MCQ Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 9

$A_n$ is the area bounded by

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

$n \epsilon N$ , then

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

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%
$\displaystyle -2 + 2\sqrt {2} < \alpha <2$
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$\displaystyle \alpha > -2 + 2\sqrt {2}$
0%
$\displaystyle \alpha > -2 - 2\sqrt {2}$
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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

The area bounded by

$y=\sec^{-1}x\, y=cosec^{-1}x \,$ and line

$x-1=0$ is

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$log(3+2 \sqrt{2}) \, - \dfrac{\pi}{2} \, sq.units$
0%
$\dfrac{\pi}{2} - log(3+2 \sqrt{2}) \,sq.units$
0%
$\pi- log_{e}3\, sq.units$
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None of these

Explanation

Integrating along

$x-axis$ , we get

$A= \displaystyle \int_{1}^{\sqrt{2}} (\ cosec^{-1}x\,- \ sec^{-1}x)dx$

Integrating along

$y-axis$ , we get

$A= 2\displaystyle \int_{0}^{\dfrac{\pi}{4}} (\ sec\,y-1)dy$

$= 2[log|\ sec\,y+\ tan\,y|-y]_{0}^{\frac{\pi}{4}}$

$= 2\left [ log|\sqrt{2}+1|-\dfrac{\pi}{4} \right ]=log(3+2\sqrt{2})- \dfrac{\pi}{2}$
1664530_1762735_ans_d521088476e448bd90c997fc27fb2004.PNG

The area of the region whose boundaries are defined by the curves

$y=2 \ cos\,x, \, y=3 \ tan\,x$ and the

$y-axis$ is

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$1+3ln\left ( \dfrac{2}{\sqrt{3}} \right )\, sq.units$
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$1+\dfrac{3}{2}ln3-3ln2\,sq.units$
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$1+\dfrac{3}{2}ln3-ln2\,sq.units$
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$ln3-ln2\, sq.units$

Explanation

Solving

$2 \cos \,x = 3 \tan \,x$ , we get

$2-2 \ sin^{2}x=3 \sin \,x \,\,\,\ \Rightarrow \ sin \,x=\dfrac{1}{2}\,\,\,\, \Rightarrow x= \dfrac{\pi}{6}$

Required Area=

$\displaystyle \int_{0}^{\dfrac{\pi}{6}} (2\ cos\,x-3 \tan\,x) dx$

$= 2 \ sin\,x - 3 log \ sec\,x|_{0}^{\frac{\pi}{6}} = 1-3ln2 + \dfrac{3}{2}ln3\, sq.units.$

1664533_1762748_ans_ffd68db432ac4b509b04fc1407f50262.PNG

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

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

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

A tangent having slope of

$-\dfrac{4}{3}$ to the ellipse

$\dfrac{x^{2}}{18}+ \dfrac{y^{2}}{32}=1$ intersects the major and minor axes at points A and B respectively. If C is the center of the ellipse , then area of the triangle ABC is

0%
12 sq. units
0%
24 sq. units
0%
36 sq. units
0%
48 sq. units

Explanation

One of the tangents of slope m to the given ellipse is

$y=mx + \sqrt{18m^{2}+ 32}$

For

$m=-\dfrac{4}{3}$ ,

we have

$y=-\dfrac{4}{3}x+ 8.$

Then points on the axis where tangents meet are A(6,0) and B(0,8).

Then area of triangle ABC is

$\dfrac{1}{2}(6)(8)= 24 units.$

The area of the region in 1st quadrant bounded by the

$y-axis, \, y=\dfrac{x}{4}, \, y=1+\sqrt{x} \, and\, y= \dfrac{2}{\sqrt{x}}$

0%
$\dfrac{2}{3}\, sq.units$
0%
$\dfrac{8}{3}\, sq.units$
0%
$\dfrac{11}{3}\, sq.units$
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$\dfrac{13}{6}\, sq.units$

Explanation

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

$\left [x + \dfrac{2x^{\frac{3}{2}}}{3}-\dfrac{x^{2}}{8} \right ]_{0}^{1}= 1+ \dfrac{2}{3}-\dfrac{1}{8}=\dfrac{37}{24}$

$A_{2}= \displaystyle \int_{1}^{4}\left ( \dfrac{2}{\sqrt{x}}-\dfrac{x}{4} \right )dx$

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

$= \left [ 8-2-4+\dfrac{1}{8} \right ]= \dfrac{17}{8}$

$\Rightarrow A=A_{1}+A_{2} \, = \dfrac{88}{24}=\dfrac{11}{3} \,sq.units.$

1664543_1762898_ans_604de44576f84998ac8b5bd87b574cf6.PNG

The area of the region bounded by

$x^2+y^2-2x-3=0$ and

$y=|x|+1$ is

0%
$\dfrac{\pi}{2} -1 \, sq.units$
0%
$2 \pi \, sq.units$
0%
$4 \pi \, sq.units$
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$\dfrac{\pi}{2} \, sq.units$

Explanation

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

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

A=

$\displaystyle \int_{1-\sqrt{2}}^{0} ( \sqrt{4-(x-1)^2} - (-x+1)dx + \displaystyle \int_{0}^{1} ( \sqrt{4-(x-1)^2} - (x+1))dx$

$= \dfrac{x-1}{2} \sqrt{4-(x-1)^{2}}+ \dfrac{4}{2} \sin^{-1} \dfrac{x-1}{2}+ \dfrac{x^{2}}{2}-x\left |_{1-\sqrt{2}}^{0}+ \dfrac{x-1}{2} \sqrt{4-(x-1)^2}+ \dfrac{4}{2} \sin^{-1} \dfrac{x-1}{2}- \dfrac{x^{2}}{2}-x \right |_{0}^{1}$

$-\left ( -1- \dfrac{\pi}{2}+\dfrac{3}{2}- \sqrt{2}-1+\sqrt{2} \right )- \dfrac{3}{2}=\dfrac{\pi}{2}-1 \, sq.units$

1664566_1762980_ans_33e29b849ac94bbc98ba07389cd9b528.PNG

The value of the parameter

$a$ such that the area bounded by

$y=a^{2}x^{2}+ax+1,$ coordinate axes and the line

$x=1$ attains its least value, is equal to

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

Explanation

$a^{2}x^{2} + ax+1$ is clearly positive for all real values of

$x$ . Area under consideration.

$A= \displaystyle \int_{0}^{1} (a^{2}x^{2} + ax+1) dx$

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

$= \dfrac{1}{6} (2a^{2}+3a+6)$

$= \dfrac{1}{6}\left ( 2\left ( a^{2}+\dfrac{3}{2}a+ \dfrac{9}{16} \right )+6-\dfrac{18}{16} \right )$

$= \dfrac{1}{6}\left ( 2\left ( a+\dfrac{3}{4} \right )^2+6-\dfrac{18}{16} \right )$

Which is clearly minimum for

$a= -\dfrac{3}{4}$

Which of the following have the same bounded area

0%
$f(x)= \ sin x, g(x)= \ sin^{2} x,$ where $0 \leq x \leq 10 \pi$
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$f(x)= \ sin x, g(x)= |\ sin x|,$ where $0 \leq x \leq 20 \pi$
0%
$f(x)= |\ sin x|, g(x)= \ sin^{3} x,$ where $0 \leq x \leq 10 \pi$
0%
$f(x)= \ sin x, g(x)= \ sin^{4} x,$ where $0 \leq x \leq 10 \pi$

Explanation

We know that the area bounded by

$y= \ sin x$ and

$x-axis$ for

$x \in [0, \pi]$ is

$2$ sq.units.
Then the area bounded by

$y= \ sin x$ and

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

$4$ sq.units for

$x \in [0, 2\pi]$
Then for

$x \in [0, 10\pi]$ , the area bounded in

$20$ sq.units.

The area bounded by

$y= \ sin x$ and

$y= | \ sin x|$ for

$x \in [0, 2 \pi]$ is

$4$ sq.units.
Then for

$x \in [0, 20 \pi]$ , the area bounded is

$40$ sq.units

The area bounded by

$y= \ sin x$ and

$y= \ sin^{3}x$ for

$x \in [0, 2\pi]$ is

$4$ sq.units.
Then for

$x \in [0, 10 \pi]$ , the area bounded in

$20$ sq.units.

Similarly, the area bounded by the

$y= \ sin x$ and

$y = \ sin^{4}x$ for

$x \in [0, 10 \pi]$ is

$20$ sq.units.
1973542_1763027_ans_3af4b07b0028485e9edd7e4bcc47cea0.jpg

Let

$A(k)$ be the are bounded by the curves

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

$y=kx+2$ .

0%
The range of $A(k)$ is $\left [ \dfrac{10 \sqrt{5}}{3},\infty \right )$
0%
The range of $A(k)$ is $\left [ \dfrac{20 \sqrt{5}}{3},\infty \right )$
0%
If function $k \rightarrow A(k)$ is defined by for $k \in [-2,\infty)$ , then $A(k)$ is many-one function.
0%
The value of $k$ for which area is minium is 1.

Explanation

Let

$y=kx+2$ passes through fixed point

$(0,2)$ for different values of

$k$ .
Also, it is obvious that minimum of

$A(k)$ occurs when

$k=0$ , as when line is rotated from this position about point

$(0,2)$ the increased part of area is more than the decreased part of area.

$\therefore$ Minimum area =

$2 \displaystyle \int_{0}^{\sqrt{5}} (2-(x^{2}-3))dx$

$2 \displaystyle \int_{0}^{\sqrt{5}} (5-x^{2})dx$

$= 2 \left [ 5x- \dfrac{x^{3}}{3} \right ]_{0}^{\sqrt{5}}$

$= 2 \left [ 5\sqrt{5}- \dfrac{5 \sqrt{5}}{3} \right ]$

$= \dfrac{20 \sqrt{5}}{3} \, sq.units$

The area of the closed figure bounded by

$x=-1, \, x=2$ and

$y= -x^2+2, \, x\leq 1$ and

$y= 2x-1 ,\, x>1$ and the abscissa axis is

0%
$\dfrac{16}{3} \, sq.units$
0%
$\dfrac{10}{3} \, sq.units$
0%
$\dfrac{13}{3} \, sq.units$
0%
$\dfrac{7}{3} \, sq.units$

Explanation

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

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

$= \dfrac{16}{3} \, sq.units$

1664561_1762975_ans_e9553a20d3d3437c9847cb58528a128b.PNG

The area bounded by the circles

$x^{2}+y^{2}=1, x^{2}+y^{2}=4$ and the pair of lines

$\sqrt{3}\left(x^{2}+y^{2}\right)=4 x y,$ is equal to

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

Explanation

${\textbf{Step - 1: Draw the diagram}}{\text{.}}$

${\text{At first if we see two given circles , both have a center at (0,0)}}{\text{.}}$

${\text{Radius of 1st one, namely }}{{x}^2} + {{y}^2} = 1,{\text{ is 1}}{\text{.}}$

${\text{Radius of 2nd one, }}{{x}^2} + {{y}^2} = 4{\text{ , is 2}}{\text{.}}$

${\text{Given pair of lines}} \Rightarrow \sqrt 3 \left( {{{x}^2} + {{y}^2}} \right) = 4{xy}{\text{.}}$

$\Rightarrow \left( {\sqrt 3 } \right){{x}^2} - 4{xy} + \left( {\sqrt 3 } \right){{y}^2} = 0$

$\sqrt 3 {{x}^2} - 3{xy} - {xy} + + \sqrt 3 {{y}^2} = 0$

$\Rightarrow \left( {{x} - \sqrt 3 {y}} \right)\left( {\sqrt 3 {x} - {y}} \right).$

${\text{So two lines }x} - \sqrt 3 {y} = 0{\text{ and }}\sqrt 3 {x} - {y} = 0.$

${\textbf{Step - 2: Find angle between pair of lines}}{\textbf{.}}$

${\text{The standard equation of pair of lines is a}}{{x}^2} + 2{hxy} + {b}{{y}^2} = \left( {{x} + {my}} \right)\left( {{px} + {qy}} \right) = 0$

${\text{There the angles between them, is }}\theta = {\tan ^{ - 1}}\dfrac{{2\sqrt {{{x}^2} - {ab}} }}{{\left| {\left. {{a} + {b}} \right|} \right.}}$

$\Rightarrow \theta = {\tan ^{ - 1}}\dfrac{2}{{2\sqrt 3 }}$

$\Rightarrow \theta = \dfrac{\pi }{6}$

${\textbf{Step - 3: Find the area}}{\textbf{.}}$

${\text{As per formula area of sectors of two circles is }}\dfrac{\theta }{{2\pi }} \times \pi \left( {{{\text{R}}^2} - {{r}^2}} \right)$

${\text{Here R}} = 2,{\text{ and }r} = 1{\text{, }}\theta = \dfrac{\pi }{6}$

${\text{So area}} = \dfrac{{\dfrac{\pi }{6}}}{{2\pi }} \times \pi \left( {{2^2} - {1^2}} \right)$

$= \dfrac{\pi }{{6 \times 2}} \times \left( {4 - 1} \right)$

$= \dfrac{{\pi \times 3}}{{6 \times 2}}$

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

${\textbf{Thus, the area bounded is }}\boldsymbol{\mathbf{\dfrac{\pi }{4}}{\textbf{ unit}}{^2}}{\text{ }}.$

The area enclosed by the curves

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

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

0%
$(\pi - 2)a^2$ sq. units
0%
$(4 -\pi)a^2$ sq. units
0%
$\dfrac{\pi a^2}{3}$ sq. units
0%
None of these

The sequence

$S_0,S_1,S_2$ .... forms a G.P with common ratio

0%
$\dfrac{e^\pi}{2}$
0%
$e^{-\pi}$
0%
$e^{\pi}$
0%
$\dfrac{e^{-\pi}}{2}$

The area of the loop of the curve,

$ay^2 =x^2 (a - x)$ is

0%
$4a^2$ sq. units
0%
$\dfrac{8a^2}{15}$ sq. units
0%
$\dfrac{16 a^2}{9}$ sq. units
0%
none of these

Explanation

$ay^2 = x^2 (a -x) \Rightarrow y = \pm \ x \sqrt{\dfrac{a-x}{a}}$

Curve tracing :

$y = x \sqrt{\dfrac{a-x}{a}}$

We must have

$x \le a$
For

$0 < x \le a, \ y > 0$ and for

$x < 0, y < 0$
Also

$y = 0 \Rightarrow x = 0, a$
Curve is symmetrical about x-axis.
When

$x \to -\infty, y \to - \infty$
Also, it can be verified that

$y$ has only one point of maxima for

$0 < x < a$ .

Area

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

$\sqrt{\dfrac{a-x}{a}} = t \Rightarrow - \dfrac{x}{a}=t^2 \Rightarrow x=a(1 - t^2)$

$\Rightarrow A = 2\displaystyle \int_1^0 a(1 - t^2)t(-2at)dt$

$=\displaystyle 4a^2 \int_0^1 (t^2 - t^4)dt$

$= 4a^2 \left[\dfrac{t^3}{3} - \dfrac{t^5}{5}\right]^1_0$

$=4a^2\left[ \dfrac{1}{3} - \dfrac{1}{5}\right] = \dfrac{8a^2}{15}$ sq. units

1631815_1786459_ans_83023640d5064f5f8a7fed16bc33963a.JPG

The area enclosed by the circle

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

0%
$4\pi \ sq\ units$
0%
$2\sqrt {2 \pi}$ sq\ units$$
0%
$4\pi^{2} sq\ units$
0%
$2\pi \,sq\ units$

Explanation

Correct answer is (D); since Area

$= 4\int_{0}^{\sqrt {2}} \sqrt {2 - x^{2}}$

$= 4\left (\dfrac {x}{2} \sqrt {2 - x^{2}} + \sin^{-1} \dfrac {x}{\sqrt {2}}\right )_{0}^{\sqrt {2}} = 2\pi \ sq\ units$ .

The area of the region bounded by the curve

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

$x = 4y - 2$ is

0%
$\dfrac {3}{8} sq\ units$
0%
$\dfrac {5}{8} sq\ units$
0%
$\dfrac {7}{8} sq\ units$
0%
$\dfrac {9}{8} sq\ units$

Explanation

Given equation of curve is

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

$x = 4y - 2$ .
For intersection point, put

$x = 4y - 2$ in equation of curve, we get

$(4y - 2)^{2} = 4y$

$\Rightarrow 16y^{2} + 4 - 16y = 4y$

$\Rightarrow 16y^{2} - 20y + 4 = 0$

$\Rightarrow 4y^{2} - 5y + 1 = 0$

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

$\Rightarrow 4y(y - 1) - 1(y - 1) = 0$

$\Rightarrow (4y - 1)(y - 1) = 0$

$\therefore y = 1, \dfrac {1}{4}$
For

$y = 1, x = \sqrt {4.1} = 2$ [since, negative value does not satisfy the equation of line]
For

$y = \dfrac {1}{4}, x = \sqrt {4.\dfrac {1}{4}} = -1$ [positive value does not satisfy the equation of line]
So, the intersection points are

$(2, 1)$ and

$\left (-1, \dfrac {1}{4}\right )$

$\therefore$ Area of shaded region

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

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

$= \dfrac {1}{4} \left (\dfrac {4}{2} + 4 - \dfrac {1}{2} + 2\right ) - \dfrac {1}{4} \left (\dfrac {8}{3} + \dfrac {1}{3}\right )$

$= \dfrac {1}{4} . \dfrac {15}{2} - \dfrac {1}{4} . \dfrac {9}{3}$

$= \dfrac {45 - 18}{24}$

$= \dfrac {27}{24} = \dfrac {9}{8} sq\ units$ .

1766214_1799530_ans_a5b91da2278b4980b45666fd0096e8c5.png

The area of the region bounded by the curve

$y = \sqrt {16 - x^{2}}$ and x-axis is

0%
$8\pi \ sq. units$
0%
$20\pi \ sq. units$
0%
$16\pi \ sq. units$
0%
$256\pi \ sq. units$

Explanation

Given equation of curve is

$y = \sqrt {16 - x^{2}}$ and the equation of line is

$X-axis$ i.e.,

$y = 0$

$\therefore \sqrt {16 - x^{2}} = 0 ..... (i)$

$\Rightarrow 16 - x^{2} = 0$

$\Rightarrow x^{2} = 16$

$\Rightarrow x = \pm 4$
So, the intersection points are

$(4, 0)$ and

$(-4, 0)$ .

$\therefore$ Area of curve,

$A = \int_{-4}^{4} (16 - x^{2})^{1/2} dx$

$= \int_{-4}^{4} \sqrt {(4^{2} - x^{2})} dx$

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

$= \left [\dfrac {4}{2} \sqrt {4^{2} - 4^{2}} + 8\sin^{-1} \dfrac {4}{4}\right ] - \left [-\dfrac {4}{2} \sqrt {4^{2} - (-4)^{2}} + 8\sin^{-1} \left (-\dfrac {4}{4}\right )\right ]$

$= \left [2.0 + 8.\dfrac {\pi}{2} - 0 + 8.\dfrac {\pi}{2}\right ] = 8\pi sq\ units$ .
1766226_1799533_ans_1c2312e088be4db0a21fa09d3dd96d99.png

Area of the region in the first quadrant enclosed by the x-axis, the line

$y = x$ and the circle

$x^{2} + y^{2} = 32$ is

0%
$16\pi \ sq units$
0%
$4\pi \ sq units$
0%
$32\pi \ sq units$
0%
$24\pi \ sq units$

Explanation

We have enclosed by

$X-axis$ i.e.,

$y = 0, y = x$ and the circle

$x^{2} + y^{2} = 32$ in first quadrant.
Since,

$x^{2} + (x)^{2} = 32 [\because y = x]$

$\Rightarrow 2x^{2} = 32$

$\Rightarrow x = \pm 4$
So, the intersection point of circle

$x^{2} + (x)^{2} = 32$ and line

$y = x$ are

$(4, 4)$ and

$(-4, -4)$ .
And

$x^{2} + y^{2} = (4\sqrt {2})^{2}$
Since,

$y = 0$

$\therefore x^{2} + (0)^{2} = 32$

$\Rightarrow x = \pm 4\sqrt {2}$
So, the circle intersects the

$X-axis$ at

$(\pm 4\sqrt {2}, 0)$ .

Area of shaded region

$= \int_{0}^{4} xdx + \int_{4}^{4\sqrt {2}} \sqrt {(4\sqrt {2})^{2} - x^{2}} dx$

$= \left |\dfrac {x^{2}}{2}\right |_{0}^{4} + \left |\dfrac {x}{2} \sqrt {(4\sqrt {2})^{2} - x^{2}} + \dfrac {(4\sqrt {2})^{2}}{2} \sin^{-1} \dfrac {x}{4\sqrt {2}}\right |_{4}^{4\sqrt {2}}$

$= \dfrac {16}{2} + \left [\dfrac {4\sqrt {2}}{2} . 0 + 16\sin^{-1} \dfrac {(4\sqrt {2})}{(4\sqrt {2})} - \dfrac {4}{2} \sqrt {(4\sqrt {2})^{2}} - 16) - 16\sin^{-1} \dfrac {4}{4\sqrt {2}}\right ]$

$= 8 + \left (16\dfrac {\pi}{2} - 2\sqrt {16} - 16\dfrac {\pi}{4}\right )$

$= 8 + [8\pi - 8 - 4\pi] = 4\pi sq\ units$ .
1766237_1799538_ans_f2789a26df784ef1b2455f43c09ece88.png

The area of the region bounded by the circle

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

0%
$2\pi sq\ units$
0%
$\pi sq\ units$
0%
$3\pi sq\ units$
0%
$4\pi sq\ units$

Explanation

We have,

$x^{2} + y^{2} = 1^{2} [\because r = \pm 1]$

$\Rightarrow y^{2} = 1 - x^{2} \Rightarrow = \sqrt {1 - x^{2}}$

$\therefore$ Area enclosed by circle

$= 2\int_{-1}^{1} \sqrt {1^{2} - x^{2}} dx = 2.2\int_{0}^{1} \sqrt {1^{2} - x^{2}}dx$

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

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

$= 4. \dfrac {\pi}{4} = \pi sq\ units$ .

1768348_1799559_ans_9b9a7ca283d443fe9de4aef4b1fabf89.png

The area of the region bounded by the curve

$y = \sin x$ between the ordinates

$x = 0, x = \dfrac {\pi}{2}$ and the x-axis is

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

Explanation

Area of the region bounded by the curve

$y = \sin x$ between the ordinates

$x = 0, x = \dfrac {\pi}{2}$ and the X-axis is

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

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

$= - \left [\cos \dfrac {\pi}{2} - \cos 0\right ]$

$= -[0 - 1] = 1\ sq\ unit$ .
1766249_1799549_ans_6e4eaef210ae44aa83fc697816f29a8a.png

The area of the region bounded by the curve

$x = 2y + 3$ and the

$y$ lines.

$y = 1$ and

$y = -1$ is

0%
$4\ sq\ units$
0%
$\dfrac {3}{2}\ sq\ units$
0%
$6\ sq\ units$
0%
$8\ sq\ units$

Explanation

Required area,

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

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

$= [y^{2} + 3y]_{-1}^{1}$

$= [1 + 3 - 1 + 3]$

$= 6\ sq\ units$ .
1766029_1799566_ans_fe4845a0d5f049e5aa25753166535230.png

The area of the region bounded by parabola

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

$2y = x$ is

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

Explanation

We have to find the area enclosed by parabola

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

$2y = x$ .

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

$\Rightarrow x^{2} = 4x \Rightarrow x (x - 4) = 0$

$\Rightarrow x = 4 \Rightarrow y = 2$ and

$x = 0 \Rightarrow y = 0$
So, the intersection points are

$(0, 0)$ and

$(4, 2)$ .
Area enclosed by shaded region,

$A = \int_{0}^{4} \left [\sqrt {x} - \dfrac {x}{2}\right ] dx$

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

$= \left [2 . \dfrac {x^{\frac {3}{2}}}{3} - \dfrac {x^{2}}{4}\right ]_{0}^{4}$

$= \dfrac {2}{3} 4^{3/2} - \dfrac {16}{4} - \dfrac {2}{3} . 0 + \dfrac {1}{4} . 0$

$= \dfrac {16}{3} - \dfrac {16}{4}$

$= \dfrac {64 - 48}{12}$

$= \dfrac {16}{12} = \dfrac {4}{3} sq\ units$ .

1766246_1799546_ans_7ed18b40d96a4cb58b1a4c2bd3540a78.png

The area of the region bounded by the curve

$y = x + 1$ and the lines

$x = 2$ and

$x = 3$ is

0%
$\dfrac {7}{2} sq\ units$
0%
$\dfrac {9}{2} sq\ units$
0%
$\dfrac {11}{2} sq\ units$
0%
$\dfrac {13}{2} sq\ units$

Explanation

Required area,

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

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

$= \left [\dfrac {9}{2} + 3 - \dfrac {4}{2} - 2\right ]$

$= \dfrac {7}{2} sq\ units$ .

1766292_1799563_ans_54c72ad830824513accc982fa0d3f415.png

The area of the region bounded by the ellipse

$\dfrac {x^{2}}{25} + \dfrac {y^{2}}{16} = 1$ is

0%
$20\pi sq\ units$
0%
$20\pi^{2} sq\ units$
0%
$16\pi^{2} sq\ units$
0%
$25\pi sq\ units$

Explanation

We have,

$\dfrac {x^{2}}{5^{2}} + \dfrac {y^{2}}{4^{2}} = 1$

Here,

$a = \pm 5$ and

$b = \pm 4$

And

$\dfrac {y^{2}}{4^{2}} = 1 - \dfrac {x^{2}}{5^{2}}$

$\Rightarrow y^{2} = 16\left (1 - \dfrac {x^{2}}{25}\right )$

$\Rightarrow y = \sqrt {\dfrac {16}{25}} (25 - x^{2})$

$\Rightarrow y = \dfrac {4}{5} \sqrt {(5^{2} - x^{2})}$

$\therefore$ Area enclosed by ellipse,

$A = 4 . \dfrac {4}{5} \int_{-5}^{5}\sqrt {5^{2} - x^{2}} dx$

$= \dfrac {16}{5} \int_{0}^{5} \sqrt {5^{2} - x^{2}} dx$

$= \dfrac {16}{5} \left [\dfrac {x}{2} \sqrt {5^{2} - x^{2}} + \dfrac {5^{2}}{2} \sin^{-1} \dfrac {x}{5} \right ]_{0}^{5}$

$= \dfrac {16}{5}\left [\dfrac {5}{2} \sqrt {5^{2} - 5^{2}} + \dfrac {5^{2}}{2} \sin^{-1} \dfrac {5}{5} - 0 - \dfrac {25}{5} . 0\right ]$

$= \dfrac {16}{5} \left [\dfrac {25}{2} . \dfrac {\pi}{2}\right ]$

$= \dfrac {16}{5} . \dfrac {25\pi}{4}$

$= 20\pi sq\ units$ .
1766279_1799552_ans_7407a568e66c49e286a740329ed72b40.png

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

Circle is

$x^{2}+y^{2}=4$ with centre

$(0,0)$ and radius

$2$ .

se required area is shaded area of circle from

$(x=0\ to\ x=2)$

$=\displaystyle\int_{0}^{2}\sqrt{4-x^{2}}dx=\displaystyle\int_{0}^{2}\sqrt{2^{2}-x^{2}}dx$

$=\left.\left(\dfrac{x}{2}\sqrt{2^{2}-x^{2}}+\dfrac{4}{2}\sin^{-1}\dfrac{x}{2}\right)\right]_{0}^{2}$

$\displaystyle\int \sqrt{a^{2}-x^{2}}dx=\dfrac{x}{a}\sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{x}{a}$

$=\dfrac{2}{2}\sqrt{2^{2}-2^{2}}+\dfrac{4}{2}\sin^{-1}\dfrac{2}{2}-\dfrac{0}{2}\sqrt{2^{2}-0^{2}}-\dfrac{4}{2}\sin^{-1}\dfrac{0}{2}$

$=1\sqrt{0}+2\sin^{-1}1-0-2\sin^{-1}0$

$=2\times \dfrac{\pi}{2}-2\times 0=\pi\ sq.units$

so answer is

$(A)\pi$ unit .....

As circle of radius

$2$ . so area of circle

$=\pi r^{2}$

$\pi \times 2^{2}=4\pi$

As we need area of only

$1$ quadrant

$\Rightarrow \dfrac{1}{4}\times$ area of circle

$=\dfrac{1}{4}\times 4\pi =\pi\ sq.units$ .

Option

$(A)$

1882278_1858934_ans_822824360dae4638a4e561ea850f26c9.png

Area of the region bounded by the curve

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

$y=3$ is

0%
$2$
0%
$\dfrac{9}{4}$
0%
$\dfrac{9}{3}$
0%
$\dfrac{9}{2}$

Explanation

(1) Curve

$y^{2}=4x$ is right handed parabola

(2) line

$y=3\Rightarrow$ points in line area

$(0,3), (1,3), (2,3), (3,3), (-1,3)(-2,3)$

(3) Pt where line meets

$y^{2}=4x$ is solved by. solving the equation

as we know

$y=3$ so

$x=\dfrac{3^{2}}{4}\Rightarrow \dfrac{9}{4}$

so pt is

$\left(\dfrac{9}{4},3\right)$

(If we solve along

$'y'$ axis)

[Here to solve along

$'x'$ axis, it is tough as to get equation of line in terms of

$x$ is tough.

(4) So area is area under parabola from

$y=0$ to

$y=3$ .

= Area of shade

$=\displaystyle\int_{0}^{3}|x|dy=\int_{0}^{3}\dfrac{y^{2}}{4}dy$

$=\left.\dfrac{1}{4}\times \displaystyle\int_{0}^{3}y^{2}dx=\dfrac{1}{4}\times \dfrac{y^{3}}{3}\right]_{0}^{3}=\dfrac{1}{4}\times \left(\dfrac{3^3}{3}-\dfrac{0^3}{3}\right)$

$=\left(\dfrac{1}{4}\times \dfrac{27}{3}\right)-0=\dfrac{9}{4}sq.units$ So Answer is

$(b)\dfrac{9}{4}$ .

Option

$(B)$

1879481_1858940_ans_59feba845ffd4607aea3079f7e6f35d6.png

I: The area bounded by the curves

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

$\mathrm{Y}$ -axis is

$\sqrt{2}-1$ sq. units.
II: The area bounded by

$y=\cos x,\ y=x+1,\ y=0$ is 3/2 sq. units.

Which of the above statement is correct?

0%
Only I
0%
Only II
0%
Both I and II
0%
Neither I nor II.

Explanation

$I)$
The required area will be

$\int_{0} ^{\frac{\pi}{4}} cos(x)-sin(x) .dx$

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

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

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

$II)$
Required area will be

$\int_{-1} ^{0} x+1 .dx+\int_{0} ^{\frac{\pi}{2}} cos(x).dx$

$=[\dfrac{x^{2}}{2}+x]_{-1} ^{0} +[sin(x)]_{0} ^{\frac{\pi}{2}}$

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

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

Arrangement of the following areas between the curves is descending order:

$\mathrm{A}:y^{2}=4x,\ x^{2}=4y$

$\mathrm{B}.\ y=x,\ y=x^{3}$

$\mathrm{C}.\ y^{2}=8x,y=2x$

$\mathrm{D}.\ y=\sqrt{x},\ y=x^{2}$

0%
A,B,C,D
0%
A,C,B,D
0%
D,B,C,A
0%
D,C,B,A

Explanation

$\frac{16}{3}a^2=16/3\:sq\:units.$

$(a=1)$
B

$1/2\:sq\:units.$
C

$\overset {2 }{ \underset { 0 }{ \int } } (2x-\sqrt{8x})dx$ .

$4-\frac{2}{3}\sqrt{8} 2\sqrt{2}$

$\left | 4-\frac{4}{3} \times 4\right |=4/3\:sq\:units.$
D

$\frac{16}{3}(1/16)=1/3\:sq\:units.$

$(a=1/4)$
So, descending order is A,C,B,D

Match the following:

List-I	List-II
Area of the region bounded by $y=\|5\sin x\|$ from $\mathrm{x}=0$ to $x=4\pi$ and x-axis	a. 3/2
The area bounded by $\mathrm{y}=$ cosx in $[0,2\pi]$ and the $\mathrm{X}$ -axis	b. $\sqrt{2}-1$
The area bounded by $y=sinx, y=cosx$ and the y-axis	c. 4
The area bounded by $y = cos x , y = x +1, y=0$	d. 40

The correct match is

0%
a,b,c,d
0%
d,b,a,c
0%
d,c,b,a
0%
d,a,b,c

Explanation

$y=|5\sin x|$

$\overset {4\pi }{ \underset { 0 }{ \int } } |5\sin x|dx=5\times4\overset {\pi }{ \underset { 0 }{ \int } } \sin x=40\:sq\:units.$
2.

$y=\cos x$

$\leftarrow [0,2\pi]$

$4\overset {\pi/2 }{ \underset { 0 }{ \int } } cos x\:dx=4\:sq\:units.$
3.

$\overset {\pi /4}{ \underset { 0 }{ \int } } (\cos x-\sin x)dx=(\sqrt{2}-1)sq\:units.$
4.

$1+1/2=3/2sq\:units.$
d,c,b,a

The area bounded by the

$y=\left| \sin { x } \right|$ , x-axis and the lines

$\left| x \right| =\pi$ is

0%
$2$ square units
0%
$1$ square units
0%
$4$ square units
0%
None of these

Explanation

Required Area

$=A_1+A_2=2A_2$

$\displaystyle =2\left[\int^{\pi}_0 y\,dx=\int^{\pi}_0\,|\sin x|dx\right]$

$\displaystyle \Rightarrow 2\int^{\pi}_0\sin x\,dx=2[-\cos x]^{\pi}_0$

$\Rightarrow 2[-\cos\pi+\cos 0]=2[-(-1)+1]$

$\Rightarrow 2\times (1+1)=2\times 2=4$ Sq. unit

The area (in square units) bounded by the curves

$y=\sqrt{x},\ 2y-x+3=0$ ,

$X-$ axis, and lying in the first quadrant is:

0%
$36$
0%
$18$
0%
$\dfrac{27}{4}$
0%
$9$

Explanation

Given,

$y=\sqrt{x},\ 2y-x+3=0$

from above we have,

$2\sqrt x-x+3=0$

$2\sqrt{x}=x-3$

$4x=x^{2}-6x+9$

$x^{2}-10x+9=0$

$x=9,\ x=1$

now,

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

$=\left[y^{2}+3y-\frac{y^{3}}{3}\right]_{0}^{3}=9+9-9=9$

Match the following:

List-I	List-II
Area of region bounded by $y=2x-x^{2}$ and $x-$ axis	a. $\dfrac13$
2. Area of the region $\{(x, y):x^{2}\leq y\leq\|x\|\}$	b. $\dfrac12$
3. Area bounded by $y=x$ and $y=x^{3}$	c. $\dfrac23$
4. Area bounded by $y=x\|x\|$ , ${x}$ -axis and ${x}=-1,\ {x}=1$	d. $\dfrac43$

The correct match for

$1\ 2\ 3\ 4$ is

0%
$1-b,2- c.3- d,4- a$
0%
$1-c,2- d,3- a,4- b$
0%
$1-d,2- a,3- b,4- c$
0%
$1-a,2- b,3- c,4- d$

Explanation

$y=2x-x^2$

The values of

$x$ where

$y$ is

$0$ are

$x = 0,2$

Now, the area of the region bounded by

$y=2x-x^2$ and

$x-$ axis is

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

$=4-\dfrac83 = \dfrac43\ \ \rightarrow d$

$x^2≤y≤|x|$

We have to find the area of the region bounded by

$y=x^2$ and

$y = |x|$ for

$x\in[-1,1]$

$\displaystyle\int_{-1}^1-(x^2-|x|)\ dx = -2\int_0^1(x^2-x)\ dx$

$= -2\left[\dfrac{x^3}3 - \dfrac{x^2}2\right]_0^1 = \dfrac13\ \ \rightarrow a$

We have to find the area of the region bounded by

$y=x$ and

$y = x^3$ for

$x\in[-1,1]$

$\displaystyle\int_{-1}^1(x-x^3)\ dx = 2\int_0^1(x-x^3)\ dx$

$=2\left[\dfrac{x^2}2 - \dfrac{x^4}4\right]_0^1 = \dfrac14\ \ \rightarrow b$

We have to find the area of the region bounded by

$y=x|x|$ and

$y = 0$ for

$x\in[-1,1]$

$\displaystyle\int_{-1}^1(x|x|)\ dx = 2\int_0^1(x^2)\ dx$

$=2\left[\dfrac{x^3}3\right]_0^1 = \dfrac23\ \ \rightarrow c$

Hence, option C.

$A$ : The area bounded by

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

$B$ : The area bounded by

$y=1-|x|$ and X-axis

$C$ : The area enclosed between the curve

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

${y}=\sqrt3{x}$
The descending order of

$A, B, C$ is

0%
$A,C,B$
0%
$C,B,A$
0%
$A,B,C$
0%
$C,A,B$

Explanation

$2\displaystyle\int_0^3\sqrt{3x}dx$

$\Rightarrow\dfrac{4\sqrt{3}}{{3}}\left|x^{\frac32}\right|_{0}^{3}$

$=\dfrac{4\sqrt{3}}{{3}}\times\sqrt{3}\times3$

$=12\:sq\:units.$

$\dfrac12\times2\times1=1\:sq\:units.$

$\displaystyle\overset {\sqrt{3} }{ \underset { 0 }{ \int } }(\sqrt3x-x^2)dx=\dfrac{\sqrt3x^2}{2}-\dfrac{x^3}{3}\left.\right|_{0}^{\sqrt{3}}$

$=\dfrac{3\sqrt3}2-\dfrac{3\sqrt{3}}{3}$

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

So, descending order
A > B > C.

The area enclosed by the curves

$y=sinx+$ cosx and

$y=|cosx-$ sinx

$|$ over the interval

$[0, \frac{\pi}{2}]$ is

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

Explanation

$\displaystyle \cos { x } > \sin { x\quad , } \forall x\epsilon \left( 0, \frac { \pi }{ 4 } \right)$ , and

$\displaystyle \cos { x } < \sin { x\quad , } \forall x\epsilon \left( \frac { \pi }{ 4 } , \frac { \pi }{ 2 } \right)$

$y_{1}=\sin x+\cos x$

$y_{2}=\left| \cos { x } -\sin { x } \right|$

$\Rightarrow$ Area

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

$=$

$\int_{0}^{ \pi/4 }$

$(( sin x+cosx)-(cos x-sin x))dx$

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

$=4-2\sqrt{2}$

The area bounded by the parabolas

$\mathrm{y}^{2}=5\mathrm{x}+6$ and

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

0%
$\displaystyle \frac{19}{5}$
0%
$\displaystyle \frac{21}{5}$
0%
$\displaystyle \frac{23}{5}$
0%
$\displaystyle \frac{27}{5}$

Explanation

$y^{2}=5x+6$ and

$x^{2}=y$
Then

$x^{4}=5x+6$
Or

$x^{4}-5x-6=0$

$x=-1$ is an obvious root.
Then

$x^{4}-5x-6=0$

$(x+1)(x-2)(x^{2}+x+3)=0$
Now

$x^{2}+x+3=0$ has no real roots since

$D<0$ .
Hence we have to find the area between the curves for x=-1 to x=2.
Therefore

$Area$

$=|\int_{-1} ^{2}x^{2}-\sqrt{5x+6}.dx|$

$=|[\dfrac{x^{3}}{3}-\dfrac{2}{15}(5x+6)^{\frac{3}{2}}]_{-1} ^{2}|$

$=|\dfrac{8}{3}-\dfrac{128}{15}-[\dfrac{-1}{3}-\dfrac{2}{15}]|$

$=|\dfrac{9}{3}-\dfrac{126}{15}|$

$=\dfrac{126-45}{15}$

$=\dfrac{81}{15}$

$=\dfrac{27}{5}$ .

The area of the smaller part bounded by the semicircle

$y=\sqrt{4-x^{2}},\ y=x\sqrt{3}$ and x-axis is

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

Explanation

The shaded is the smaller part
$\overset{1}{\underset{0}{\int }}\sqrt{3}x\:dx+\overset{2}{\underset{1}{\int }}\sqrt{4-x^2}dx$
$\dfrac{\sqrt{3}x^2}{2}\left|_{0}^{1}+\left ( x/2\sqrt{4-x^2} +4/2\sin ^{-1}(x/2)\right ) \right|_{1}^{2}$
$\dfrac{\sqrt{3}}{2}+2\sin ^{-1}(1)-\dfrac{\sqrt{3}}{2}-2\sin ^{-1}(1/2)$
$2(\pi /2-\pi /6)$
$2\left ( \dfrac{3-1}{6} \right )\pi=\dfrac{2\pi }{3}sq\:units.$

The ratio of the areas into which the circle

$x^{2}+y^{2}=64$ is divided by the parabola

$y^{2}=12x$ is:

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

Explanation

$x^2+y^2 = 64$ ... (i)

$y^2 = 12x$ ... (ii)
Substituting (ii) in (i), we get

$x^2 + 12x -64 = 0$

$x^2 + 16x -4x -64 = 0$

$(x-4)(x+16) = 0$

$x =4$ is the feasible solution.

$y^2 = 12x$

$y = \pm \sqrt{48} = \pm 4\sqrt{3}$
Angle made at the center by the two points

$= 2 tan^{-1}(4\sqrt{3}/4) = \cfrac{2\pi}{3}$

Area of Region inside Parabola

$=$ Area of sector

$+$ area b/w parabola and Straight line from origin (as shown in figure)

$= _0 ^4\int (\sqrt{12x} - \sqrt{3}x)dx$

$= 8 \sqrt{3}$
So Total area inside parabola

$= \cfrac{2 \pi/3}{2 \pi} \pi 64 + \cfrac{16 \sqrt{3}}{3}$

$= \cfrac{16}{3}(4 \pi + \sqrt{3})$

Area of remaining part of circle

$= 64 \pi - \cfrac{16}{3}(4 \pi + \sqrt{3})$
Ratio

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

The area bounded by the parabola

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

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

0%
$\displaystyle \frac{16}{3}a^{2}$
0%
$\displaystyle \frac{8}{3}a^{2}$
0%
$\displaystyle \frac{4}{3}a^{2}$
0%
$\displaystyle \frac{2}{3}a^{2}$

Explanation

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

$\mathrm{y}^{2}=-4\mathrm{a}(\mathrm{x}-\mathrm{a})$
The area bounded by the parabola in the 2nd quadrant,

$\overset{0}{\underset{-a}{\int}}\sqrt{4a(x+a)}dx$
Since the figure is symmetric, multiply the given area by 4.

$4\times \: 2\sqrt{a}\overset{0}{\underset{-a}{\int}}\sqrt{(x+a)}dx$

$\displaystyle \frac{2}{3}\times 8\sqrt{a}(x+a)^{3/2}|_{-a}^{0}$

$\displaystyle \frac{2}{3}\times 8\times a^2=\displaystyle \frac{16}{3}a^2sq\:units$ .

Let

$\displaystyle \mathrm{f}(\mathrm{x})=\min\{x+1,\ \sqrt{1-x}\}$ , then the area bounded by

$\mathrm{y}={f}({x})$ and

${x}$ -axis is:

0%
$\dfrac76$
0%
$\dfrac56$
0%
$\dfrac16$
0%
$\dfrac{11}{6}$

Explanation

$\displaystyle \mathrm{f}(\mathrm{x})=\min\{x+1,\ \sqrt{1-x}\}$

$y=x+1$ and

$y=\sqrt{1-x}$ curves intersect at

$(0,1)$ and

$f(x) = x+1$ for

$x\in (-1,0)$

$f(x) = \sqrt{1-x}$ for

$x\in (0,1)$

$\therefore$ Area under the curve and x-axis is

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

$\displaystyle\left[ \frac { x^{ 2 } }{ 2 } +x \right]^0_{-1} -\left[ \frac { \left( 1-x \right) ^{ 3/2 } }{ 3/2 } \right] ^{ 1 }_{ 0 }\\ =\dfrac { 1 }{ 2 } +\dfrac { 2 }{ 3 } =\dfrac { 7 }{ 6 }$

$\therefore$ required area is

$\displaystyle \frac { 7 }{ 6 }$

Area bounded by the curve

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

$\mathrm{x}=2\mathrm{a}$ is:

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

Explanation

Given,

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

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

Let

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

$dx=4a \sin ^2\theta$

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

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

$=8a^2 \int_{0}^{\tfrac{\pi}{2}} \sqrt{\sin ^6\theta }\sin \theta d\theta$

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

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

The area bounded by the curves

$\mathrm{y}=2^{\mathrm{x}}$ ,

$\mathrm{y}=2\mathrm{x}-\mathrm{x}^{2}$ between the lines

$\mathrm{x}=0,\ \mathrm{x}=2$ is

0%
$\displaystyle \frac{3}{\log 2}-\frac{4}{3}$ sq. units
0%
$\displaystyle \frac{3}{\log 2}+\frac{4}{3}$ sq.units
0%
$3-4\log 2$ sq. units
0%
$\displaystyle \frac{4}{3}-\frac{3}{\log 2}$ sq.units

Explanation

$\mathrm{y}=2^{\mathrm{x}}$ ,

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

$\overset{2}{\underset{0}{\int}}2^{\mathrm{x}}dx-\overset{2}{\underset{0}{\int}}(2x-x^2)dx$

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

$\dfrac{3}{\log 2}-\left [ 4-\dfrac{8}{3} \right ]$

$=\left ( \dfrac{3}{\log 2}-4/3 \right )sq\:units.$

The area of the region bounded by the curves

$y=ex\log x$ and

$y=\displaystyle \frac{\log x}{ex}$ is:

0%
$\displaystyle \frac{e^{2}-5}{4e}$
0%
$e-\displaystyle \frac{5}{4}$
0%
$\displaystyle \frac{e}{4}-5$
0%
$\displaystyle \frac{e}{4}-\frac{1}{4e}$

Explanation

$\left | \overset {1 }{ \underset { 1/e}{ \int } } ex\log x-\dfrac{\log x}{ex}dx\right |$

$\left |e\overset {1 }{ \underset { 1/e}{ \int } } x\log x-\dfrac{1}{e}\overset {1 }{ \underset { 1/e}{ \int } } \frac{\log x}{x}dx\right |$

$\left |e \left ( log x \frac{x^2}{2}-\frac{1}{4}x^2\right) \left |_{1/e}^{1}-\dfrac{1}{e}\dfrac{(\log x)^2}{2} \right |_{1/e}^{1}\right|$

$\left |e \left (\dfrac{-1}{4}-\left (\dfrac{(-1)e^{-2}}{2} -\dfrac{1}{4}e^{-2} \right ) -1/e\left (0-\left ( \dfrac{1}{2} \right ) \right ) \right )\right|$

$\dfrac{-e}{4}+\left ( \dfrac{3}{4}e^{-1} \right )+\dfrac{2}{4e}$

$\left |\dfrac{-e}{4} +\dfrac{5}{4e} \right |=\dfrac{e^2-5}{4e}sq\:units.$

The area bounded by two branches of the curve

$(y-x)^{2}=x^{3} \& x=1$ equals

0%
$3/5$
0%
$5/4$
0%
$6/5$
0%
$4/5$

Explanation

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

$y = x^{3/2} +x$ and

$y = -x^{3/2} +x$
Area 1

$= _0 ^1\int ( x^{3/2} +x) dx = \cfrac{2x^{5/2}}{5} + \cfrac{x^2}{2} |_0 ^1 = 2/5 + 1/2$
Area 2

$= _0 ^1\int ( -x^{3/2} +x) dx = \cfrac{-2x^{5/2}}{5} + \cfrac{x^2}{2} |_0 ^1 = -2/5 +1/2$
Total Area

$=$ Area 1 - Area 2

$=$

$\cfrac{4}{5}$

Area bounded by

$x^{2}=4ay$ and

$y=\displaystyle \frac{8a^{3}}{x^{2}+4a^{2}}$ is:

0%
$\displaystyle \frac{a^{2}}{3}(6\pi-4)$
0%
$\displaystyle \frac{\pi a^{2}}{3}$
0%
$\displaystyle \frac{a^{2}}{3}(6\pi+4)$
0%
$0$

Explanation

points of intersection

$x^{2}=4ay$ ,

$y=\displaystyle \frac{8a^{3}}{x^{2}+4a^{2}}$

$y(4ay+4a^{2})=8a^3$

$\Rightarrow 4ay^2+4a^{2}y-8a^3=0$

$y^2+ay-2a^2=0$

$y=-2a;y=a$
there is only 1 point of intersection for

$y=a$
So, they don't enclose area
So, zero

For which of the following values of

$\mathrm{m}$ , is the area of the region bounded by the curve

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

$y=mx$ equals 9/2 sq. unit

0%
-4
0%
-2
0%
2
0%
4

Explanation

$-x^2+x=mx$

$-x+1= m$

$x=1-m$
So, area

$\overset { 1-m}{ \underset { 0 }\: { \int } } \left [(x-x^2)-mx \right ]dx$

$\overset { 1-m}{ \underset { 0 }\: { \int } } \left [(1-m)x-x^2 \right ]dx=9/2$

$=\dfrac{(1-m)x^2}{2}-\dfrac{x^3}{3}=9/2 \Rightarrow$

$\dfrac{| (1-m)|^3}{6}=9/2$

$=\dfrac{| (1-m)|^3} {3}=9$

$\left | (1-m) \right |=3$

$\Rightarrow1-m=\pm 3$

$m=4; m=-2$

Area of the figure bounded by the lines

$y=\sqrt{x},x\in [0,1],\ y=x^{2},\ x\in[1,2]$ and

$y=-x^{2}+2x+4,x\in0,2]$ is:

0%
$\dfrac{10}{7}$
0%
$\dfrac{3}{5}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{19}{3}$

Explanation

As shown in the figure, the area bounded

$= \int _0 ^2 (-x^{2}+2x+4) dx - \int _0 ^1 \sqrt{x} dx - \int _1 ^2 x^2 dx$

$= \cfrac{19}{3}$

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Applications Of Integrals Quiz 9 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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