Application of Integrals Test

Result

Application of Integrals Test - 53

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The function $$\displaystyle \mathrm{f}(\mathrm{x})=\max$$ $$\{x^{2},(1-x)^{2},2x(1-x) \forall 0\leq x \leq 1\}$$ then area of the region bounded by the curve $$\mathrm{y}=\mathrm{f}(\mathrm{x})$$ , $$\mathrm{x}$$-axis and $$\mathrm{x}= 0,\ \mathrm{x} =$$ 1 is equals

A
$$\dfrac{27}{17}$$

B
$$\dfrac{17}{27}$$

C
$$\dfrac{18}{17}$$

D
$$\dfrac{19}{17}$$

Solution

Solving $$y={ x }^{ 2 }$$ and $$y=2x\left( 1-x \right) $$ we get $$\displaystyle x=0,\frac { 2 }{ 3 } $$

And solving $$y=(1-{ x })^{ 2 }$$ and $$y=2x\left( 1-x \right) $$, we get $$\displaystyle x=1,\frac { 1 }{ 3 } $$

Therefore
The required area $$\displaystyle A=\int _{ 0 }^{ 1 }{ f\left( x \right) dx } =\int _{ 0 }^{ \frac { 1 }{ 3 } }{ { \left( 1-x \right) }^{ 2 } } dx+\int _{ \frac { 1 }{ 3 } }^{ \frac { 2 }{ 3 } }{ 2x } \left( 1-x \right) dx\quad +\int _{ \frac { 2 }{ 3 } }^{ 1 }{ { x }^{ 2 }dx } $$

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

$$\displaystyle =\frac { 19 }{ 81 } +\frac { 13 }{ 81 } +\frac { 19 }{ 81 } =\frac { 17 }{ 27 } $$
Question 2
1 / -0

The ratio in which the area bounded by the curves $$y^2=12x $$ and $$x^2=12y$$ is divided by the line x $$=$$ 3 is

A
15 : 16

B
15 : 49

C
1 : 2

D
None of these

Solution

$$y^2 =12 x$$ $$x^2 =12 y$$
$$\displaystyle \int_0^3 \sqrt{12 x}-\int_0^3 \dfrac{x^2}{12} =\int_0^3 \sqrt{12 x} -\dfrac{x^2}{12}$$
$$\displaystyle \dfrac{2\sqrt{12}x^{\dfrac{3}{2}}}{3} - \dfrac{x^3}{36}\int_0^3$$
$$\displaystyle \dfrac{\sqrt{12}3\sqrt{3}}{3} - \dfrac{27}{36}$$
$$12 - \dfrac{3}{4} =\dfrac{45}{4}$$
Same as $$\dfrac{8}{4} $$ for remaining part
$$\therefore ratio =\dfrac{15}{49}$$
Question 3
1 / -0

Find the area enclosed between the curves $$y^2- 2ye^{sin^{- 1}x} + x^2- 1 +[x] +e^{2sin^{- 1}x} = 0$$
and line x = 0 and $$x=\frac{1}{2}$$ is (where [.] denotes greatest integer function)

A
$$\dfrac{\sqrt{3}}{4}+\dfrac{\pi }{6}$$

B
$$\dfrac{\sqrt{3}}{2}+\dfrac{\pi }{6}$$

C
$$\dfrac{\sqrt{3}}{4}-\dfrac{\pi }{6}$$

D
$$\dfrac{\sqrt{3}}{2}-\dfrac{\pi }{6}$$

Solution

$$\left (y-e^{sin^-1 x} \right )+x^2+[x]-1=0$$
$$\therefore x\in \left ( 0, 1/2 \right )$$;[x] =0
$$\Rightarrow y-e^{sin^- 1x} = \pm \sqrt{1- x^2}$$
$$\Rightarrow y=e^{sin^- 1x}\pm \sqrt{1- x^2}$$
$$\Rightarrow \overset {\frac{1}{2}}{\underset { 0 }{ \int } } y dx =\overset {\frac{1}{2}}{\underset { 0 }{ \int } } \left |e^{sin^- 1x}+ \sqrt{1+ x^2}- \left (e^{sin^- 1x}- \sqrt{1- x^2} \right ) \right |dx$$
$$\Rightarrow A = 2\overset {\frac{1}{2}}{\underset { 0 }{ \int } } \sqrt{1- x^2}dx$$ ; Let $$x =sin\, \theta$$ so $$dx =cos\, \theta \,d\, \theta$$
$$\Rightarrow A = \overset {\frac{\pi }{6}}{\underset { 0 }{ \int } } 2\,cos^2\,\theta\,d\, \theta=\left [ \theta +\dfrac{sin\, 2\, \theta }{2} \right ]_0^{\pi /6}$$
$$\Rightarrow A ={\pi /6}+\dfrac{\sqrt{3}}{4}$$
Question 4
1 / -0

If $$A_1$$ is the area bounded by $$y= \cos x, y = \sin x$$ & $$x=0$$ and $$A_2$$ the area bounded by $$y = \cos x , y = \sin x , y = 0$$ in $$(0,\frac{\pi}{2})$$ then $$\displaystyle \dfrac{A_1}{A_2}$$ equals to :

A
$$\dfrac{1}{2}$$

B
$$\dfrac{1}{\sqrt2}$$

C
$$1$$

D
None of these

Solution

$$\displaystyle A_{1}=\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x)dx =(\sin x +\cos x)_{0}^{\frac{\pi}{4}}$$
$$\displaystyle =\left ( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} -1\right ) =(\sqrt2-1)$$
$$\displaystyle A_{2}=\int^{0}_{\frac{\pi}{4}}\sin x dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos x dx =\left [ -\cos x \right ]_{0}^{\frac{\pi}{4}}+(\sin x)_{\frac{\pi}{4}}^{\frac{\pi}{2}}$$
$$\displaystyle =\left ( 1-\frac{1}{\sqrt{2}} \right )+\left ( 1-\frac{1}{\sqrt{2}} \right )=\left ( 2-\sqrt{2} \right )=\sqrt{2}(\sqrt{2}-1)$$
Then, $$A_{1}:A_{2}=1:\sqrt{2}=\dfrac{1}{\sqrt{2}}$$
Question 5
1 / -0

The area bounded by the curves $$y = sin^{-1} |sin x|$$ and $$y = (sin^{-1} | sin x|)^{2},$$ where $$0\leq x\leq 2\pi ,$$ is:

A
$$\dfrac{1}{3}+\dfrac{\pi ^{2}}{4}$$ sq. units

B
$$\dfrac{1}{6}+\dfrac{\pi ^{3}}{8}$$ sq. units

C
$$2$$ sq. units

D
$$\displaystyle \frac{4}{3}+\pi ^{2}\frac{\pi -3}{6} sq.units$$

Solution

$$y=sin^{-1}\left |sin x \right | =\begin{cases} x & 0\leq x< \pi /2 \\ \pi-x & \pi /2 \leq
x< \pi \\ x-\pi & \pi \leq x< 3\pi /2 \\ 2\pi-x & 3\pi /2 \leq x< 2\pi \ \end{cases}$$
$$y=\left (sin^{-1}\left |sin x \right | \right )^{2}=\begin{cases} x^{2} & 0\leq x< \pi /2
\\ \left (\pi-x \right )^{2} & \pi /2 \leq x< \pi \\ \left (x-\pi \right )^{2} & \pi \leq
x< 3\pi /2 \\ \left (2\pi-x \right )^{2} & 3\pi /2 \leq x< 2\pi \ \end{cases}$$

Therefore required area $$\displaystyle=4\int_{0}^{1}\left (x-x^{2} \right )^{2}\mathrm{d} x+\int_{1}^{\frac{\pi }{2}}\left (x^{2}-x \right )\mathrm{d} x$$
$$\displaystyle =\frac{4}{3}+\pi ^{2}\frac{\pi -3}{6} sq.units$$
Question 6
1 / -0

If $$\left| z- (4 + 4i) \right| \geq 4$$, then area of the region bounded by the locii of $$z,\; iz,\; - z$$ and $$-iz$$ is:

A
$$4(4-\pi )$$

B
$$16(4-\pi )$$

C
$$16(\pi -1)$$

D
$$4(\pi -1)$$

Solution

Locus of $$|z-(4+4i)|=4$$ is a circle with center at $$(4,4)$$ and radius $$4$$ is Complex plane.
Hence, locus of $$|z-(4+4i)| \geq 4$$ is all points either on or outside the circle with radius $$4$$ and center $$(4,4)$$.
Similarly, locus of $$|-z-(4+4i)| \geq 4$$ is all points on or outside the circle with radius $$4$$ and center $$(-4,-4)$$.
Locus of $$|iz-(4+4i)| \geq 4$$ is all points on or outside the circle with radius $$4$$ and center $$(-4,4)$$.
Finally, locus of $$|-iz-(4+4i)| \geq 4$$ is all points on or outside the circle with radius $$4$$ and center $$(4,-4)$$.
Hence, the area bounded by locus of all four will be the area enclosed by the four circles in argand plane as shown in the figure.
Area bounded$$=$$ area of shaded region
$$= 64 - \pi r^{2}$$
$$=64-16\pi =16(4-\pi )$$
Question 7
1 / -0

If the area bounded by the curve $$y = f(x)$$, the coordinate axes and the line $$x = x_1$$ is given by $$x_1e^{x_1}$$. Then $$f(x)$$ equals

A
$$e^x$$

B
$$xe^x$$

C
$$xe^x-e^x$$

D
$$xe^x+e_x$$

Solution
Question 8
1 / -0

If the area bounded by the curve $$|y|=sin^{-1}|x|$$ and $$x=1$$ is $$a(\pi+b)$$, then the value $$a-b$$ is:

A
1

B
2

C
3

D
4

Solution

From the graph, the area bounded is $$Area = \displaystyle 2\int_0^1 sin^{-1}x dx =\left[2x sin^{-1}x \right]_0^1+\int_0^1\frac{-2x}{\sqrt{1-x^2}}dx$$
$$=\pi + \left[ 2\sqrt{1-x^2}\right]_0^1=\pi-2$$
$$\therefore a=1$$ and $$b=-2$$
$$\therefore a-b=3$$
Question 9
1 / -0

Area of the region bounded by the curve $$y = x^{2}$$ and $$y = sec^{-1} [sin^{2}x]$$ (where [ . ] denotes the greatest integer function) is

A
$$\frac{\pi}{3}\sqrt{\pi}$$

B
$$\frac{2\pi\sqrt{\pi}}{3}$$

C
$$\frac{4\pi\sqrt{\pi}}{3}$$

D
$$\frac{6\pi\sqrt{\pi}}{3}$$

E
$$\frac{3\pi\sqrt{\pi}}{2}$$

Solution

$$y=\sec^{-1}[-\sin^{2}x]\Rightarrow \sec^{-1}(-1)$$
$$\Rightarrow y=\pi, x \neq n\pi$$
$$A=2\int_{0}^{\sqrt{\pi}}(\pi - x^{2})dx=\frac{4}{3}\pi\sqrt{\pi}$$
Question 10
1 / -0

The area bounded by $$y=\sec^ {-1}{x}, y= \text{cosec}^{-1}{x}$$ and the line $$x-1=0$$ is:

A
$$\ln(3+2\sqrt{2})-\displaystyle \frac{\pi}{2}$$

B
$$\displaystyle \frac{\pi}{2}+\ln(3+2\sqrt{2})$$

C
$$\pi - \ln3$$

D
$$\pi + \ln3$$

Solution

$$\displaystyle A = \displaystyle {\int}_0^{\pi /4} \sec y dy + {\int}_{\pi /4}^{\pi /2} \text{cosec} y\ dy - 1\times \frac{\pi}{2}$$

$$\displaystyle=[\ln(\sec y + \tan y)]^{\pi /4}_0 + \left[\ln\tan\frac{y}{2}\right]^{\pi /2}_{\pi /4}-\frac{\pi}{2}$$

$$\displaystyle=[\ln(\sqrt{2}+1)-\ln1] + \ln \tan\frac{\pi}{4}- \ln\tan \frac{\pi}{8}-\frac{\pi}{2}$$

$$\displaystyle=\ln(\sqrt{2}+1)-\ln(\sqrt{2}-1)-\frac{\pi}{2}$$

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

$$\displaystyle=\ln(3+2\sqrt{2})-\frac{\pi}{2}$$

Hence, option A is the correct answer.