MCQ Questions for Class 12 Commerce Maths Application Of Integrals Quiz 8

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 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 curve

$y={ e }^{ x }$ and the lines y = |x - 1|, x = 2 is given by :

0%
${ e }^{ 2 }+1$
0%
4 ${ e }^{ 2}-1$
0%
${ e }^{ 2 }-2$
0%
$e - 2$

Explanation

Given,

$y=e^x,y=|x-1|,x=2$

$\rightarrow e^x=|x-1|$

$\rightarrow e^x-(1-x)=0$

intervals:

$|x-1|=0\Rightarrow x=1$

$x=2$

Area is given by,

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

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

$=\left [ e^x+\dfrac{x^2}{2}-x \right ]_0^1+\left [ e^x+\dfrac{x^2}{2}-x \right ]_1^2$

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

$=e^2-2$

Area bounded by the curve

$y^2(2a-x)=x^3$ and the line

$x=2a$ , is

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

Explanation

Given,

$y^2(2a-x)=x^3$

$y=\sqrt{\dfrac{x^3}{2a-x}}$

let,

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

$dx=4a\sin \theta \cos \theta d\theta$

$\Rightarrow A =\int_{0}^{2a}\sqrt{\dfrac{x^3}{2a-x}}dx$

$A=\int_{0}^{\tfrac{\pi }{2}}\sqrt{\dfrac{8a^3\sin ^6\theta }{2a\cos ^2\theta }}4a\sin \theta \cos \theta d\theta$

$=8a^2\int_{0}^{\tfrac{\pi }{2}}\sin ^4\theta d\theta$

$=8a^2\int_{0}^{\tfrac{\pi }{2}}\sin ^2\theta (1-\cos ^2\theta )$

$=8a^2\int_{0}^{\tfrac{\pi }{2}}\left ( \dfrac{1-\cos 2\theta }{2} \right )-\dfrac{1}{4}\sin ^22\theta d\theta$

$=8a^2\left [ \dfrac{\pi }{4}-0-\dfrac{1}{4}\left ( \dfrac{\pi }{4}-0 \right ) \right ]$

$=8a^2\left [\dfrac{\pi }{4}-\dfrac{\pi }{16} \right ]=\dfrac{3}{2}\pi a^2$

1460214_1411590_ans_3543f3bca2ac4d61a8d7c4405564e4d4.PNG

Area of the region bounded by

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

$y=-\cos^{-1}{\left| \cos{x}\right|}$ in the interval

$[0,2\pi]$ is equal to

0%
$\cfrac{\pi^2}{2}$
0%
$\pi^2$
0%
$\cfrac{\pi^2}{4}$
0%
$2\pi^2$

The area bounded by the circles

$x^{2} + y^{2} = r^{2}, r = 1, 2$ and the rays given by

$2x^{2} - 3xy - 2y^{2} = 0, y > 0$ is

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

The area bounding by

$y = 2 - |2 - x|$ and y =

$\frac { 3 }{ |x| }$ is :

0%
$\frac { 4+3\ell n3 }{ 2 }$
0%
$\frac { 4-3\ell n3 }{ 2 }$
0%
$\frac { 3 }{ 2 } +\ell n3$
0%
$\frac { 1 }{ 2 } +\ell n3$

The area enclosed between the curves

$y=ax^2$ and

$x=ay^2(a>0)$ is

$1$ sq.unit, then the value of

$a$ is

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

The area inside the parabola

$5x^2-y=0$ but outside the parabola

$2x^2-y+9=0$ , is

0%
$12\sqrt 3$
0%
$6\sqrt 3$
0%
$8\sqrt 3$
0%
$4\sqrt 3$

Explanation

From given, we have,

$5x^2-y=0$

$2x^2-y+9=0$

we have boundaries,

$5x^2=2x^2+9$

$3x^2=9$

$x=\pm \sqrt 3$

$A=\int _{-\sqrt{3}}^{\sqrt{3}}5x^2-\left(2x^2+9\:\right)$

$\int _{-\sqrt{3}}^{\sqrt{3}}5x^2-\left(2x^2+9\:\right)$

$=\int _{-\sqrt{3}}^{\sqrt{3}}3x^2dx-\int _{-\sqrt{3}}^{\sqrt{3}}9dx$

$=6\sqrt{3}-18\sqrt{3}$

$=-12\sqrt{3}$

$\therefore Area =|-12\sqrt{3}|=12\sqrt{3}$

Area bounded by the curve

$y= sin^{-1}x, y-axis$ and

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

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

Explanation

Given,

$y=\sin ^{-1}x,y=\cos ^{-1}x,y-axis$

$\Rightarrow \sin y=x, \cos y=x$

$\Rightarrow \sin y=\cos y \Rightarrow \tan y=1$

$\therefore y=\tan ^{-1}1=\dfrac{\pi }{4}$

$Area=\int _0^{\frac{\pi }{4}}\cos \left(y\right)-\sin \left(y\right)dy$

$=\int _0^{\frac{\pi }{4}}\cos \left(y\right)dy-\int _0^{\frac{\pi }{4}}\sin \left(y\right)dy$

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

$=\sqrt{2}-1$

The area of the figure bounded by the curves

$y=\ln nx$ &

${(\ln nx)}^{2}$ is

0%
$e+1$
0%
$e-1$
0%
$3-e$
0%
$1$

The area of the region bounded by x = 0, y = 0, x = 2, y = 2,

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

$y\ge \ell n$ x, is

0%
6 - 4 $\ell n$ 2
0%
4 $\ell n$ 2 - 2
0%
2 $\ell n$ 2 - 4
0%
6 - 2 $\ell n$ 2

The area is bounded by

$x+x_1, y=y_1$ and

$y=-(x+1)^2$ . Where

$x_1, y_1$ are the values of

$x, y$ satisfying the equation

$sin^{-1} x +sin^{-1} y = -\pi$ will be (nearer to origin)

0%
$1/3$
0%
$3/2$
0%
$1$
0%
$2/3$

Explanation

( Wrong question,

$x=x$ , instead of

$x+x_{1}$ )

Given that:

$x =x_{1} \\$

$y =y_{1} \\$

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

$\text { & } x_{1}, y_{1} \text { satisfy } \sin ^{-1} x+\sin ^{-1} y=z$

We know:

$-\frac{z}{2} \leq \sin ^{-1} x \leq \frac{z}{2}$

$\frac{-z}{2} \leq \sin ^{-1} y \leq \frac{z}{2}$

$\therefore \quad-z \leq \sin ^{-1} x+\sin ^{-1} y \leq z \\$

$\text { So, for } \\$

$\qquad \sin ^{-1} x+\sin ^{-1} y=-z \\$

$\Rightarrow \sin ^{-1} x=-\frac{\pi}{2} \text { & } \sin ^{-1} y=-z / 2 \\$

$x =-1 \text { and } y=-1 \\$

$\text { area required } =1-|\int_{1}^{0}-(x+1)^{2} d x \mid \\$

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

$=1- |\frac{-\left.\left(x^{3}+x+x^{2}\right)\right|_{-1} ^{0}}{3}\\$

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

$=1-\frac{1}{3} \\$

$=2 / 3 sq \text { . Units }$

The area bounded by curves

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

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

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

Explanation

First we form a table for different values of

$x$ and

$y$ for both equations as shown .

Here length of

$AC=2$ units and distance(height) between vertices

$D$ and line

$AC=1$ units and between vertices

$B$ and line

$AC=1$ unit

$\because$ Area of triangle

$=\dfrac{1}{2}\times b\times h$

$\therefore\,$ Area of

$ABCD=2\times\,Area\,of\,\triangle{ABC}$

$=2\times\dfrac{1}{2}\times 2\times 1=2$ sq.units.

1473040_1441059_ans_7c3fa67759614487804f23883b3cc814.png

The area of the triangle formed by the lines joining the vertex of the parabola

$x^2 = 12y$ to the ends of its latus rectum is-

0%
$16$ sq. units
0%
$12$ sq. units
0%
$18$ sq. units
0%
$24$ sq. units

Explanation

$\textbf{Step 1: Make an appropriate diagram}$

$\text{Above figure shows a parabola of the form }x^2=4ay$

$\text{Having vertex at } V(0, 0)$

$\text{Focus at } S(0, a)$

$\text{Latus rectum is a line } AB \text{ passing through focus } S \text{ and perpendicular to the axis of the parabola }$

$\text{Thus coordinates of end points of latus rectum is } A(-2a,a) \text{ and } B(2a,a)$

$\textbf{Step 2: Find coordinates of end points of the triangle for given parabola}$

$\text{Given parabola is }x^2=12y$

$\Rightarrow 4a=12$

$\therefore a=3$

Therefore the coordinates of triangle are,

$A(-6,3),V(0,0),B(6,3)$

Area of

$\triangle VAB=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$=\dfrac{1}{2}|0(3-3)+(-6)(3-0)+6(0-3)|$

$=\dfrac{1}{2}|-36|$

$=18sq.units$ \

$\textbf{Hence, Option 'C' is correct}$

The area enclosed between the curves

$y=log_e(x+e), x=log_e\left(\dfrac{1}{y}\right)$ and the x-axis is?

0%
$2e$
0%
$e$
0%
$4e$
0%
None of these

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

$y=x^{2}-1$ , the tangent at the point (2,3) to it and the y-axis is:

0%
$\frac{14}{3}$
0%
$\frac{56}{3}$
0%
$\frac{8}{3}$
0%
$\frac{32}{3}$

Explanation

Equation of tangent at

$(2,3)$ on

$y=x^2-1$ , is

$y=(4x-5).......(i)$

$\therefore$ Required shaded area

$= ar(\triangle ABC)- \int\limits_{-1}^3 \sqrt{y+1}dy$

$= \dfrac12. (8).(2)- \dfrac23 ((y+1)^{3/2})_{-1}^3$

$= 8- \dfrac{16}{3}= \dfrac83$
1934900_1519662_ans_16da7f0c982a4ee6bd272627c3628935.png

The area bounded by the curve

$y \le x^2 + 3x , 0 \le y \le 4, \, 0 \le x \le 3$ , is

0%
$\dfrac{59}{6}$
0%
$\dfrac{57}{4}$
0%
$\dfrac{59}{3}$
0%
$\dfrac{57}{6}$

Explanation

$y = x^2 + 3x = 4 \Rightarrow x^2 + 3x - 4 = 0$

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

$x = 1$
area

$= \displaystyle \int_0^1 (x^2 + 3x ) dx + 2(4) = \dfrac{1}{3} + \dfrac{3}{2} + 8 = \dfrac{11}{6} + 8 = \dfrac{59}{6}$
1242405_1611195_ans_f65d501894ac4c6494b556c2dd4896c3.png

The area of the region bounded by the curve

$y=\phi(x),y=0$ and

$x=10$ is

0%
$\dfrac {81}{4}$
0%
$\dfrac {79}{4}$
0%
$\dfrac {73}{4}$
0%
$19$

The area bounded by the curve

$y=x$

$X-$ axis and the lines

$x=-1$ and

$x=1$ is

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

If the area (in sq. units) of the region

$\{(x, y): y^2\le 4x, x+y\le 1, x\geq 0, y\ge 0 \}$ is

$a\sqrt{2}+b$ , then

$a-b$ is equal to?

0%
$\dfrac{8}{3}$
0%
$\dfrac{10}{3}$
0%
$6$
0%
$-\dfrac{2}{3}$

Explanation

$\{(x, y): y^2\leq 4x, x+y\leq 1, x\geq 0, y\geq 0\}$

$A\displaystyle\int^{3-2\sqrt{2}}_02\sqrt{x}dx+\dfrac{1}{2}(1-(3-2\sqrt{2}))(1-(3-2\sqrt{2}))$

$=\dfrac{2[x^{3/2}]_0^{3-2\sqrt{2}}}{3/2}+\dfrac{1}{2}(2\sqrt{2}-2)(2\sqrt{2}-2)$

$=\dfrac{8\sqrt{2}}{3}+\left(-\dfrac{10}{3}\right)$

$a=\dfrac{8}{3}$ ,

$b=-\dfrac{10}{3}$

$a-b=6$ .

1250606_1615212_ans_0544942d7a0743bba3d6c4382879139f.png

Let

$f(x, y)=\{(x, y): y^2 \leq 4x, 0\leq x\leq \lambda\}$ and

$s(\lambda)$ is area such that

$\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5}$ . Find the value of

$\lambda$ .

0%
$4\left(\dfrac{4}{25}\right)^{1/3}$
0%
$4\left(\dfrac{2}{25}\right)^{1/3}$
0%
$2\left(\dfrac{4}{25}\right)^{1/3}$
0%
$2\left(\dfrac{2}{25}\right)^{1/3}$

Explanation

$y^2=4x$

$S(I)=\left.2\displaystyle\int^{\lambda}_{0}2\sqrt{x}dx=\dfrac{4x^{3/2}}{3/2}\right]^{\lambda}_{0}=\dfrac{8}{3}\lambda^{3/2}$

$\dfrac{S(\lambda)}{S(4)}=\dfrac{2}{5}$

$\Rightarrow \dfrac{\lambda^{3/2}}{4^{3/2}}=\dfrac{2}{5}$

$\lambda =4\left(\dfrac{2}{5}\right)^{2/3}=4\left(\dfrac{4}{25}\right)^{1/3}$ .
1245676_1611399_ans_f1c829415761464d9e489eca2e72568d.PNG

Region formed by

$|x-y| \le 2$ and

$|x + y| \le 2$ is

0%
Rhombus of side is $2$
0%
Square of area is $6$
0%
Rhombus of area is $8\sqrt{2}$
0%
Square of side is $2\sqrt{2}$

Explanation

$\dfrac{\text{ABCD is a square}}{Area = 4 \times \dfrac{1}{2}\times 2 \times 2 = 8}$
side

$= 2 \sqrt{2}$
1246013_1612421_ans_edd37e29cdb14b9b951732e1528cc200.png

The region represented by

$|x - y| \le 2$ and

$|x + y| \le 2$ is bounded by a:

0%
Square of side length $2\sqrt{2}$ units
0%
Rhombus of side length $2$ units
0%
Square of area $16 \,sq$ units
0%
Rhombus of area $8\sqrt{2} sq.$ units

Explanation

$|x - y| \le 2$ and

$|x + y| \le 2$

$\Rightarrow x-y\le 2$ or

$x-y \ge 2$ and

$x+y \le 2$ and

$x+y \ge-2$

After plotting them on the graph the common region is

Square whose side is

$2\sqrt{2}$

1250475_1614697_ans_b375140c0c8b41e892fbc924f5e21f61.png

Let

$S(\alpha) = \{ (x, y) : y^2 \le x, 0 \le x \le \alpha\}$ and

$A(\alpha)$ is area of the region

$S(\alpha)$ . If for a

$\lambda , 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$ , then

$\lambda$ equals

0%
$2\left(\dfrac{4}{25}\right)^{\frac{1}{3}}$
0%
$4\left(\dfrac{4}{25}\right)^{\frac{1}{3}}$
0%
$2\left(\dfrac{2}{5}\right)^{\frac{1}{3}}$
0%
$4\left(\dfrac{2}{5}\right)^{\frac{1}{3}}$

Explanation

$\displaystyle S(\alpha) =\{(x, y) : y^2 \le x, 0 \le x \le \alpha\}$

$\displaystyle A(\alpha)= 2 \int^{\alpha}_{0} \sqrt{x}dx = 2\alpha^{\frac{3}{2}}$

$A(4) = 2 \times 4^{3/2} = 16$

$A(\lambda) = 2\times \lambda^{3/2}$

$\dfrac{A(\lambda)}{A(4)} = \dfrac{2}{5} \Rightarrow \lambda = 4. \left(\dfrac{4}{25}\right)^{1/3}$

The area (in sq. units) of the region

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

0%
$\dfrac{10}{3}$
0%
$\dfrac{9}{2}$
0%
$\dfrac{31}{6}$
0%
$\dfrac{13}{6}$

Explanation

$x^2\leq y\leq x+2$

$x^2=y; y=x+2$

$x^2=x+2$

$x^2-x-2=0$

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

$x=2, -1$
Area

$=\displaystyle\int^2_{-1}(x+2)-x^2dx=\left[\dfrac {x^2}{2}+2x-\dfrac {x^3}{3}\right]^2_{-1}$

$=2+4-\dfrac 83-\dfrac 12 +2 -\dfrac 13=\dfrac{9}{2}$ .

1250286_1614740_ans_ee037376810c4beca81b05c41299a870.png

Area of the region bounded by

$y^2\leq 4x, x+y\leq 1, x\geq 0, y\geq 0$ is

$a\sqrt{2}+b$ , then value of

$a-b$ is?

0%
$4$
0%
$6$
0%
$8$
0%
$12$

Explanation

Let P be the point common to

$x+y=1$ &

$y^2=4x$
So

$y^2=4(1-y\Rightarrow y^2+4y-4=0$

$\Rightarrow y=\dfrac{-4\pm \sqrt{16+16}}{2}$

$\Rightarrow =-2+2\sqrt{2}$
Hence

$P(3, -2\sqrt{2}, -2+2\sqrt{2})$
Hence started area

$=$ Area of region (OPN)

$+$ Area of (

$\Delta$ OPQ)

$=\displaystyle\int^{3-2\sqrt{2}}_02\sqrt{x}dx+\dfrac{1}{2}[-1-(3-2\sqrt{2})]^2$

$=\dfrac{2}{3}\cdot 2(\sqrt{2}-1)(3-2\sqrt{2})+\dfrac{1}{2}[2(\sqrt{2}-1)]^2$

$=\dfrac{4}{3}\left\{-7+5\sqrt{2}\right\}+2(3-2\sqrt{2})=\left(\dfrac{20}{3}-4\right)\sqrt{2}+6-\dfrac{28}{3}=\dfrac{8}{3}\sqrt{2}-\dfrac{10}{3}$
Hence

$a=\dfrac{8}{3}, b=\dfrac{-10}{3}$
So

$a-b=6$ .

1246466_1612999_ans_df69b9d24a5347dfaa26b7bf0e823367.png

If the area enclosed by the curves

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

$y=\lambda x$ is

$\cfrac { 1 }{ 9 }$ square units then value of

$\lambda$ is

0%
$24$
0%
$37$
0%
$48$
0%
$38$

Explanation

${ y }^{ 2 }=4\lambda x\quad \quad ;y=\lambda x$
If

$\lambda> 0$ then
Hence

$\int _{ 0 }^{ 4/\lambda }{ \left( 2\sqrt { \lambda } \sqrt { x } -\lambda x \right) } dx=\cfrac { 1 }{ 9 }$

${ \left( \cfrac { 2\sqrt { \lambda } { x }^{ 3/2 } }{ 3/2 } -\cfrac { \lambda { x }^{ 2 } }{ 2 } \right) }^{ \cfrac { 4 }{ \lambda } }=\cfrac { 1 }{ 9 } \Rightarrow \cfrac { 4 }{ 3 } \sqrt { \lambda } \cfrac { 8 }{ { \lambda }^{ 3/2 } } -\lambda \cfrac { 8 }{ { \lambda }^{ 2 } } =\cfrac { 1 }{ 9 } \Rightarrow \cfrac { 32 }{ 3\lambda } =\cfrac { 1 }{ 9 } \Rightarrow \lambda =24$
1246506_1613240_ans_01ab307bfadb4b6a986b484f6db91497.png

1246506_1613240_ans_01ab307bfadb4b6a986b484f6db91497.png

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

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

$y=\left| x+1 \right|$ , in the first quadrant is:

0%
$\cfrac { 3 }{ 2 } -\cfrac { 1 }{ \log _{ e }{ 2 } }$
0%
$\cfrac { 1 }{ 2 }$
0%
$\log _{ e }{ 2 } +\cfrac { 3 }{ 2 }$
0%
$\cfrac { 3 }{ 2 }$

Explanation

Required Area

$\int _{ 0 }^{ 1 }{ \left( \left( x+1 \right) -{ 2 }^{ x } \right) } dx={ \left( \cfrac { { x }^{ 2 } }{ 2 } +x-\cfrac { { 2 }^{ x } }{ \ln { 2 } } \right) }_{ 0 }^{ 1 }=\left( \cfrac { 1 }{ 2 } +1-\cfrac { { 2 }^{ } }{ \ln { 2 } } \right) -\left( 0+0-\cfrac { 1 }{ \ln { 2 } } \right) =\cfrac { 3 }{ 2 } -\cfrac { 1 }{ \ln { 2 } }$
1250568_1614843_ans_72dd375f71a94307894fd373cec7379c.png

1250568_1614843_ans_72dd375f71a94307894fd373cec7379c.png

If the area (in sq. units) bounded by the parabola

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

$y = \lambda x, \lambda > 0$ , is

$\dfrac {1}{9}$ , then

$\lambda$ is equal to

0%
$24$
0%
$48$
0%
$4\sqrt {3}$
0%
$2\sqrt {6}$

Explanation

$Area = \dfrac {1}{9} = \int_{0}^{\dfrac {4}{\lambda}}(\sqrt {4\lambda x} - \lambda x)dx$

$\Rightarrow \lambda = 24$ .
1247390_1615189_ans_711863040f774ff58a2aa29414fec782.png

The area bounded by the line

$y=x$ , x-axis and ordinates

$x=-1$ and

$x=2$ is?

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

The area of the region

$\left \{(x, y) : xy \leq 8, 1 \leq y\leq x^{2}\right \}$ is

0%
$16\log_{6} 2 - 6$
0%
$8\log_{6} 2 - \dfrac {7}{3}$
0%
$16\log_{6} 2 - \dfrac {14}{3}$
0%
$8\log_{6} 2 - \dfrac {14}{3}$

Explanation

$xy \leq 8$

$1\leq y\leq x^{2}$

$x^{2} . x = 8$

$x = 2$
Require Area

$= \int_{1}^{4} \left (\dfrac {8}{y} - \sqrt {y}\right )dy = \left [8ln y - \dfrac {y^{3/2}}{3/2}\right ]_{1}^{4} = 8\ ln 4 - \dfrac {2}{3} . 8 - 0 + \dfrac {2}{3} = 16 ln2 - \dfrac {14}{3}$ .
1283329_1633322_ans_6eb1e811d09d4f01bc251dd29c59bbdb.png

1283329_1633322_ans_6eb1e811d09d4f01bc251dd29c59bbdb.png

The area bounded by curve

$y=\sin { 2x } \left( x=0\quad to\quad x=\pi \right)$ and X-axis is ______

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

Explanation

As the crest and trugh are same we can write

$A=2\int _{ 0 }^{ \pi /2 }{ \sin { 2x } } dx$

$=2{ \left[ \cfrac { -\cos { 2x } }{ 2 } \right] }_{ 0 }^{ \pi /2 }$

$=(1+1)$

$A=2$

1342548_1646655_ans_715db58ed3084a2faf2d09155ec1a966.png

Area of the region bounded by the curve

$y = \cos x$ between

$x = 0$ and

$x = \pi$ is

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

Explanation

Required area enclosed by the curve

$y = \cos x, x = 0$ and

$x = \pi$

$A = \int_{0}^{\pi/2} \cos x dx + \left |\int_{\pi/2}^{\pi} \cos x dx\right |$

$=[sinx]_0^{\pi/2}+[sinx]_{\pi/2}^{\pi}$

$= \left [\sin \dfrac {\pi}{2} - \sin 0\right ] + \left |\sin \dfrac {\pi}{2} - \sin \pi \right |$

$= 1 + 1 = 2\ sq\ units$ .
1767566_1623165_ans_ef321d002610413380045956839a3fd2.png

The area bounded by the curves

$y = -x^2 + 3$ and

$y = 0$

0%
$\sqrt{3} + 1$
0%
$\sqrt{3}$
0%
$4\sqrt{3}$
0%
$5\sqrt{3}$

Explanation

$\textbf{Step -1: Finding the points of intersection.}$

$\text{The two curves are }y=-x^2+3\text{ and }y=0$

$\text{Substituting }y=0\text{ in }y=-x^2+3$

$\Rightarrow 0=-x^2+3$

$\Rightarrow x^2=3$

$\Rightarrow x=\pm\sqrt3$

$\text{The points of intersection are }(-\sqrt3,0)\text{ and }(\sqrt3,0)$

$\textbf{Step -2: Finding the area between the curves.}$

$\text{Area between the curves}=\int_{-\sqrt3}^{\sqrt3}(-x^2+3)dx$

$=2\times\int_0^{\sqrt3}(-x^2+3)dx$

$=2\times[\dfrac{-x^3}{3}+3x]_0^{\sqrt3}$

$= 2[\dfrac{-3\sqrt3}{3}+3\sqrt3]$

$=4\sqrt3\text{ sq. units}$

$\textbf{Hence , the area bounded by the curves }\mathbf{y=-x^2+3}\textbf{ and }\mathbf{y=0}\textbf{ is }\mathbf{4\sqrt3}\textbf{ sq. units.}$

The area (in sq. units) of the region

$\left\{ \left( x,y \right) \in { R }^{ 2 }:{ x }^{ 2 }\le y\le 3-2x \right\}$ , is:

0%
$\cfrac { 29 }{ 3 }$
0%
$\cfrac { 34 }{ 3 }$
0%
$\cfrac { 31 }{ 3 }$
0%
$\cfrac { 32 }{ 3 }$

Explanation

From given data in question,

Point of intersection of given two curve

$y = x^2$ and

$y = -2 x + 3$

$x^2 = -2 x + 3$

$x = -3 , 1$

Area of shaded region

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

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

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

$= 12 + 8 - \dfrac{28}{3} = \dfrac{32}{3}$ .

1477799_1697585_ans_b6a6f51c45cc4966a73e8cd79a6a0abb.png

The area bounded by

$y = \sin^2 x , x = \dfrac{\pi}{2}$ and

$x = \pi$ is

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

Explanation

Required area

$= \displaystyle \int_{\frac{\pi}{2}}^{\pi} \sin^2 x dx$

$= \displaystyle \int^{\pi}_{\tfrac{\pi}{2}} \left[\dfrac{1 - \cos 2x}{2}\right] dx$

$= \dfrac{1}{2} \displaystyle \int^{\pi}_{\tfrac{\pi}{2}} (1 - \cos 2x) dx$

$= \dfrac{1}{2} \left[x - \dfrac{\sin 2x}{2} \right]_{\tfrac{\pi}{2}}^{\pi}$

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

$= \dfrac{1}{2} \left[\dfrac{\pi}{2} \right] = \dfrac{\pi}{4}$

The area of the region bounded by the curve

$y=2x-x^2$ and the line

$y=x$ is ________ square units.

0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{7}{6}$

Explanation

We note that the region bounded by these curves is in the region

$x \in [0,1 ]$ . In this region, the curve

$y = x$ lies below the curve

$y = 2x-x^2$

So, to calculate the area of said region, we evaluatie the following integral:

$\int_0^1 2x-x^2 - x dx$

$= \int_0^1 x-x^2 dx$

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

$= \frac{1}{2} - \frac{1}{3} = \frac{1}{6}.$

Given

$f(x)=\begin{cases} x,0\le x<\dfrac { 1 }{ 2 } \\ \dfrac { 1 }{ 2 } ,x=\dfrac { 1 }{ 2 } \\ 1-x,\dfrac { 1 }{ 2 } <x\le 1 \end{cases}$ and

$g(x)=\left(x-\dfrac{1}{2}\right)^{2},x\epsilon R$ , Then the area (in sq.units) of the region bounded by the curves

$y=f(x)$ and

$y=g(x)$ between the lines

$2x=1$ and

$2x=\sqrt{3}$ , is:

0%
$\dfrac{1}{3}+\dfrac{\sqrt{3}}{4}$
0%
$\dfrac{1}{2}-\dfrac{\sqrt{3}}{4}$
0%
$\dfrac{1}{2}+\dfrac{\sqrt{3}}{4}$
0%
$\dfrac{\sqrt{3}}{4}-\dfrac{1}{3}$

Explanation

$\left\{\begin{matrix}x &,0<n<\dfrac{1}{2} \\\dfrac{1}{2} &,n=\dfrac{1}{2} \\ 1-n& ,\dfrac{1}{2}<n<1\end{matrix}\right.$

and

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

graph of the current is as

refer above image.

Required image =are of trapezium

$\displaystyle ABCD-\int^{\dfrac{\sqrt{3}}{2}}_{\dfrac{1}{2}}\left(x-\dfrac{1}{2}\right)^2dx$

$=\dfrac{1}{2}\left(\dfrac{\sqrt{3}-1}{2}\right)\left(\dfrac{1}{2}+1-\dfrac{\sqrt{3}}{2}\right)-\dfrac{1}{3}\left(\left(x-\dfrac{1}{2}\right)^3\right)^{\dfrac{\sqrt{3}}{2}}_{\dfrac{-1}{2}}$

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

1479945_1697516_ans_ca7a5e1600db45c1b097cb8bb5ac9811.png

The area of the region, enclosed by the circle

$x^2+y^2=2$ which is not common to the region bounded by the parabola

$y^2=x$ and the straight line

$y=x$ , is:

0%
$\dfrac{1}{3}(15\pi-1)$
0%
$\dfrac{1}{6}(24\pi-1)$
0%
$\dfrac{1}{6}(12\pi-1)$
0%
$\dfrac{1}{3}(6\pi-1)$

Explanation

$\textbf{Step-1: Find the radius of the circle}$

$\text{The given equation of the circle is}$

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

$\text{Comparing with the general form of equation of a circle,}$

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

$\text{We get, }$

$r^2=2$

$r=\sqrt{2}$

$\textbf{Step-2: Find the area of required region}$

$\text{Here, Required Area= Area of circle-Area of shaded region}$

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

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

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

$=\dfrac{12\pi-1}{6}$

$\textbf{Hence, Area of the required region is}$

$\mathbf{\dfrac{12\pi-1}{6}}$

The area (in sq. units) of the region

$\{(x, y) \in R^2 |4x^2 \le y \le 8x + 12\}$ is :

0%
$\dfrac{128}{3}$
0%
$\dfrac{127}{3}$
0%
$\dfrac{125}{3}$
0%
$\dfrac{124}{3}$

Explanation

Shaded are is the area between curve

$Y = 4x^2$ __(I)

and

$Y = 8x + 12$ __(II)

on solving equation (I) & (II) (to find intersection point of the curves).

$4x^2 - 8x - 12 = 0$

$4(x +1 )(x - 3) = 0$

$x = -1$ or

$x = 3$

$\therefore$ shaded Area

$= \displaystyle \int_{-1}^3 [(8x + 12) - (4x^2)] dx$

$= \displaystyle \int_{-1}^3 8x dx + \int_{-1}^3 12 dx - 4 \int_{-1}^3 x^2 dx$

$= \left[4x^2 + 12 x - \dfrac{4}{3} x^3 \right]_{-1}^3$

$=4\cdot 9+12\cdot 3-\dfrac 43\cdot 27-4+12-\dfrac 43$

$= \dfrac{128}{3}$

1476830_1697397_ans_a0298a8e0998490f86f6a7ff28f2cabc.png

The area bounded by the curve

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

$( 1 , 4)$ and y -axis and

0%
$\frac {2}{3}$ sq units
0%
$\frac {1}{3}$ sq units
0%
2 sq units
0%
None of these

$A_n$ is the area bounded by

$y = ( 1 -x^2)^n$ and coordinates axes ,

$n \epsilon N$ , then

0%
$A_n = A_{n-1}$
0%
$A_n < A_{n-1}$
0%
$A_n > A_{n-1}$
0%
$A_n =2 A_{n-1}$

The area enclosed between the curves

$y=log_{e}(x+e)\, , \,x=log_{e}\left ( \dfrac{1}{y} \right )$ and the

$x-axis$ is

0%
$2$ sq.units
0%
$1$ sq.units
0%
$4$ sq.units
0%
None of these

Explanation

$y=log_{e}(x+e)\, , \, x=log_{e}\left ( \dfrac{1}{y} \right ) \Rightarrow y=e^{-x}$

for

$y=log_{e}(x+e)$ shift the graph os

$log_{e}x\, , e$ units left hand side.

Required area =

$\displaystyle \int_{1-e}^{0}log_e(x+e)dx\,+ \displaystyle \int_{0}^{\infty }e^{-x}dx$

$= |xlog_{e}(x+e)|_{1-e}^{0}-\displaystyle \int_{1-e}^{0}\dfrac{x}{x+e}dx-|e^{x}|_{0}^{\infty }$

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

$= |x-elog(x+e)|_{0}^{1-e}-0+1$

$=1-e +e\, loge+1 = 2sq.units$

1664505_1762657_ans_669d47a9f0fb4e0e96b550f4d254cbd5.PNG

$\displaystyle \left ( \alpha ^2,\alpha - 2 \right )$ be a point interior to the region of the parabola

$\displaystyle y^2 = 2x$ bounded by the chord joining the points

$\displaystyle \left ( 2,2 \right ) and \left ( 8,-4 \right )$ then

$\alpha$ belongs to the interval

0%
$\displaystyle -2 + 2\sqrt {2} < \alpha <2$
0%
$\displaystyle \alpha > -2 + 2\sqrt {2}$
0%
$\displaystyle \alpha > -2 - 2\sqrt {2}$
0%
None of these

The area of the region enclosed by the curves

$y=x\log x$ and

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

0%
$\dfrac{7}{12}\,sq.units$
0%
$\dfrac{1}{12}\,sq.units$
0%
$\dfrac{5}{12}\,sq.units$
0%
None of these

Explanation

Curve taking :

$y = x \log_ex$

Clearly,

$x > 0$ ,

For

$0 < x < 1, x \log_ex < 0$ , and for

$x > 1, x \log_ex > 0$

Also

$x \log_ex = 0 \Rightarrow x = 1$ .

Further,

$\dfrac{dy}{dx} = 0 \Rightarrow 1 + \log_e x = 0 \Rightarrow x = 1/e$ , which is a point of minima.

Required area

$= \displaystyle \int_0^1 (2x - 2x^2)dx - \int_0^1 x \log x \ dx$

$= \left[x^2 - \dfrac{2x^3}{3}\right]_0^1 - \left[\dfrac{x^2}{2} \log x - \dfrac{x^2}{4}\right]_0^1$

$= \left( 1 - \dfrac{2}{3} \right) - \left[ 0 - \dfrac{1}{4} - \dfrac{1}{2} \underset{x \to 0}{\lim} x^2 \log x\right] = \dfrac{1}{3} + \dfrac{1}{4} = \dfrac{7}{12}$ .

1636682_1761716_ans_02d5c3792ab24cd3950c27cbdef3ec3b.JPG

Area of the region bounded by the curve

$y =e^x, y=e^{-x}$ and the straight line x= 1 given by

0%
$e-e^{-1} +2$
0%
$e-e^{-1} - 2$
0%
$e+ e^{-1} -2$
0%
None of these

The area bounded by the curve

$y = (x)$ the x-axis and the ordinate

$x= 1$ and

$x = b$ is

$(b- 1)$

$cos ( 3b + 4)$ , then

$f(x)$ is given by

0%
$(x -1 ) sin (3x +4)$
0%
$(x-1) sin (3x-4)$
0%
$-3(x -1) sin ( 3x+ 4) + cos (3x+ 4)$
0%
None of these

The area of the closed figure bounded by

$y=\dfrac{x^{2}}{2}-2x+2$ and the tangents to it at

$(1,\dfrac{1}{2})$ and

$(4,2)$ is

0%
$\dfrac{9}{8} \, sq.units$
0%
$\dfrac{3}{8} \, sq.units$
0%
$\dfrac{3}{2} \, sq.units$
0%
$\dfrac{9}{4} \, sq.units$

Explanation

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

$\dfrac{dy}{dx} = x-2 \, , x-2 \,\, , \left ( \dfrac{dy}{dx} \right )_{x=1}=-1\,\, , \left ( \dfrac{dy}{dx} \right )_{x=4}=2$

$\Rightarrow$ Tangent at

$(1, \dfrac{1}{2}$ is

$y- \dfrac{1}{2}=-1(x-1) \, or \, 2x+2y-3=0$

Tangent at

$(4,2)$ is

$y-2=2(x-4) \, or\, 2x-y-6=0$

Hence, A=

$\displaystyle \int_{1}^{\frac{5}{2}} \left ( \dfrac{x^{2}}{2}-2x+2-\dfrac{3-2x}{2} \right )dx+\displaystyle \int_{\frac{5}{2}}^{4}\left ( \dfrac{x^2}{2}-2x+2-(2x-6) \right )dx$

$=\displaystyle \int_{1}^{4}\left ( \dfrac{x^{2}}{2}-2x+2 \right )dx- \displaystyle \int_{1}^{\frac{5}{2}}\left ( \dfrac{3-2x}{2} \right )dx- \displaystyle \int_{\frac{5}{2}}^{4}(2x-6)dx$

$= \left ( \dfrac{x^{3}}{6}-x^{2}+2x \right )_{1}^{4}-\dfrac{1}{2}(3x-x^{2})_{1}^{\frac{5}{2}}-(x^{2}-6x)_{\frac{5}{2}}^{4}$

$= =\left ( \dfrac{63}{6}-15+6 \right )-\dfrac{1}{2}\left ( 3 \times \dfrac{3}{2}-\left ( \dfrac{25}{4}-1 \right ) \right )-\left ( \left ( 16-\dfrac{25}{4} \right )-6\left ( 4-\dfrac{5}{2} \right ) \right )$

$= \dfrac{3}{2}- -\dfrac{1}{2}\left ( \dfrac{9}{2}-\dfrac{21}{4} \right )-\left ( \dfrac{39}{4}-6\left ( \dfrac{3}{2} \right ) \right )$

$\dfrac{9}{8} \, sq.units$

1664553_1762970_ans_fc94508629084221a8ebd2d99b2a1b7a.PNG

CBSE Questions for Class 12 Commerce Maths Application Of Integrals Quiz 8 - 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 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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