MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 1

The area of the region bounded by the curves

$y = x^2$ and

$y = |x|$ is

0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{5}{6}$
0%
$\dfrac{5}{3}$

Area bounded by the curves

$y = \sin x ,$ tangent drawn to it at

$x = 0$ and the line

$x = \frac { \pi } { 2 }$ is equal to

0%
$\frac { \pi ^ { 2 } - 4 } { 2 }$ sq.units
0%
$\frac { \pi ^ { 2 } - 4 } { 4 }$ sq.units
0%
$\frac { \pi ^ { 2 } - 2 } { 4 }$ sq.units
0%
$\frac { \pi ^ { 2 } - 2 } { 2 }$ sq.units

The area enclosed by the line y = x + 1, X- axis and the lines x = -3 and x = 3 is

0%
5
0%
7
0%
10
0%
9

Area bounded by curve

$y = (x - 1)(x - 2)(x - 3)$ and x-axis between lines

$x = 0, x = 3$

0%
$5/2$
0%
$11/4$
0%
$1$
0%
$3$

As shown in the figure of an ellipse

$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ . The area of shaded region is .......
.

0%
$12\pi$
0%
$3(\pi -2)$
0%
$4(\pi -2)$
0%
$12(\pi -2)$

The area bounded by the curves

$x+2|y|=1$ and

$x=0$ is?

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

Explanation

$x+2\left| y \right| =1$ and

$x=0$

$y=+ve$

$x+2y=1$

$y=-ve$

$x-2y=1$

Area of triangle

$=\cfrac{1}{2}\times base\times height\\=\cfrac{1}{2}\times 1\times 1\\=\cfrac{1}{2}\text{ }unit$

Area included between

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

$y = \dfrac { 8 a ^ { 3 } } { x ^ { 2 } + 4 a ^ { 2 } }$ is

0%
$\dfrac { a ^ { 2 } } { 3 } ( 6 \pi - 4 )$
0%
$\dfrac { a ^ { 2 } } { 3 } ( 4 \pi + 3 )$
0%
$\dfrac { a ^ { 2 } } { 3 } ( 8 \pi + 3 )$
0%
None of these

The area of the figure formed by

$a|x|+b|y|+c=0$ , is

0%
$\dfrac{c^{2}}{|ab|}$
0%
$\dfrac{2c^{2}}{|ab|}$
0%
$\dfrac{c^{2}}{2|ab|}$
0%
$None\ of\ these$

Find the area of bounded by

$y=\sin x$ from

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

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

0%
$\dfrac{\sqrt 2-1}{\sqrt2}$
0%
$\dfrac12$
0%

$\dfrac 14$
0%
None of these

Explanation

The area bounded is given as

$\displaystyle \int _{\pi/4}^{\pi/2}\sin xdx\\\left.-\cos x \right|_{\pi/4}^{\pi/2}\\-\cos \dfrac {\pi}2+\cos \dfrac{\pi}4\\\dfrac{\sqrt 2-1}{\sqrt2}$

The area of the region by curves

$y=x\log x$ and

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

0%
$1/12$
0%
$3/12$
0%
$7/12$
0%
$None\ of\ these$

The area of the region
A=

$\left [ \left ( x,y \right ) :0\leq y\leq x\left | x \right |+1and -1\leq x\leq 1\right ]$ . in sq. units, is:

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

The area bounded by x-axis the curve

$y=f(x)$ and the lines

$x =1,x=b$ equal to

$\left( \sqrt { \left( { b }^{ 2 }+1 \right) } -\sqrt { 2 } \right) for\quad all\quad b>1,then\quad f(x)$

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

The area bounded by the parabolas

$y = {\left( {x + 1} \right)^2}\;and\;y = {\left( {x - 1} \right)^2}$ and the line

$y = \frac{1}{4}$ is

0%
${\text{4sq}}{\text{.}}\;{\text{units}}$
0%
$\frac{1}{6}\;{\text{sq}}{\text{.}}\;{\text{units}}$
0%
$\frac{4}{3}\;{\text{sq}}{\text{.}}\;{\text{units}}$
0%
$\frac{1}{3}\;{\text{sq}}{\text{.}}\;{\text{units}}$

The area enclosed by parabola

${ y }^{ 2 }=64x$ and its latus-rectum is

$\lambda$ then

$3\lambda$ then

$3\lambda$ =

0%
2048
0%
2408
0%
2804
0%
2084

the volume of a solid obtained by revolving about y-axis enclosed between the ellipse

${x}^{2}+9{y}^{2}=9$ and the straight line

$x+3y=3$ in the first quadrant is

0%
$3\pi$
0%
$4\pi$
0%
$6\pi$
0%
$9\pi$

The area of the region described by

$A= (x,y):x^{2}+y^{2}\leq 1$ and

$y^{2}\leq 1-x$ is:

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

Explanation

$A=\displaystyle \frac{1}{2}\times \pi +2\int_{0}^{1}\sqrt{1-x}dx$

$=\displaystyle \frac{\pi }{2}+\displaystyle \frac{4}{3}$

What is the area of the region bounded by the lines

$x=y,y=0$ and

$x=4$ ?

0%
$4$ square units
0%
$8$ square units
0%
$12$ square units
0%
$16$ square units

Explanation

The region bounded by

$x=y,y=0$ and

$x=4$ is the triangle

$OAB$ as shown in the figure.
Co-ordinate of

$A$ is

$(4,4)$
Length of

$AB$ is

$4$ unit.

Since, the bounded area is a triangle. So, we have
Area of the bounded region

$=\dfrac{1}{2}\times AB\times OB=\dfrac { 1 }{ 2 }\cdot 4\cdot 4=8$
Option B is correct

The area of the region bounded by the parabola

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

$(2,\ 3)$ and the

$\mathrm{x}$ -axis is

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

Explanation

Equation of tangent at

$(2,3)$

$:$

$2y=x+4$

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

$=\displaystyle \int_{0}^{3}[(y-2)^{2}-2y+5]\text{ }dy$

$=\left[\dfrac{(y-2)^{3}}{3}-y^{2}+5y\right]_{0}^{3}$

$=9$

The parabolas

$y^{2}=4x$ and

$x^{2}=4y$ divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If

$S_{1},S_{2},S_{3}$ are respectively the areas of these parts numbered from top to bottom(Example:

$S_1$ is the area bounded by

$y=4$ and

$x^{2}=4y$ ); then

$S_{1},S_{2},S_{3}$ is

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

Explanation

Since

$y^{2} = 4x$ and

$x^{2} = 4y$
are symmetric about the line y = x
Now, Area bounded between

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

$\int_{0}^{4}\left ( 2\sqrt{x} - x \right )dx$

$= \dfrac{8}{3}$

$A_{s_{2}} = \dfrac{16}{3}$ and

$A_{s_{1}} = A_{s_{3}} = \dfrac{16}{3}$

$A_{s_{1}}:A_{s_{2}}:A_{s_{3}} :: 1:1:1$

The area of the region bounded by the curves

$x+2y^{2}=0$ and

$x+3y^{2}=1$ is equal to

0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{5}{3}$
0%
$\displaystyle \frac{1}{3}$

Explanation

Solving the equations

$x+2{y}^{2}=0$ and

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

we get points of intersection

$(-2,1)$ and

$(-2,-1)$
The bounded region is show in figure.
The required area

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

$\displaystyle =2\int _{ 0 }^{ 1 }{ \left( 1-{ y }^{ 2 } \right) } dy=2\left[ y-\frac { { y }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }=2\times \frac { 2 }{ 3 } =\frac { 4 }{ 3 } .$

The area bounded by the curves

$\mathrm{y}=$ cosx and

$\mathrm{y}=$ sinx between the ordinates

$\mathrm{x}=0$ and

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

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

Explanation

Required area

$=\displaystyle \int_0^{\frac{3\pi}{2}}|sinx-cosx|dx$

$= \displaystyle \int_0^{\frac{\pi}{4}}(\cos x-\sin x)dx+ \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}(\sin x-\cos x)dx+ \int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}}(\cos x-\sin x)dx$

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

$= 4\sqrt{2}-2$

The area of the region above the x-axis bounded by the curve

$y=tan x, 0 \leq x\leq \frac {\pi}{2}$ and the tangent to the curve at

$x=\frac {\pi}{4}$ is :

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

The area bounded between the parabolas

$4 x^{2}=y$ and

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

$\mathrm{y}=2$ is:

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

Explanation

The given curves

$y=4x^2$ and

$\displaystyle y=\dfrac {1}{9}x^2$

$\displaystyle Area(\text{yellow region})=2\int_0^2\left(3\sqrt y-\frac {\sqrt y}{2}\right)dy=2\left[\frac {5y\sqrt y}{3}\right]_0^2$

$\displaystyle =2\times \frac {5}{3}\times 2\sqrt 2=\frac {20\sqrt 2}{3}$

The area enclosed by the curves

$y^{2} = x$ and

$y = |x|$ is

0%
$\dfrac{2}{3}$
0%
$1$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$

Explanation

Required area

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

$\displaystyle ={ \left[ \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }=\frac { 2 }{ 3 } -\frac { 1 }{ 2 } =\frac { 1 }{ 6 }$

The area (in sq. units) of the region described by

$\left \{(x, y);y^2 \leq 2x \; {\text {and}} \;\;y\geq 4x-1\right \}$ is

0%
$\dfrac {7}{32}$
0%
$\dfrac {5}{64}$
0%
$\dfrac {15}{64}$
0%
$\dfrac {9}{32}$

Explanation

Solving the curve

$y^2=2x$ and

$y=4x-1$ we get the point of intersection
of the curves which are

$\left(\cfrac{1}{2},1\right)$ and

$\left(\cfrac{1}{8},-\cfrac{1}{2}\right)$

Hence required area is,

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

$\displaystyle =\frac{1}{4}\int_{-\frac{1}{2}}^1 \left(y+1-2y^2\right) dy=\frac{1}{4}\left(\frac{y^2}{2}+y-\frac{2}{3}y^3\right)_{-\frac{1}{2}}^1 =\cfrac{9}{32}$

Note: By plotting the curves on graph it could be understood very easily.

The area (in square units) of the region bounded by the curves

$y + 2x^2 = 0$ and

$y + 3x^2 = 1$ , is equal to

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{3}{5}$
0%
$\displaystyle \frac{3}{4}$

Explanation

Solving both the equations we get point of intersection both the curves as

$(1,-2)$ and

$(-1,-2)$
Thus required are is given by,

$\displaystyle A =\int_{-1}^{1}(1-3x^2-(-2x^2))dx=\cfrac{4}{3}$

The area (in sq. units) of the region

$\{(x, y):x \geq 0, x+y \leq 3, x^2 \leq 4y$ and

$y\leq 1+\sqrt{x}\}$ is.

0%
$\displaystyle\frac{59}{12}$
0%
$\displaystyle\frac{3}{2}$
0%
$\displaystyle\frac{7}{3}$
0%
$\displaystyle\frac{5}{2}$

Explanation

$C_1: x≥0$

$C_2: x+y≤3$

$C_3: x^2≤4y$

$C_4: y≤1+\sqrt x$

Solving these curves, we get their intersection points as shown in the figure.

Now, we have to find the area of the shaded region, which is formed using the constraint.

$Area = \displaystyle\int_0^1(1+\sqrt x) \ dx + \int_1^2(3-x)\ dx - \int_0^2(x^2)\ dx$

$\Rightarrow Area = \left[x+\cfrac23x^{3/2}\right]_0^1+[3x-x^2]_1^3 - \left[\cfrac{x^3}3\right]_0^2 = \cfrac52$

The area (in sq. units) of the region described by

$A = \left \{(x, y)| y \geq x^{2} - 5x + 4, x + y \geq 1, y \leq 0\right \}$ is:

0%
$\dfrac {17}{6}$
0%
$\dfrac {13}{6}$
0%
$\dfrac {19}{6}$
0%
$\dfrac {7}{2}$

Explanation

$y \geq x^2-5x+4= (x-\dfrac{5}{2})^2-\dfrac{9}{4}$

$y+\dfrac{9}{4} \geq (x-\dfrac{5}{2})^2$
This is a parabola with vertex at

$\left(\dfrac{5}{2},-\dfrac{9}{4}\right)$
And other two curves are

$x + y \geq 1$ and

$y \leq 0$
Draw it. And From the figure we can see
Area of the shaded region AED= Area of triangle ADE+ area of the region EBD
Area of triangle ABD

$\displaystyle = \dfrac{1}{2}\times2\times2=2$

Area of region EBD

$\displaystyle \left|\int_{3}^{4} ydx\right| =\left| \int_{3}^{4} (x^2-5x+4)dx \right|$

$=\left| \left[\dfrac {x^3}{3} - \dfrac {5x^2}{2} +4x\right]^4_3 \right|$

$= \left |\dfrac{263}{3}-45\right|=\dfrac{7}{6}$

Hence, Total Area

$=2+\dfrac{7}{6}=\dfrac{19}{6}$

The area(in sq. units) of the smaller portion enclosed between the curves,

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

$y^2=3x$ , is.

0%
$\displaystyle\frac{1}{\sqrt{3}}+\frac{4\pi}{3}$
0%
$\displaystyle\frac{1}{2\sqrt{3}}+\frac{\pi}{3}$
0%
$\displaystyle\frac{1}{2\sqrt{3}}+\frac{2\pi}{3}$
0%
$\displaystyle\frac{1}{\sqrt{3}}+\frac{2\pi}{3}$

Explanation

Here,

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

$y^2=3x$

So finding the co-ordinates of

$x$

$\Rightarrow$

$x^2+3x-4=0$

$(x+4)(x-1)=0$

$x=-4, x=1$

Area

$=\left(\displaystyle\int^1_0\sqrt{3}\cdot \sqrt{x}dx+\displaystyle\int^2_1\sqrt{4-x^2}\cdot dx\right ) \times 2$

$=\left(\displaystyle \sqrt{3}\left(\displaystyle\frac{x^{3/2}}{3/2}\right)^1_0+\left(\displaystyle\frac{x}{2}\sqrt{4-x^2}+2sin^{-1}\frac{x}{2}\right)^2_1\right)\times 2$

$=\left(\sqrt{3}\left(\displaystyle\frac{2}{3}\right)+\left\{2\cdot \displaystyle\frac{\pi}{2}-\left(\displaystyle\frac{\sqrt{3}}{2}+\frac{\pi}{3}\right)\right\}\right)\times 2$

$=\left(\displaystyle\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}+\frac{2\pi}{3}\right)\times 2$

$=\left(\displaystyle\frac{1}{2\sqrt{3}}+\frac{2\pi}{3}\right)\times 2=\displaystyle\frac{1}{\sqrt{3}}+\frac{4\pi}{3}$ .

The area bounded by the curves

${ y }=\sqrt { { x } } ,2{ y }+3={ x }$ and

$x$ -axis in the

${1}^{st}$ quadrant is

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

Explanation

Given curves are

${ y }=\sqrt { { x } } ,2{ y }+3={ x }$

$\Rightarrow y^{2}=2y+3\Rightarrow y=3,-1$

Then, the curves intersection in the first quadrant at

$y=3, x=9$

To find the area between the curves

$y=\sqrt { x } ,2y+3=x$ and

$x$ -axis in the

${1}^{st}$ quadrant
The required area is given by shaded portion

$OABO$

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

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

$\displaystyle=\left( \frac { 2 }{ 3 } .27 \right) -\frac { 1 }{ 2 } \left\{ \left( \frac { 81 }{ 2 } -27 \right) -\left( \frac { 9 }{ 2 } -9 \right) \right\}$

$\displaystyle=18-\frac { 1 }{ 2 } \left( 18 \right) =9$

The area enclosed between the curves

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

$\mathrm{x}=\mathrm{a}\mathrm{y}^{2} (\mathrm{a}>0)$ is 1 sq. unit, then the value of a is

0%
$1/\sqrt{3}$
0%
$1/2$
0%
$1$
0%
$1/3$

Explanation

Points of intersection of

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

$\mathrm{x}=\mathrm{a}\mathrm{y}^{2}$ are

$(0, 0)$ and

$\left(\displaystyle \frac{1}{\mathrm{a}}, \displaystyle \frac{1}{\mathrm{a}}\right)$ .
Hence area is,

$\displaystyle \int_{0}^{1/\mathrm{a}}\left(\sqrt{\frac{\mathrm{x}}{\mathrm{a}}}- ax^{2} \right)\mathrm{dx} =1\Rightarrow \mathrm{a}=\frac{1}{\sqrt{3}} (\mathrm{a}\mathrm{s} \ \mathrm{a}>0)$ .

The area of the region between the curves

$\mathrm{y}=\sqrt{\dfrac{1+\sin \mathrm{x}}{\cos \mathrm{x}}}$ and

$\mathrm{y}=\sqrt{\dfrac{1-\sin \mathrm{x}}{\cos \mathrm{x}}}$ bounded by the lines

$\mathrm{x}=0$ and

$\displaystyle \mathrm{x}=\frac{\pi}{4}$ is

0%
$\displaystyle \int_{0}^{\sqrt{2}-1}\frac{\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} \mathrm{d}\mathrm{t}$
0%
$\displaystyle \int_{0}^{\sqrt{2}-1}\frac{4\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} \mathrm{d}\mathrm{t}$
0%
$\displaystyle \int_{0}^{\sqrt{2}+1}\frac{4\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} dt$
0%
$\displaystyle \int_{0}^{\sqrt{2}+1}\frac{\mathrm{t}}{(1+t^{2})\sqrt{1-\mathrm{t}^{2}}} dt$

Explanation

$\displaystyle \int_{0}^{\pi/4}(\sqrt{\dfrac{1+\sin \mathrm{x}}{\cos \mathrm{x}}}-\sqrt{\dfrac{1-\sin \mathrm{x}}{\cos \mathrm{x}}})\mathrm{d}\mathrm{x}$

$=\int _{0}^{\pi /4}(\sqrt{\dfrac{1+tan\dfrac{x}{2}}{1-tan\dfrac{x}{2}}}-\sqrt{\dfrac{1-tan\dfrac{x}{2}}{1+tan\dfrac{x}{2}}})dx=\int \dfrac{(1+tan\dfrac{x}{2})-(1-tan\dfrac{x}{2})}{\sqrt{1-tan^{2}\dfrac{x}{2}}}dx$

$=\displaystyle \int_{0}^{\pi/4}\dfrac{2\tan\dfrac{\mathrm{x}}{2}}{\sqrt{1-\tan^{2}\frac{\mathrm{x}}{2}}} dx =\displaystyle \int_{0}^{\sqrt{2}-1}\dfrac{4\mathrm{t}}{(1+\mathrm{t}^{2})\sqrt{1-\mathrm{t}^{2}}} dt$ as

$\displaystyle \tan\dfrac{\mathrm{x}}{2}=\mathrm{t}$ .

Area of the region bounded by the curve

$\mathrm{y}=\mathrm{e}^{\mathrm{x}}$ and lines

$\mathrm{x}=0$ and

$\mathrm{y}=\mathrm{e}$ is:

0%
$\mathrm{e}-1$
0%
$\displaystyle \int_{1}^{\mathrm{e}}\ln(\mathrm{e}+1-\mathrm{y}) dy$
0%
$\displaystyle \mathrm{e}-\int_{0}^{1}\mathrm{e}^{\mathrm{x}}\mathrm{d}\mathrm{x}$
0%
$\displaystyle \int_{1}^{\mathrm{e}}\ln ydy$

Explanation

Required Area

$=\displaystyle \int_{1}^{\mathrm{e}}\ln ydy$

$=(\mathrm{y}\ln \mathrm{y}-\mathrm{y})_{1}^{\mathrm{e}}=(\mathrm{e}-\mathrm{e})-\{-1\}=1$ .

Also,

$\displaystyle \int_{1}^{\mathrm{e}}\ln ydy =\displaystyle \int_{1}^{\mathrm{e}}\ln(\mathrm{e}+1-\mathrm{y}) dy$

Further the required area

$=\displaystyle \mathrm{e}\times 1-\int_{0}^{1}\mathrm{e}^{\mathrm{x}}\mathrm{d}\mathrm{x}$

The area bounded by the curve

$y = f\left( x \right)$ , above the

$x$ -axis, between

$x = a$ and

$x = b$ is:

0%
$\displaystyle\int _{ f\left( a \right) }^{ b }{ ydy }$
0%
$\displaystyle\int _{ a }^{ f\left( b \right) }{ xdx }$
0%
$\displaystyle\int _{ a }^{ b }{ xdy }$
0%
$\displaystyle\int _{ a }^{ b }{ ydx }$

Explanation

We need to calculate the area bounded by

$y=f(x)$ between

$x=a$ and

$x=b$

So, we need to find the area of the region inside the curve

$y=f(x)$ having limit points

$a$ to

$b$ and it is given by

$\displaystyle \int_{a}^{b} f(x)dx$

$=\displaystyle \int_{a}^{b} ydx$

Hence, option D is correct.

The area bounded by the

$x-$ axis, the curve

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

$x=1$ and

$x=b$ is equal to

$\left(\sqrt{{b}^{2}+1}-\sqrt{2}\right)$ for all

$b>1$ , then

$f\left(x\right)$ is

0%
$\sqrt{x-1}$
0%
$\sqrt{x+1}$
0%
$\sqrt{{x}^{2}+1}$
0%
$\dfrac{x}{\sqrt{{x}^{2}+1}}$

Explanation

$\int_{1}^{b}{f\left(x\right)dx}=\sqrt{\left({b}^{2}+1\right)}-\sqrt{2}$
Differentiating both sides w.r.t

$b$ , we get

$f\left(b\right)=\dfrac{b}{\sqrt{\left({b}^{2}+1\right)}}$
Hence

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

Area enclosed between the curves

$y=8-{x}^{2}$ and

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

0%
$32/3$
0%
$64/3$
0%
$30/4$
0%
$9$

Explanation

$y=8-{ x }^{ 2 }\quad y={ x }^{ 2 }$

Intersection points are

${ x }^{ 2 }=8-{ x }^{ 2 }$

$2{ x }^{ 2 }=8$

${ x }^{ 2 }=4\quad x=-2,2$

$\int _{ -2 }^{ 2 }{ \left( 8-{ x }^{ 2 }-{ x }^{ 2 } \right) dx }$

$\int _{ -2 }^{ 2 }{ \left( 8-2{ x }^{ 2 } \right) dx }$

$2{ \left( 8x-\cfrac { 2{ x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }$

$=2\left( 8\left( 2 \right) -\cfrac { 2\left( 8 \right) }{ 3 } \right)$

$=2\left( 16-\cfrac { 16 }{ 3 } \right)$

$=2\left( \cfrac { 32 }{ 13 } \right)$

$=\cfrac { 64 }{ 3 }$ sq. units

1122444_1037384_ans_6edebbde645b4f50ac93480d5e1613cb.png

If area bounded by the curves

$x=at^2$ and

$y=ax^2$ is

$1$ , then a

$=$ __________.

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

Explanation

$x = ay^2$ and

$y = ax^2$

$\Rightarrow a^3x^3 = 1 \Rightarrow (x,y) = (\dfrac{1}{a}, \dfrac{1}{a})$

The area bounded by the curves is computed by:

$\displaystyle \int_{0}^{1/a} |ax^2 - \sqrt{\dfrac{x}{a}}| dx = 1 \Rightarrow \dfrac{1}{3a^2} = 1 \Rightarrow a = \dfrac{1}{\sqrt{3}}$

So option

$C$ is the correct answer.

Calculate the area of the shaded region in the figure, where

$\square ABCD$ is a square with side 8 cm each.

$(\pi =3.14)$

0%
$36.48 \,cm^2$
0%
$25.40 \,cm^2$
0%
$15 \,cm^2$
0%
$65 \,cm^2$

Explanation

Area of square =

$Side^2$
=

$8^2$
=

$64$ sq cm
Area of shaded portion = Double area of segment formed by diagonal of the square
Area of two quadrants =

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

$\dfrac{1}{2} \times 3.14 \times 8^2$

$100.48$ sq cm
Area of square - Area of two triangles formed by radii and the cord
So,
Area of shaded portion =

$100.48-64$
=

$36.48 sq$ cm

The area included between the parabolas

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

$y=\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } }$ is

0%
${ a }^{ 2 }\left( 2\pi +\dfrac { 2 }{ 3 } \right)$
0%
${ a }^{ 2 }\left( 2\pi -\dfrac { 8 }{ 3 } \right)$
0%
${ a }^{ 2 }\left( \pi +\dfrac { 4 }{ 3 } \right)$
0%
${ a }^{ 2 }\left( \pi -\dfrac { 4 }{ 3 } \right)$

Explanation

Firstly we have to find the point of intersection.

$\dfrac { { x }^{ 2 } }{ 4a } =\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } }$

On solving the equation We get $x^{2}=4a^{2}$ and $x^{2}= -8a^{2}$ (which is not possible)

We have $x=2a$ and $x=-2a$

Now area between these 2 curves is

$\displaystyle \int _{ -2a }^{ 2a }{ \left(\dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ 4a } \right)}dx$

$\displaystyle =\int _{ -2a }^{ 2a }{ \dfrac { 8{ a }^{ 3 } }{ { x }^{ 2 }+4{ a }^{ 2 } } dx } -\int _{ -2a }^{ 2a }{ \dfrac { { x }^{ 2 } }{ 4a } } dx$

$=\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { x }{ 2a } } -\dfrac { { x }^{ 3 } }{ 12a }$

After putting the limits we get

$\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { 2a }{ 2a } } -\dfrac { { 8a }^{ 3 } }{ 12a } -(\dfrac { 8{ a }^{ 3 } }{ 2a } \tan ^{ -1 }{ \dfrac { -2a }{ 2a } } -\dfrac { { -8a }^{ 3 } }{ 12a } )$

$=\pi { a }^{ 2 }-\dfrac { 4 }{ 3 } { a }^{ 2 }$

Find the area of the region bounded by the curve

$y^2=4x$ and the line

$x=3$ .

0%
$4\sqrt{3}$
0%
$8\sqrt{3}$
0%
$6$
0%
$2\sqrt{3}$

Explanation

Required area = area of OACO = 2 ( Area of OAB)

$\displaystyle =2\int_{0}^{3} 2\sqrt{x} dx$

$=\dfrac{8}{3} \times (3)^{3/2}=8\sqrt{3}$ units.

1392650_890350_ans_03107aa1322c40bb92cfd4dbd07985b6.png

The area enclosed between the

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

$y=|x|$ is

0%
$\dfrac {1}{3}$
0%
$\dfrac {2}{3}$
0%
$1$
0%
$\dfrac {1}{6}$

Explanation

$y = x - \left( 1 \right)\,\,and\,{y^2} = x - \left( 2 \right)$

$solving\,1\,and\,2$

${y^2} = y$

${y^2} - y = 0$

$y\left( {y - 1} \right) = 0$

so,

$y = 0,1$

thus points of intersection are

$(0,0)$ ,

$(1,1)$

Area enclosed=

$\int_0^1 {\left( {{y_1} - {y_2}} \right)dx}$

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

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

$= \dfrac{2}{3} - \dfrac{1}{2} = \dfrac{1}{6}$

1024017_1052357_ans_6fb10a833bd54d91aa29f9b4d684ee9b.jpg

Points of inflexion of the curve

$y = x^4 - 6x^3 + 12x^2 + 5x + 7$ are

0%
$(1, 19); (1, 12)$
0%
$(1, 19); (2, 33)$
0%
$(1, 2); (2, 1)$
0%
$(1, 7); (2, 6)$

Explanation

$y=x^{4}-6x^{3}+12x^{2}+5x+7$

$y(x)=4x^{3}-18x^{2}+24x+5$

$y(x)=12x^{2}-36x+24$

$y(x)=0$

$12x^{2}-36x+24=0$

$x^{2}-3x+2=0$

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

$x(x-2)-1(x-2)=0$

$(x-1)-1(x-2)=0$

$(x-1)(x-2)=0$

$x=1, 2$

Inflection point of a function is where the function changes from concave up to concave down or vice-versa

$x<1, f(x)>0$

$1<x<2, f(x)<0$

$x>2, f(x)>0$

$\because f(x)$ changes sign

$\therefore$ At

$x=1, 2y=f(x)$ has inflection point

$x=1, y=f(1)=19$

$x=2, y=f(2)=33$

Point of inflection

$(1,19); (2,33)$

The value of

$a$ for which the area between the curves

${y^2} = 4ax$ and

${x^2} = 4ay$ is

$1\,sq.\,unit$ , is-

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

Explanation

$\begin{array}{l}{y^2} = 4ax\\y = \sqrt {4ax} \\{x^2} = 4ay\\y = \dfrac{{{x^2}}}{{4a}}\\area = \int_0^{4a} {\sqrt {4ax} dx} - \int_0^{4a} {\dfrac{{{x^2}}}{{4a}}dx} \\\left. {2\sqrt a \times \dfrac{2}{3}{{\left( x \right)}^{3/2}}} \right]_0^{4a} - \left. {\dfrac{{{x^3}}}{{3\left( {4a} \right)}}} \right]_0^{4a}\\ = \dfrac{{32{a^2}}}{3} - \dfrac{{16{a^2}}}{3}\\ = \dfrac{{16{a^2}}}{3} = 1\\a = \dfrac{{\sqrt 3 }}{4}\end{array}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is false and Reason are correct

Explanation

The given curves are

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

$\left(1\right)$
and

$y=\lambda x+1$ .........

$\left(2\right)$
Solving eqns

$\left(1\right)$ and

$\left(2\right)$ we get

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

$\alpha,\beta$ are the roots of the quadratic , then

$\alpha+\beta=\lambda-2,\alpha\beta=-4$
Hence, Required area

$S\left(\lambda\right)=\left|\int_{\alpha}^{\beta}{\left(\lambda x+1\right)-\left({x}^{2}+2x-3\right)dx}\right|$

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

$=\left|4\left(\beta-\alpha\right)+\dfrac{\left(\lambda-2\right)}{2}\left({\beta}^{2}-{\alpha}^{2}\right)-\dfrac{1}{3}\left({\beta}^{3}-{\alpha}^{3}\right)\right|$

$=\sqrt{{\left(\beta+\alpha\right)}^{2}-4\beta\alpha}\left|\left\{4+\dfrac{\left(\lambda-2\right)}{2}\left(\beta+\alpha\right)-\dfrac{1}{3}\left\{{\left(\alpha+\beta\right)}^{2}-\alpha\beta\right\}\right\}\right|$

$=\dfrac{1}{6}{\left\{{\left(\lambda-2\right)}^{2}+16\right\}}^{\frac{3}{2}}$
For least value of

$S\left(\lambda \right),\lambda-2=0$

$\therefore \lambda=2$

Hence, the assertion and reason are correct and Reason is the correct explanation of assertion.

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 curves

$3x^2 + 5y = 32$ and

$y = \left|x-2\right|$ is

0%
25
0%
17/2
0%
33/2
0%
33

Explanation

$\begin{array}{l} 3{ x^{ 2 } }+5y=32 \\ Draw\, a\, parabola\, .\, symmetrical\, about\, negative\, x-axis \\ Also, \\ y=\left| { x-2 } \right| \\ \Rightarrow y=\left( \begin{array}{l} x-2,\, \, \, \, \, x\ge 2 \\ 2-x,\, \, \, \, \, x<2 \end{array} \right) \\ y=x-2,\, \, x\ge 2\, \, represents\, a\, line,\, cutting\, the\, x-axis\, at\, \, A\left( { 2,0 } \right) \, \, and\, the\, parabola\, C\left( { 3,1 } \right) \\ And,\, y=2-x,\, x<2\, \, represents\, a\, line,\, cutting\, the\, parabola\, at\, \left( { -2,4 } \right) \\ Shaded\, area\, AEYCA \\ =\int _{ -2 }^{ 2 }{ \left[ { \left( { \frac { { 32-32{ x^{ 2 } } } }{ 5 } } \right) -\left( { 2-x } \right) } \right] }dx+\int _{ 2 }^{ 3 }{ \left[ { \left( { \frac { { 32-32{ x^{ 2 } } } }{ 5 } } \right) -\left( { x-2 } \right) } \right] }dx \\ =\int _{ -2 }^{ 3 }{ \left( { \frac { { 32-32{ x^{ 2 } } } }{ 5 } } \right) dx- }\int _{ -2 }^{ 2 }{ \left( { 2-x } \right) dx+ }\int _{ 2 }^{ 3 }{ \left( { x-2 } \right) dx } \\ =\frac { 1 }{ 5 } \left[ { 32-x } \right] _{ -2 }^{ 3 }-\left[ { 2x-\frac { { { x^{ 2 } } } }{ 2 } } \right] _{ -2 }^{ 2 }-\left[ { \frac { { { x^{ 2 } } } }{ 2 } -3x } \right] _{ 2 }^{ 3 } \\ =\frac { 1 }{ 5 } \left[ { 32\times 3-27+64-8 } \right] -\left( { 4-2+4+2 } \right) -\left( { \frac { 9 }{ 2 } -6-2+4 } \right) \\ =\frac { 1 }{ 5 } \left[ { 96-35+64 } \right] -\left( 8 \right) -\left( { \frac { 9 }{ 2 } -4 } \right) \\ =\frac { { 125 } }{ 5 } -4-\frac { 9 }{ 2 } \\ =25-4-\frac { 9 }{ 2 } \\ =\frac { { 42-9 } }{ 2 } \\ =\frac { { 33 } }{ 2 } \, \, sq.units. \\ Hence,\, option\, C\, is\, the\, correct\, answer. \end{array}$
1233440_1074715_ans_e94c91a35c034f15bc10526e46ffcfce.PNG

1233440_1074715_ans_e94c91a35c034f15bc10526e46ffcfce.PNG

The area of the figure bounded by

$f\left(x\right)=\sin{x}, g\left(x\right)=\cos{x}$ in the first quadrant is:

0%
$2\left(\sqrt{2}-1\right)$ sq.unit
0%
$\sqrt{3}+1$ sq.unit
0%
$2\left(\sqrt{3}-1\right)$ .sq.unit
0%
none of these.

Explanation

We know that

$\sin{x}\ge \cos{x}$ for

$x\in\left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right]$
and

$\sin{x}\le \cos{x}$ for

$x\in\left[0,\dfrac{\pi}{4}\right]$

$\therefore$ the required area is

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

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

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

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

The area under the curve

$y=2x^3+4x^2$ between

$x=2,x=4$ is

0%
$192.6$
0%
$198.6$
0%
$88.3$
0%
$172.3$

Explanation

The area under the curve is given as

$\displaystyle \int _2^4 2x^3+4x^2 dx\\\displaystyle \int _2^4 2x^3dx+\int _2^4 4x^2dx\\\left.2\dfrac{x^4}{4}\right|_2^4+\left.4\dfrac{x^3}{3}\right|_2^4\\64\times 2-8+\dfrac{4^4}3-\dfrac{32}{3}\\120- \dfrac{224}{3}=192.6$

If the curves

$y=x^3+ax$ and

$y=bx^2+c$ pass through the point

$(-1, 0)$ and have common tangent line at this point, then the value of

$a+b$ is?

0%
$0$
0%
$-2$
0%
$-3$
0%
$-1$

Explanation

As the curve pass through the point

$P(-1,0)$

$0 = -1-a$

$\implies a = -1$

$0 = b+c$

Common tangent at this point

$\cfrac{dy}{dx} = 2bx$ and

$\cfrac{dy}{dx} = 3x^2+a = 3-1 = 2$

$2bx = 2$

$-2b = 2$

$b = -1$

$a+b = -2$

The area of the plane region bounded by the curves

$x + 2 y ^ { 2 }= 0 \text { and } x + 3 y ^ { 2 } = 1$

0%
$\dfrac { 4 } { 3 }$
0%
$\dfrac { 5 } { 3 }$
0%
$\dfrac { 2 } { 3 }$
0%
$\dfrac { 1 } { 3 }$

Explanation

$\begin{array}{l} 1-3{ y^{ 2 } }=-2{ y^{ 2 } }\Rightarrow { y^{ 2 } }=1\, \therefore y=\pm 1 \\ y=-1\, \, \Rightarrow x=-2 \\ y=1\, \, \, \, \Rightarrow x=-2 \\ The\, \, bounded\, \, region\, \, is\, as\, under \\ The\, \, desired\, \, area\, \, =2\int _{ 0 }^{ 1 }{ \left[ { \left( { 1-3{ y^{ 2 } } } \right) -\left( { -2{ y^{ 2 } } } \right) } \right] }dy \\ =22\int _{ 0 }^{ 1 }{ \left[ { \left( { 1-{ y^{ 2 } } } \right) } \right] }dy=2\left[ { y-\cfrac { { { y^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 } \\ =2\times \cfrac { 2 }{ 3 } =\cfrac { 4 }{ 3 } sq.\, units \end{array}$

Hence, this is the answer.

1228664_1322975_ans_854f2a7caa0844b895d39756148df21a.JPG

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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