Application of Integrals Test

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Application of Integrals Test - 31

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Weekly Quiz Competition

Question 1
1 / -0

The area bounded by the curves $$y=\sin x,y=\cos x$$ and $$y-$$axes in first quadrant is:

A
$$\sqrt {2}-1$$

B
$$\sqrt {2}$$

C
$$\sqrt {2}+1$$

D
None of the above

Solution

The area bounded by the curves $$y=\sin x, y= \cos x$$ and $$y-$$ axis in the first quadrant is,

$$A =\displaystyle \int_{0}^{\pi/4} (\cos x - \sin x) dx$$

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

$$ = \left(\sin \dfrac{\pi}{4} + \cos\dfrac{ \pi}{4}\right) - (\sin 0 + \cos 0) $$

$$ = \sqrt{2} - 1$$
Question 2
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Consider the function $$f\left( x \right) = \left| {x - 1} \right| + {x^2},\,\,where\,\,x \in R$$
What is the area of the region bounded by X-axis, the curve $$y = f\left( x \right)$$ and the two ordinates $$x = \frac{1}{2}\,\,\,and\,\,\,x = 1$$.

A
$$\frac{5}{{12}}sq\,unit$$

B
$$\frac{5}{6}sq\,unit$$

C
$$\frac{7}{6}sq\,unit$$

D
$$2\,sq\,units$$

Solution

$$f\left( x \right) = \left| {x - 1} \right| + {x^2}\,\,\,\,\,x \in R$$
$$f\,\,\,\,x \ge 1$$
$$f\left( x \right) = x - 1 + {x^2}$$
$$f\,\,\,\,\,\,x < 1$$
$$f\left( x \right) = 1 - x - {x^2}$$
$$\int_{\frac{1}{2}}^1 {f\left( x \right)dx} = Area\,\,\,\,\,\,\,\,\,\,\, \to here\,f\left( x \right) = 1 + {x^2} - x$$
$$\int_{\frac{1}{2}}^1 {\left( {1 + {x^2} - x} \right)dx} \Rightarrow {\left. x \right|^1}_{\frac{1}{2}} + {\left. {\frac{{{x^3}}}{3}} \right|^1}_{\frac{1}{2}} - {\left. {\frac{{{x^2}}}{2}} \right|^1}_{\frac{1}{2}}$$
$$\frac{1}{2} + \frac{1}{3}\left\{ {\frac{7}{8}} \right\} - \frac{1}{2}\left\{ {1 - \frac{1}{4}} \right\} = \frac{{12 + 7 - 9}}{{24}} = \frac{5}{{12}}$$
Question 3
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The area bounded by $$y = cos \, x , \, y = x + 1 , \, y = 0$$ is

A
$$\dfrac{3}{2}$$

B
$$\dfrac{2}{3}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{5}{2}$$

Solution

Area $$=$$Area of $$\triangle{AOB}+$$Area of region $$AOC$$

$$=\dfrac{1}{2}\times 1\times 1+\displaystyle\int_{0}^{\dfrac{\pi}{2}}{\cos{x}\,dx}$$
$$=\dfrac{1}{2}+\left[\sin{x}\right]_{0}^{\dfrac{\pi}{2}}$$

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

$$=\dfrac{1}{2}+1=\dfrac{3}{2}$$
Question 4
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The area enclosed between the curve $$y^2 = x \, and \, y = |x|$$ is

A
$$\dfrac{1}{6}$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{2}{3}$$

D
$$1$$

Solution

$$y=\mid x\mid$$$$\left\{ x\quad ;\quad x\ge 0\\ -x\quad ;x<0 \right\} $$
to find $$P$$
$$y^2=x$$ and $$y=x$$
$$x^2=x\\x(x_1)=0$$
$$x=0,1\\P=1$$
$$A=\int _{ 0 }^{ 1 }{ \sqrt { x } -x } dx\\ A={ [\cfrac { 2{ x }^{ 3/2 } }{ 3 } -\cfrac { { x }^{ 2 } }{ 2 } ] }_{ 0 }^{ 1 }\\ A={ [\cfrac { 2 }{ 3 } -\cfrac { 1 }{ 2 } ] }\\ A=\cfrac { 1 }{ 6 }$$
Question 5
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The area bounded by the curve f(x) = x + sin x and its inverse function between the ordinates $$x = 0 \, and \, x = 2 \pi$$ is

A
$$4 \pi$$

B
$$8 \pi$$

C
$$4$$

D
$$8$$

Solution

R.E.F image
$$f(x) = x + \sin x$$
So area between $$f(x)$$ & $$f^{-1} (x)$$ is double
the area between $$ f(x) \& \,y = x $$
Area (A) = $$\displaystyle\int_0^{2\pi} (x + \sin x)^{dx} \, \int_0^{2 \pi} x dx$$
$$= \displaystyle \int_0^{2 \pi} \sin x \, dx$$
$$= \displaystyle\int^{\pi}_0 \, \sin x dx - \int_{\pi}^{2 \pi} \sin x dx$$
$$= [-\cos x]_0^{\pi} - [-\cos x ]_{\pi}^{2 \pi}$$
$$= 2 + 2 = 4$$
Actual Area = $$2 \times A = 8$$
Option D is correct.
Question 6
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Ratio in which curve $$\left| y \right| + x = 0$$ divides the area bounded by curve $$y = {\left( {x + 2} \right)^2}$$ and coordinate axes, is-

A
$$\frac{{11}}{5}$$

B
$$\frac{{11}}{16}$$

C
$$\frac{{20}}{3}$$

D
$$\frac{{5}}{8}$$

Solution

$$\begin{array}{l}\int\limits_{ - 2}^0 {{{\left( {x + 2} \right)}^2}dx} \\ = \int\limits_{ - 2}^0 {\left( {{x^2} + 2x + 4} \right)dx} \\ = \frac{{{x^3}}}{3} + {x^2} + 4x\\ = - \left[ { - \frac{8}{3} + 4 - 3} \right]\\ = - \left[ {4 - \frac{{32}}{3}} \right]\\ = - \left[ {\frac{{12 - 32}}{3}} \right]\\ = - \left[ { - \frac{{20}}{3}} \right] = \frac{{20}}{3}\end{array}$$
Question 7
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Consider the functions $$f(x)$$ and $$g(x)$$, both defined from $$R \rightarrow R$$ and are defined as $$f(x)=2x-x^{2}$$ and $$g(x)=x^{n}$$ where $$n \in N$$. If the area between $$f(x)$$ and $$g(x)$$ in first quadrant is $$1/2$$ then $$n$$ is not a divisor of :

A
$$12$$

B
$$15$$

C
$$20$$

D
$$30$$

Solution

$$ Area=\int\limits_{a}^{b}{\left( upper\,curve \right)-\left( lower\,curve \right)dx} $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,=\int\limits_{0}^{1}{\left( f\left( x \right)-g\left( x \right) \right)dx} $$
$$ \,\,\,\,\,\,\,\,\,\,\,=\int\limits_{0}^{1}{\left( 2x-{{x}^{2}}-{{x}^{n}} \right)dx} $$
$$ \,\,\,\,\,\,\,\,\,=\left[ x-\frac{{{x}^{3}}}{3}-\frac{{{x}^{n+1}}}{n+1} \right]_{0}^{1} $$
$$ \,\,\,\,\frac{1}{2}=1-\frac{1}{3}-\frac{1}{n+1} $$
$$ \Rightarrow \frac{1}{2}=\frac{2}{3}-\frac{1}{n+1} $$
$$ \Rightarrow \frac{1}{n+1}=\frac{2}{3}-\frac{1}{2} $$
$$ \Rightarrow \frac{1}{n+1}=\frac{1}{6} $$
$$ \Rightarrow n+1=6 $$
$$ n=5 $$
$$ Hence\,n\,is\,not\,a\,divisor\,of \ 12 $$
Question 8
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The area bounded by $$y=x^2, y=[x+1], x \leq 1 $$ and the y-axis is, where $$[.]$$ is greatest integer function

A
$$\dfrac{1}{3}$$

B
$$\dfrac{2}{3}$$

C
$$1$$

D
$$\dfrac{7}{3}$$

Solution

Since, $$0\leq x \leq 1$$
Therefore, $$y=[x+1]=1$$
Hence, required area = $$\int_{0}^{1} \sqrt{y} \ dy$$
$$=\dfrac{2}{3}[y^{3/2}]_{0}^{1}$$

$$=\dfrac{2}{3}$$ sq. units
Question 9
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The area between the curves y = tanx, y = cotx and x - axis in the interval $$[0,\pi / 2]$$ is

A
$$log 2$$

B
$$log 3$$

C
$$log \sqrt{2}$$

D
None of these

Solution

$$A=2\int _{ 0 }^{ \pi /4 }{\tan{x} } dx$$
$$ A=2{ [-\log \cos { x } ] }_{ 0 }^{ \pi /4 }$$
$$ A= 2[-\log \cos { \pi/4 } ] $$ $$ -[-\log \cos { 0 } ] $$
$$A=-2\log \sqrt {1/2}$$
$$A=\log 2$$
Question 10
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The area of the region formed by $$x^2+y^2-6x-4y+12\leq 0$$, $$y\leq x$$ and $$x\leq \dfrac{5}{2}$$ is?

A
$$\left(\dfrac{1}{16}+\sin^{-1}\dfrac{\pi}{6}\dfrac{5\sqrt{3}}{7}\right)$$ sq. units

B
$$\left(\dfrac{1}{8}+\dfrac{5\sqrt{3}}{7}+\dfrac{\pi}{6}\right)$$ sq. units

C
$$\left(\dfrac{1}{8}+\dfrac{5\sqrt{3}}{8}+\dfrac{\pi}{6}\right)$$ sq. units

D
None of these

Solution

$$ {{x}^{2}}+{{y}^{2}}-6x-4y+12\le 0 $$
$$ \Rightarrow {{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\le 1 $$
$$ \operatorname{Re}quired\,Area=\int\limits_{2}^{5/2}{xdx-\int\limits_{2}^{5/2}{2+\sqrt{1+{{\left( x-3 \right)}^{2}}}dx}} $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left[ \frac{{{x}^{2}}}{2} \right]_{2}^{5/2}-\left[ 2x \right]_{2}^{5/2}-\int\limits_{2}^{5/2}{\sqrt{1+{{\left( x-3 \right)}^{2}}}dx} $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \frac{25}{8}-2 \right)-\left( 5-4 \right)-\left[ \left( \frac{x-3}{2} \right)\sqrt{1+{{\left( x-3 \right)}^{2}}}+\frac{1}{2}{{\sin }^{-1}}\left( x-3 \right) \right]_{2}^{5/2} $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{25}{8}-2-1-\left[ \left( \frac{-1}{4}\sqrt{1-\frac{1}{4}}+\frac{1}{2}{{\sin }^{-1}}\left( \frac{-1}{2} \right) \right)-\left( 0+\frac{1}{2}{{\sin }^{-1}}\left( -1 \right) \right) \right] $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{8}-\left[ \left( \frac{-\sqrt{3}}{8}-\frac{\pi }{12} \right)-\left( 0-\frac{\pi }{4} \right) \right] $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1}{8}+\frac{\sqrt{3}}{8}+\frac{\pi }{12}-\frac{\pi }{4} $$
$$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{1+\sqrt{3}}{8}-\frac{\pi }{6} $$