MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 7

Find area bounded by curves

$\left \{ (x,y):y\geq x^{2}andy=\mid x\mid \right \}$

0%
$\dfrac {5}{3}$
0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{3}$
0%
$\dfrac {1}{9}$

Explanation

$y=\mid x\mid$

$=\left\{ x\quad ;\quad x\ge 0\\ -x\quad ;x<0 \right\}$

$P$ and

$Q$ are

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

$x=0,1$

$Q=1$

similarly

$p=-1$

$A=\int _{ 0 }^{ 1 }{ x } -{ x }^{ 2 }dx\\ A={ [\cfrac { { x }^{ 2 } }{ 2 } -\cfrac { { x }^{ 3 } }{ 3 } ] }_{ 0 }^{ 1 }\\ A=\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 3 } \\ A=\cfrac { 1 }{ 3 }$

The area enclosed between the curves

$y^2 = x$ and

$y=|x|$ is

0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{6}$
0%
$\dfrac {1}{8}$
0%
$\dfrac {1}{16}$

Find the area bounded by the curve

$y = sin\;x$ with x-axis between

$x = 0$ to x = 2

$\pi$ .

0%
4 sq. unit
0%
8 s q . unit
0%
4 $\pi 4$ sq. unit
0%
8 $\pi 4$ sq. unit

Explanation

Consider the problem

We have to draw the graph of

$y=sin\;x$

Therefore,

Required area

$=Area\; OABO+Area \;BCDB$

$\begin{array}{l} =\int _{ 0 }^{ \pi }{ \sin xdx+\left| { \int _{ \pi }^{ 2\pi }{ \sin xdx } } \right| } \\ ={ \left[ { -\cos x } \right] _{ 0 } }^{ \pi }+\left| { { { \left[ { -\cos x } \right] }_{ \pi } }^{ 2\pi } } \right| \\ =\left[ { -\cos \pi +\cos 0 } \right] +\left| { -\cos 2\pi +\cos \pi } \right| \\ =1+1+\left| { \left( { -1-1 } \right) } \right| \\ =2+\left| { -2 } \right| \\ =2+2 \\ =4units \end{array}$

1144271_1149713_ans_f7cb5b19637b4987bc1bd31de31cb3db.png

The area enclosed between the curve

$y = |x|^{3}$ and

$x = y^{3}$ is

0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{8}$
0%
$\dfrac {1}{16}$

The area between the curves y=

$x^{2}$ and

$y=\frac{2}{1+x^{2}}$ is-

0%
$\pi -\frac{1}{3}$
0%
$\pi -2$
0%
$\pi -\frac{2}{3}$
0%
$\pi +\frac{2}{3}$

Explanation

we have,

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

$y=\dfrac{2}{{{x}^{2}}+1}\,\,.......\,\,\left( 2 \right)$

Taking limit then,

By equation (1) and (2) to, and get,

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

Put x=1 and we get,

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

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

$1=1$

Now, taking’

$x=-1$ and we get,

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

$1=1$

Then, the limit of this function of $1\,\,to\,\,-1$ .

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

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

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

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

$=2\left[ 2\dfrac{\pi }{4}-2\times 0-\dfrac{1}{3} \right]$

$=\pi -\dfrac{2}{3}$

Hence, this is the answer.

If the area enclosed between

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

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

$\cfrac{1}{3}$ sq. units, then

$m,n$ can be roots of (where

$m,n$ are non zero real numbers)

0%
$2{x}^{2}+5x+1=0$
0%
$2{x}^{2}+3x-2=0$
0%
$2{x}^{2}+3x+2=0$
0%
$2{x}^{2}+5x-1=0$

Explanation

$y=mx^{2}$

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

$x=nm^{2}x^{1/3}$

$x = \left( \dfrac{1}{nm^{2}} \right)^{1/3}$

$y=m \dfrac{1}{(nm^{2})^{2/3}}$

$y= m\dfrac{1}{n^{2/3} m^{4/3}}$

$y=\dfrac{1}{(n^{2}m)^{1/3}}$

$(1/nm^{2})^{1/3}$

Area

$= \int_{0} \sqrt{\dfrac{x}{n}}- mx^{2}$

$=\left[ \right]^{(1/nm^{2})^{1/3}}$

$=\dfrac{2}{3\sqrt{n}} \left( \dfrac{1}{nm^{2}} \right)^{\dfrac{1}{3}. \dfrac{3}{2}}- \dfrac{m}{3} \left( \dfrac{1}{nm^{2}} \right)^{3 \times \dfrac{1}{3}}$

$=\dfrac{2}{3\sqrt{n}} \dfrac{1}{\sqrt{n}m}- \dfrac{m}{3} \dfrac{1}{nm^{2}}$

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

$\dfrac{1}{3}= \dfrac{1}{3nm}$

$nm=1$

Product of roots

$=1$

In the given figure, a square

$OABC$ has been inscribed in the quadrant

$OPBQ$ . If

$OA=20cm$ then the area of the shaded region is

$\left[take\pi=3.14\right]$

0%
$214cm^{2}$
0%
$228cm^{2}$
0%
$222cm^{2}$
0%
$242cm^{2}$

Explanation

Area of shaded region

$=Area\ of \left[OPBQ\right]-Area\ of \left[OABC\right]$

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

$\left[ \because OA=a=20cm\\ r=OB=OA\sqrt { 2 } =20\sqrt { 2 } cm \right]$

$=\dfrac{1}{4}\pi {\left(20\sqrt{2}\right)}^{2}-{20}^{2}$

$=200\pi-400=228\ {cm}^{2}$

The area bounded by

$y=x^2,x=y^2$ is

0%
$1$
0%
$\dfrac 16$
0%
$\dfrac 34$
0%
$None\ of\ these$

Explanation

$y=x^2\\y^2=x\implies y=\sqrt x$

The curves intersect at

$(0,0)$ and

$(1,1)$

Area between the curves is given by

$\displaystyle \int _0^1 \sqrt x-x^2dx\\\left.\dfrac 23 x^{3/2}+\dfrac{x^3}{3} \right|_0^1\\\dfrac 23+\dfrac13=1$

The area of the plane region bounded by the curve

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

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

0%
$-\dfrac{4}{3}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{2}{3}$
0%
None of these

Explanation

The given curves are

$x+2y^2 = 0$ and

$x+3y^2 = 1$ .

The first curve is

$x+2y^2 = 0$ which implies

$2y^2 = -x$ and hence

$y^2 = -x/2$ , which is a parabola.

Second curve

$x+3y^2 = 1$ which is again a parabola as

$y^2 = -1/3 (x-1)$ .

Hence we are required to find the area between the two parabolas.

On solving the equations of the two parabolas, we get the points of intersection as

$(-2, 1)$ and

$(-2, -1)$ .

Hence we set

$x = -2$ and

$y = 1, -1$ .

Hence the required area is

$\int(-2y^2 -1 + 3y^2)dy$ , where the integral runs from 0 to 1.

$\int(y^2 -1) dy$ , where the integral runs from 0 to 1.

$(1/3 -1)$

$\dfrac{2}{3}$ .

Hence the area of the region bounded by the curves is

$\dfrac{4}{3}$ .

The angle between the curves

$y=sin x$ and

$y=cos x$ is :

0%
$tan^{-1}(2\sqrt{2})$
0%
$tan^{-1}(3\sqrt{2})$
0%
$tan^{-1}(3\sqrt{3})$
0%
$tan^{-1}(5\sqrt{2})$

The area bounded by

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

$|x|+|y|=1$ is

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

The area bounded by the curve

$y=\cos x$ and

$y=\sin x$ between the ordinates

$x=0$ and

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

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

Explanation

$\displaystyle \int_0^{3\pi/2}|\sin x-\cos x|dx----(1)$

$\displaystyle \int_0^{\pi / 4}(\cos x-\sin x)dx +\displaystyle \int_{\pi /4}^{5\pi /4}(\sin x-\cos x)dx$

$\rightarrow \ \displaystyle \int_{5\pi /4}^{3\pi/2}(\cos x-\sin x)dx [\sin x+\cos x]_0^{\pi /4}$

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

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

$4\sqrt {2}-2$

1353696_1204390_ans_5e40305cc335490e861a2734d42bd1f2.png

The area bounded by

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

$a > 0$ and

$b > 0$ is

0%
$\dfrac { 1 } { 2 } a b$
0%
$ab$
0%
$2ab$
0%
$4ab$

Explanation

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

$a > o,b>o$

We should consider only 1st quadrant

are of

$\triangle =\dfrac{1}{2}\times a\times b$

$=\dfrac{1}{2}ab$ .

1330751_1209606_ans_3e4bc054d9f44fd3b263c05d4c12742f.png

Area bounded by the curve

$y = \sin ^ { - 1 } x , y - a x i s$ and

$y = \cos ^ { - 1 } x$ is equal to

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

The area of the figure bounded by the parabola

$x = - 2 y ^ { 2 } \text { and } x = 1 - 3 y ^ { 2 }$ is ?

0%
$1 / 6$
0%
$2/3$
0%
$3/2$
0%
$4/3$

Explanation

$x = - 2{y^2} \to \left( i \right)\,\,\& \,\,x = 1 - 3{y^2} \to \left( {ii} \right)$

Equating equation (i) & (ii)

$\begin{array}{l} -2{ y^{ 2 } }=1-3{ y^{ 2 } } \\ -2{ y^{ 2 } }+3{ y^{ 2 } }=1 \\ { y^{ 2 } }=1 \\ y\pm 1 \end{array}$

So, area of region bounded by

$\begin{array}{l} A\Rightarrow 2\int _{ 0 }^{ 1 }{ \left[ { \left( { 1-3{ y^{ 2 } } } \right) -\left( { -2{ y^{ 2 } } } \right) } \right] dy } \\ =2\int _{ 0 }^{ 1 }{ \left( { 1-{ y^{ 2 } } } \right) dy } \\ =2\left[ { y-\frac { { { y^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 }\Rightarrow 2\left[ { \left( { 1-\frac { 1 }{ 3 } } \right) -\left( { 0-\frac { 0 }{ 3 } } \right) } \right] \\ =2\left[ { \frac { 1 }{ 1 } -\frac { 1 }{ 3 } } \right] \Rightarrow 2\left[ { \frac { 2 }{ 3 } } \right] =\frac { 4 }{ 3 } \, \, sq.\, \, units \end{array}$

1213656_1213322_ans_c8f70f105c6b41f6b1d5bee087fcfb69.PNG

Area bounded by

$y=x^2$ and line

$y=x$

0%
2/3
0%
1/6
0%
1/3
0%
1/4

The area bounded by the curve

$y=(x+1)^2,y=(x-1)^2$ and the line

$y=0$ is

0%
$\dfrac{1}{6}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{3}$

Explanation

R.E.F image

$\rightarrow y = (x+1)^{2}$ is obtained

by shifting origin to (-1,0)

$x^{2} = y,$ for

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

Similarly (1,0)

As graph is symmetric

about y-axis, area A

would be,

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

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

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

Area common to the cutve

$y=\sqrt {9-x^{2}}$ and

$x^{2}-y^{2}=6x$ is:

0%
$\dfrac {\pi +\sqrt {3}}{4}$
0%
$\dfrac {\pi -\sqrt {3}}{4}$
0%
$3\left(\pi +\dfrac {\sqrt {3}}{4}\right)$
0%
$None\ of\ these$

The area enclosed by the curves

$y=x^{2},y=x^{3},x=0$ and

$x=p$ , where

$p > 1$ , is

$\dfrac{1}{6}$ . then

$p$ equals

0%
$8/3$
0%
$16/3$
0%
$4/3$
0%
$2$

The area (in sq units) of the region

$\{ (x,y):{ y }^{ 2 }\ge 2x$ and

${ x }^{ 2 }+{ y }^{ 2 }\le 4x ,\chi \ge 0, Y\ge 0\}$

0%
$\pi -\frac { 4 }{ 3 }$
0%
$\pi -\frac { 8 }{ 3 }$
0%
$\pi -\frac { 4\sqrt { 2 } }{ 3 }$
0%
$-\frac { 2\sqrt { 2 } }{ 3 }$

Let

$f\left( x \right)$ be a non-negative continuous function such that the area bounded by the curve

$y= f\left( x \right)$ , x-axis and the ordinates

$x=\cfrac { \pi }{ 4 }$ ,

$x=\beta >\cfrac { \pi }{ 4 }$ is

$\left( \beta \sin { \beta } +\cfrac { \pi }{ 4 } \cos { \beta } +\sqrt { 2 } \beta -\cfrac { \pi }{ 2 } \right)$ . Then

$f\left( \cfrac { \pi }{ 2 } \right)$ is

0%
$\left( 1-\dfrac { \pi }{ 4 } -\sqrt { 2 } \right)$
0%
$\left( 1-\dfrac { \pi }{ 4 } +\sqrt { 2 } \right)$
0%
$\left( \cfrac { \pi }{ 4 } +\sqrt { 2 } -1 \right)$
0%
$\left( \cfrac { \pi }{ 4 } -\sqrt { 2 } +1 \right)$

Explanation

Given that,

$\displaystyle\int^{\beta}_{\pi/4}f(x)dx=\beta \sin\beta +\dfrac{\pi}{4}\cos\beta +\sqrt{2}\beta -\dfrac{\pi}{2}$

Differentiating w.r.t x and we get

$f(\beta)=\beta \cos\beta +\sin\beta -\dfrac{\pi}{4}\sin\beta +\sqrt{2}-0$

$f\left(\dfrac{\pi}{2}\right)=\left(1-\dfrac{\pi}{4}\right)\sin\dfrac{\pi}{2}+\sqrt{2}$

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

$=1-\dfrac{\pi}{4}+\sqrt{2}$

Hence, this is the answer.

1196176_1242263_ans_c6c3c5a1067a4dcda008f7510650fc50.png

The area bounded by the curve

$xy^{2}=1$ and the lines

$x=1$ ,

$x=2$ is

0%
$4\left( {\sqrt 2 - 1} \right)$
0%
$4\left( {\sqrt 2 + 1} \right)$
0%
$2\left( {\sqrt 2 - 1} \right)$
0%
$2\left( {\sqrt 2 + 1} \right)$

The area bounded by the curves

$y=\sqrt{-x}$ and

$x=-\sqrt{-y}$ , where

$x,y\le0$ , is equal to

0%
$\dfrac{2}{3}sq. unit$
0%
$\dfrac{1}{3}sq. unit$
0%
$\dfrac{1}{2}sq. unit$
0%
$Cannot\ be\ determined$

Explanation

Given the curve

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

$x=-\sqrt{-y}$

Now, the area bound by the both curve

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

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

change

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

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

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

$=\dfrac{1}{3}\ sq.\ unit$

Hence, this is the answer.

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

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

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

The area of the region 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%
$6$ sq. units

The area enclosed between the curves

${y^2} = x$ and

$y = |x|\;is$ -

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

Explanation

POint of intersection P

$y^2=x\\x^2=x\\x=1\quad and\quad 0$

Area enclosed

$=\int_0^1{\sqrt x-x}\\ \Rightarrow\cfrac{2}{3}x^{2/3}-\cfrac{x^2}{2}]^1_0\\ \Rightarrow\cfrac{2}{3}-\cfrac{1}{2}\\ \Rightarrow\cfrac{4-3}{6}=\cfrac{1}{6}$

1210430_1244484_ans_8007b410710247088874f0c08edb0a18.png

Let

$T$ be the triangle with vertices

$\left (0,0\right), \left (0,{c}^{2}\right)\ and \left (c,{c}^{2}\right)$ and let

$R$ be the region between

$y=cx$ and

$y={x}^{2}\ where c>0$ then

0%
Area $\left( R \right) =\dfrac { { c }^{ 3 } }{ 6 }$
0%
Area of $R=\dfrac {e^{x}}3{3}$
0%
$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=3$
0%
$\lim_{c \rightarrow 0}\dfrac {Area(T)}{Area(R)}=\dfrac {3}{2}$

The area of the figure bounded by the curves

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

$x - y - 1 = 0$ is

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

The area common to the parabola

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

$\quad y={ x }^{ 2 }+4$

0%
$\dfrac { 2 }{ 3 } sq.\quad units\quad$
0%
$\dfrac { 3 }{ 2 } sq.\quad units\quad$
0%
$\dfrac { 32 }{ 3 } sq.\quad units\quad$
0%
none of these.

Explanation

Find the P.O.I of the parabolas equate the equations

$y = 2x^2$ and

$y = x^2 + y$ we get

$2x^2 = x^2 + 4$

$x^2 = 4$

$x = \pm 2$

$y = 8$

$\therefore$ POI are

$A(-2, 8)$ &

$C(2, 8)$

$\therefore$ the required are ABCD

$A = \displaystyle \int_{-2}^2 (y_1 - y_2) dx$ (where

$y_1 = x^2 + 4 \, \& \, y_2 = 2x^2$ )

$= \displaystyle \int_{-2}^2 (x^2 + 4 - 2x^2) dx = \int_{-2}^2 (4 - x^2) dx = \left(4x - \dfrac{x^3}{3} \right)^2_{-2}$

$\left[4(2) - \dfrac{(2)^3}{3} \right] - \left[4(-2) - \dfrac{(-2)^3}{3} \right] = \left[8 - \dfrac{8}{3} \right] - \left[8 + \dfrac{8}{3} \right]$

$= \dfrac{32}{3}$ sq. units

The are boundede by the curve

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

$y^{2}=4x-3$ is

$k$ , them the value of

$9k$ is

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

If area bounded by to curves

$y^2 = 4ax$ and y=mx is

$\dfrac{a^2}{3}$ , then the value of m is

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

Explanation

Given,

$y^2=4ax,y=mx$ curves intersect at

$\left ( 4\dfrac{a}{m^2},\dfrac{a}{m} \right )$

Area enclosed is,

$A=\displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx$

$\therefore \displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx=\dfrac{a^2}{3}$ .... given

$\left[2\sqrt a \ \dfrac{x^{\tfrac 32 }}{\tfrac 32 }- m \dfrac {x^2} 2 \right]^{4\frac{a}{m^2}}_0=\dfrac{a^2}{3}$

$\dfrac {4\sqrt a} 3 {({4\frac{a}{m^2}})^{\tfrac 3 2}}-\dfrac m 2 {(4\frac{a}{m^2})}^2-0-0=\dfrac{a^2}{3}$

$\dfrac{32 a ^2} {3m^3}-\dfrac{8 a ^2} {m^3}=\dfrac{a^2}{3}$

$\dfrac{8}{3}\dfrac{a^2}{m^3}=\dfrac{a^2}{3}$

$m^3=8$

$\therefore m=2$

1490097_1252277_ans_4d83f8396de6430c9fd8d17b40c99fc3.png

The area bounded by curves

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

$y = |x-2|$ is

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

The area of the plane region bounded by the curves

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

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

0%
$\cfrac { 5 }{ 3 } sq.unit$
0%
$\cfrac { 1 }{ 3 } sq.unit$
0%
$\cfrac { 2 }{ 3 } sq.unit$
0%
$\cfrac { 4 }{ 3 } sq.unit$

Explanation

The given curves are

$x+2y^2=0$ &

$x+3y^2=1$

$1$ st curve

$\Rightarrow x^2+2y^2=0$

$\Rightarrow x^2=-2y^2$

$\therefore y^2=-x/2$ (parabola)

$2nd$ curve

$\Rightarrow x+3y^2=1$

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

$y^2=(1-x)/3$ (parabola)

So, area between parabolas is required

Solving

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

$\Rightarrow -3x=2-2x$

$\Rightarrow -3x+2x=2$

$\Rightarrow -x=2$

$=x=-2$

$\therefore y=1, -1$

$(-2, -1)$ &

$(-2, 1)$ are points of intersection.

Required area

$=2|\displaystyle\int^1_0(-2y^2-1+3y^2)dy|$

$=2|\displaystyle\int ^1_0(y^2-1)dy|=2\left[y-\dfrac{y^3}{3}\right]^1_0$

$=2|\left(\dfrac{1}{3}-1\right)|=2\times 2/3=4/3$ square units.

1177766_1261095_ans_ff0a3acea7eb4efe95d87816d34a25f5.jpg

The area bounded by the curves

$y=x(x-3)^{2}$ and

$y=x$ is (in sq.units) is

0%
$28$
0%
$32$
0%
$4$
0%
$8$

The area enclosed by the curves

$xy^{2}=a^{2}(a-x)$ and

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

0%
$(\pi-2)a^{2}\ sq.units$
0%
$(4-\pi)a^{2}\ sq.units$
0%
$(\pi a^{2}/3\ sq.units$
0%
$\dfrac{\pi+a^{2}}{4}\ sq.units$

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

$\left\{(x,y):y^{2}\le 2x\ and\ y\ge 4x-1\right\}$ is

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

The area of the region bounded by the curves

$y = \sin x$ and

$y = \cos x ,$ and lying between the lines

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

$x = \dfrac { 5 \pi } { 4 } ,$ is

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

The area bounded by the curves

$y = \log _ { e } x$ and

$y = \left( \log _ { e } x \right) ^ { 2 }$ is

0%
$3 - e$
0%
$e-3$
0%
$\frac { 1 } { 2 } ( 3 - e )$
0%
$\frac { 1 } { 2 } ( e - 3 )$

Explanation

$\begin{array}{l} y=\left( { { { \log }_{ e } }x } \right) \, \, \, \, \, \, \, ----\left( 1 \right) \\ y={ \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, \, \, \, -----\left( 2 \right) \\ For\, the\, po{ { int } }\, of\, { { int } }er\sec tion \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, =\left( { { { \log }_{ e } }x } \right) \\ { \left( { { { \log }_{ e } }x } \right) ^{ 2 } }\, \, -\left( { { { \log }_{ e } }x } \right) =0 \\ \left( { { { \log }_{ e } }x } \right) \left[ { \left( { { { \log }_{ e } }x } \right) \, -1 } \right] \, \\ \left( { { { \log }_{ e } }x } \right) =0\, \, and\, \left( { { { \log }_{ e } }x } \right) =1 \\ x=1\, \, and\, \, x=e \\ area\, bounded \\ =\left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ consider \\ I=\int { \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } \\ { \log _{ e } }x=t \\ x={ e^{ t } } \\ dx={ e^{ t.dt } } \\ Now, \\ I=\int { { e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} dt } \\ ={ e^{ t } }\left\{ { { t^{ 2 } }-t } \right\} -\left\{ { \left( { 2t-1 } \right) { e^{ t } }-2{ e^{ t } } } \right\} \, \, \, \, \left[ { by\, { { int } }ergration\, by\, parts } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-t-2t+1+2 } \right] \\ ={ e^{ t } }\left[ { { t^{ 2 } }-3t+3 } \right] \\ =x\left[ { \log { { \left( x \right) }^{ 2 } } -3\log \left( x \right) +3 } \right] \\ Now, \\ \left| { \int _{ 1 }^{ e }{ \left\{ { { { \left( { { { \log }_{ e } }x } \right) }^{ 2 } }\, -\left( { { { \log }_{ e } }x } \right) } \right\} dx } } \right| \, \, \\ =\left| { \left[ { x\left\{ { loh{ { \left( x \right) }^{ 2 } } } \right\} -3\log \left( x \right) +3 } \right] _{ 1 }^{ e } } \right| \\ =\left| { e\left( { 1-3+3 } \right) -1\left( { 0-0+3 } \right) } \right| \\ =\left| { e-3 } \right| \\ =3-e \end{array}$

Hence, the option

$B$ is the correct answer.

The area bounded by the curves

$y=x(x-3)^{2}$ and

$y=x$ is (in

$sq.units$ ) is

0%
$28$
0%
$32$
0%
$4$
0%
$8$

Area bounded by

$y=2x^{2}$ and

$y=\dfrac{4}{(1+x^{2})}$ will be (in sq units)

0%
$(2\pi+4/3)$
0%
$(2\pi-4/3)$
0%
$4/3-2\tan^{-1}2+\pi/2$
0%
$4/3-8\tan^{-1}2+2\pi$

Explanation

$\begin{array}{l} Po{ { int } }\, of\, intersection\, : \\ 2{ x^{ 2 } }=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ \Rightarrow { x^{ 4 } }+{ x^{ 2 } }-2=0 \\ \Rightarrow { x^{ 4 } }+2{ x^{ 2 } }-{ x^{ 2 } }-2=0 \\ \Rightarrow \left( { { x^{ 2 } }-1 } \right) \left( { { x^{ 2 } }+2 } \right) =0 \\ Then \\ Area\, bounded\, by\, y=2{ x^{ 2 } }\, and\, y=\dfrac { 4 }{ { 1+{ x^{ 2 } } } } \\ 2\int _{ 0 }^{ 1 }{ \left( { \dfrac { 4 }{ { 1+{ x^{ 2 } } } } -2{ x^{ 2 } } } \right) }dx=2\left[ { 4{ { \tan }^{ -1 } }x-\dfrac { { 2{ x^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 } \\ =2\left[ { 4\times \dfrac { \pi }{ 4 } -\dfrac { 2 }{ 3 } } \right] \\ =2\left[ { \pi -\dfrac { 2 }{ 3 } } \right] =2\pi -\dfrac { 4 }{ 3 } \\ Hence,\, option\, \, \, B\, \, is\, the\, correct\, answer. \end{array}$
1216600_1308671_ans_cc248d4277894ce1ba61262826dac1b3.PNG

1216600_1308671_ans_cc248d4277894ce1ba61262826dac1b3.PNG

The area enclosed between the curve

$y=\log_{e}\left(x+e\right)$ and the coordinate axes is

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

Explanation

$A=\int_{1-e}^{0} ln(x+e)dx$

$=\int_{1-e}^{0} x^{0} ln (x+e)dx$

$(x ln(x+e))-\int_{1-e}^{0}\frac{x}{x+e}dx$

$-\int_{1-e}^{0}\frac{x+e-e}{x+e}dx$

$=\int_{1-e}^{0}(\frac{e}{x+e})dx$

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

1204240_1296377_ans_331eb95141d54c64b9d4aaa84e0594ed.jpg

If a curve

$y = a \sqrt { x } +$ bx passes through the point

$( 1,2 )$ and the area bounded by the curve, line

$x = 4$ and

$x$ axis is

$8$ square units, then

0%
$a = 3 , b = - 1$
0%
$a = 3 , b = 1$
0%
$a = - 3 , b = 1$
0%
$a = - 3 , b = - 1$

The area bounded by the circle

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

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

$y=x$ in first quadrant is

$\dfrac{2}{3}+k\pi$ , where

$k=$

0%
$\dfrac{5}{7}$
0%
$2$
0%
$\dfrac{3}{5}$
0%
$3$

The area of the region bounded by the curves

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

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

The region in the

$xy$ - plane is bounded by curve

$y=\sqrt {(25-x^2)}$ and the line

$y=0$ . If the point

$(a,a+1)$ lies in the interior of the region, then

0%
$a \in \left( { - 4,3} \right)$
0%
$a \in (- \infty, -1) \cup (3, \infty)$
0%
$a \in (-1,3)$
0%
None of these

The area of the region formed by

${ x }^{ 2 }+{ y }^{ 2 }-6x-4y+12\le 0,y\le x\quad and\quad x\quad \le \quad \dfrac { 5 }{ 2 } is$

0%
$\frac { \pi }{ 6 } -\frac { \sqrt { 3}+1 }{ 8 }$
0%
$\frac { \pi }{ 6 } +\frac { \sqrt { 3}+1 }{ 8 }$
0%
$\frac { \pi }{ 6 } -\frac { \sqrt { 3}-1 }{ 8 }$
0%
none of these

Area of the region bounded by

$x^{2}+y^{2}-6y\leq 0$ and

$3y\leq x^{2}$ is

0%
$\frac{9\pi }{2}-12$
0%
$\frac{9\pi }{4}-6$
0%
$9\pi$ -24
0%
$\frac{9\pi }{2}+6$

The area enclosed by the curves y = cosx - sin x and y = [socx - sin x] and between x = 0 and

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

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

The area (in sq. units) in the first quadrant bounded by the parabola,

$y = x^2 + 1$ , the tangent to it at the point

$(2, 5)$ and the coordinate axes is:-

0%
$\dfrac{14}{3}$
0%
$\dfrac{187}{24}$
0%
$\dfrac{37}{24}$
0%
$\dfrac{8}{3}$

The area bounded by the parabola

${{\text{y}}^{\text{2}}}{\text{ = 4}}\;{\text{ax}}\;{\text{and}}\;{{\text{x}}^{\text{2}}}\;{\text{ = }}\;{\text{4ay}}\;$ is

0%
$\dfrac{{8{a^2}}}{3}$
0%
$\dfrac{{16{a^2}}}{3}$
0%
$\dfrac{{32{a^2}}}{3}$
0%
$\dfrac{{64{a^2}}}{3}$

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 7 - 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 7 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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