .

this is the required answer.

The area of the region bounded by the curve

$y = x^{2}$ and

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

0%
$\dfrac {16}{3}sq. units$
0%
$\dfrac {8}{3}sq. units$
0%
$\dfrac {4}{3}sq. units$
0%
$\dfrac {2}{3}sq. units$

Explanation

Here area,

$\int _{ 0 }^{ 2 }{ (4x-x^{ 2 }-x^{ 2 })dx }$

$=\int _{ 0 }^{ 2 }{ (4x-2x^{ 2 })dx }$

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

$=\dfrac { 8 }{ 3 }$

The area of the figure bounded by the parabolas

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

$x = 1 - 3y^{2}$ is

0%
$\dfrac {4}{3}$ square units
0%
$\dfrac {2}{3}$ square units
0%
$\dfrac {3}{7}$ square units
0%
$\dfrac {6}{7}$ square units

Explanation

Given curves are

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

$x=1-3y^{2}$

$-2y^{2}=1-3y^{2}$

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

$\therefore$ Curves intersect at

$(-2, \pm 1)$
Area

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

$=\int_{-1}^{1} (1-y^{2})\ dy$

$=\int_{-1}^{0} (1-y^{2})\ dy +\int_{0}^{1} (1-y^{2})\ dy$

Substituting

$y=-y$ in first integral we get,

Area

$=\int_{0}^{1} (1-y^{2})\ dy +\int_{0}^{1} (1-y^{2})\ dy$

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

$=2\left|y-\dfrac{y^{3}}{3}\right|_{0}^{1}$

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

Hence, area bounded by the given curves is

$\dfrac{4}{3}$ square units.

The area formed by triangular shaped region bounded by the curves

$y=\sin { x } ,y=\cos { x }$ and

$x=0$ is

0%
$\left( \sqrt { 2 } -1 \right)$ sq unit
0%
$1$ sq unit
0%
$\sqrt { 2 }$ sq units
0%
$\left( \sqrt { 2 } +1 \right)$ sq units

Explanation

$\left( \because y=\sin { x } \text{ and } y=\cos { x } \text{ meet at } x={ \pi }/{ 4 } \right)$

Then, the upper limit will be

$\dfrac{\pi}{4}$ .

Required Area

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

$=\left[\sin x+\cos x\right]_{0}^{\pi/4}$

$=\left( \sqrt { 2 } -1 \right)$ sq unit

The area of the figure bounded by the curves

$y = |x - 1|$ and

$y = 3 - |x|$ is

0%
$2\ sq. units$
0%
$3\ sq. units$
0%
$4\ sq. units$
0%
$1\ sq. units$

Explanation

Area of the curve figure bounded by curves

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

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

$y=x-1\longrightarrow (1)\\ y=1-x\longrightarrow (2)\\ y=3-x\longrightarrow (3)\\ y=3+x\longrightarrow (4)$

For : (B)

$y=1-x\\ y=3-x\\ x=-1\\ y=2$

Area required = Area of rectangle ABCD

$=(AB)(AD)$

$=\sqrt { 8 } \times \sqrt { 2 } \\ =4\quad sq.\quad units$

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 3 - 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 3 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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