Application of Integrals Test

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Application of Integrals Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

The area of the triangle formed by the positive X-axis and the normal and tangent to the circle $$x^{2}+y^{2}=4$$ at $$(1,\sqrt{3})$$ in sq. units is:

A
$$\sqrt{3}$$

B
$$\displaystyle \frac{1}{\sqrt{3}}$$

C
$$2\sqrt{3}$$

D
$$3\sqrt{3}$$

Solution

the tangent of $$x^{2}+y^{2}=4$$ at $$(1,\sqrt{3}) \:is\: \sqrt{3}y+x=4$$ and that of normal is $$y=\sqrt{3}x$$
So, the area of
$$\triangle ^{le}=1/2\times4\times\sqrt{3}=2\sqrt{3}\:sq\;units$$
Question 2
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I : The area bounded by $$ x=2\cos\theta,\ y=3\sin\theta$$ is $$ 36\pi$$ sq. units.
II: The area bounded by $$ x=2\cos\theta,\ y=2\sin\theta$$ is $$ 4\pi$$ sq.units.
Which of the above statement is correct?

A
Only I

B
Only II

C
Both I and II

D
Neither I nor II.

Solution

$$I$$:    $$\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{3} \right )^2=1$$
           $$Area=\pi \times 2 \times3$$
             $$=6\pi sq\:units.$$
$$I$$      $$\left ( \dfrac{x}{2} \right )^2+\left ( \dfrac{y}{2} \right )^2=1$$
              $$\Rightarrow x^2+y^2=4$$
              $$\pi (2)^2$$
              $$=4\pi\:sq\:units.$$
Question 3
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The area between the curves $$y=\tan x$$,$$y=\cot x$$ and $$\mathrm{x}$$-axis $$(0\displaystyle \leq x\leq\frac{\pi}{2})$$ is

A
$$\log 2$$

B
$$2 \log 2$$

C
$$\displaystyle \frac{1}{2}$$ $$\log2$$

D
$$1$$

Solution

$$y=\tan x$$,$$y=\cot x$$ and $$\mathrm{x}$$-axis $$(0\displaystyle \leq x\leq\frac{\pi}{2})$$
$$\overset {\pi/4}{ \underset { 0}{ \int } } \tan x\:dx+\overset {\pi/2}{ \underset { \pi/4}{ \int } } (\cot x)dx$$
$$-\log |\cos x||_{0}^{\pi /4}+\log|\sin x||_{\pi /4}^{\pi /2}$$
$$-\log(1/\sqrt{2})+\log(1)-\log(1/\sqrt{2})$$
$$\log 2$$
Question 4
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I: The area bounded by the line $$\mathrm{y}=\mathrm{x}$$ and the curve $$y=x^{3}$$ is $$1/_{2}$$ sq. units.
II: The area bounded by the curves $$y=x^{3}$$ and $$ y=x^{2}$$and the ordinates $$x=1$$, $$x=2$$ is $$\frac{7}{12}$$ sq. units.
Which of the above statement is correct?

A
Only I

B
Only II

C
Both I and II

D
Neither I nor II.

Solution

$$SI$$:           $$\overset {1}{ \underset { 0 }{ \int } } (x-x^3)dx=\left (\dfrac{x^2 }{2}-\dfrac{x^4}{4} \right )\bigg|_{0}^{1}$$
                                                  $$=2(\dfrac12-\dfrac14)=2(1/4)sq\:units$$
                                                                 $$=\dfrac12sq\:units.$$
$$SII:$$          $$\overset {2}{ \underset { 1 }{ \int } } (x^3-x^2)dx$$
                                $$=\dfrac{x^4}{4}-\dfrac{x^3}{3}\bigg|_{1}^{2}$$
                                  $$=\left ( \dfrac{2^4}{4}-\dfrac{2^3}{3} \right )-\left ( \dfrac{1}{4}-\dfrac{1}{3} \right )$$
                                  $$=\dfrac{15}{4}-\dfrac{7}{3}=\dfrac{45-28}{12}=\dfrac{17}{12}sq\:units.$$

only $$I$$ is true
Question 5
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Assertion(A): The area bounded by $$y^{2}=4x$$ and $$y=x$$ is $$\displaystyle \frac{8}{3}$$ sq. units.

Reason(R): The area bounded by $$y^{2}=4ax$$ and $$y=mx$$ is $$\displaystyle \frac{8a^{2}}{3m^{3}}$$ sq. units.

A
Both A and R are true and R is the correct explanation of A.

B
Both A and R are true but R is not the correct explanation of A.

C
A is true but R is false.

D
A is false but R is true.

Solution

$$\overset {4a/m^2 }{ \underset { 0 }{ \int } } (\sqrt{4x}-x)dx$$
$$\frac{2}{3}\times2a^{1/2}x^{3/2}-\frac{x^2}{2}|_{0}^{4a/m^2}$$
$$=8a^2/3m^3$$ (R is true)
A is true by inspection
So, A,R are true and R explains A.
Question 6
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Area bounded by $$\displaystyle \mathrm{f}(\mathrm{x})=\max.(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})$$ $$\displaystyle \forall 0\leq x\leq\frac{\pi}{2}$$ and the co-ordinate axis is equal to:

A
$$\displaystyle \frac{1}{\sqrt{2}}$$ sq.units

B
$$\sqrt{2}$$ sq.units

C
$$2$$ sq.units

D
$$1 $$ sq. unit

Solution

$$\displaystyle

\mathrm{f}(\mathrm{x})=\max.(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{x},\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{x})$$
$$\displaystyle \forall 0\leq x\leq\frac{\pi}{2}$$
$$\overset {\pi/4}{ \underset { 0}{ \int } } \cos x\:dx+\overset {\pi/2}{ \underset { \pi/4}{ \int } } \sin x\:dx$$
$$\sin x|_{0}^{\pi /4}-cos x|_{0}^{\pi /4}$$
$$=(1/\sqrt{2})-[0-1/\sqrt{2}]$$
$$=\sqrt{2}sq\:units.$$
Question 7
1 / -0

The area lying in the first quadrant between the curves $$x^{2}+y^{2}=\pi^{2}$$ and $$y=\sin x$$ and y- axis is

A
$$\displaystyle \frac{\pi^{3}-8}{4}$$ sq. units

B
$$\displaystyle \frac{\pi^{3}+8}{4}$$ sq. units

C
$$4(\pi^{3}-8)$$ sq. units

D
$$\displaystyle \frac{\pi-8}{4}$$ sq. units

Solution

Area of the circle with radius $$\pi$$ is $${\pi}^{3}$$
Then, area of the arc of circle in the $first quadrant is $$\dfrac{\pi^3 }{4}sq\:units$$ ....... $$(i)$$
Also, the area of the curve $$y=\sin x$$ is $$\overset {\pi }{ \underset { 0 }{ \int } } \sin x\:dx=2\:sq\:units$$ ....... $$(ii)$$
So, the required area is obtained by taking out $$(ii)$$ from $$(i)$$ i.e. $$(i)-(ii)$$
Thus, the reuqired area
$$=\dfrac{\pi^3 }{4}-2 = \dfrac{\pi ^3-8}{4}sq\:units.$$
Question 8
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The area of the portion of the circle $$x^{2}+y^{2}=1$$, which lies inside the parabola $${y}^{2}=1-x$$ is

A
$$\displaystyle \frac{\pi}{2}-\frac{2}{3}$$

B
$$\displaystyle \frac{\pi}{2}+\frac{2}{3}$$

C
$$\displaystyle \frac{\pi}{2}+\frac{4}{3}$$

D
$$\displaystyle \frac{\pi}{2}-\frac{4}{3}$$

Solution

The given curve intersect at $$(0,1)$$ and $$(0,-1)$$
Therefore, the required area $$\displaystyle =\left[\frac { 1 }{ 2 } \pi { r }^{ 2 }\right]_{0}^{1}+2\int _{ 0 }^{ 1 }{ \left( 1-{ y }^{ 2 } \right) dy } $$
$$\displaystyle =\frac { 1 }{ 2 } \pi { \left( 1 \right) }^{ 2 }+2{ { \left[ y-\frac { { y }^{ 3 } }{ 3 } \right] } }_{ 0 }^{ 1 }=\frac { \pi }{ 2 } +\frac { 4 }{ 3 } $$
Question 9
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Area of the region bounded by $$y=e^{x},y=e^{-x},x=0$$ and $$x=1$$ in sq. units is:

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

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

C
$$\left(\sqrt{e}+\dfrac{1}{\sqrt{e}}\right)^{2}$$

D
$$\left(\sqrt{e}-\dfrac{1}{\sqrt{e}}\right)^{2}$$

Solution

$$y=e^{x},y=e^{-x}$$
Then, area bounded by these two curves s given by
$$A=\displaystyle \overset{1}{\underset{0}{\int}}(e^{x}-e^{-x})dx=\overset{1}{\underset{0}{\int}}e^{x}dx-\overset{1}{\underset{0}{\int}}e^{-x}dx$$
$$=(e^1-1)+1(e^{-1}-1)$$
$$=e+e^{-1}-2$$
$$=\left(\sqrt{e}-\dfrac{1}{\sqrt{e}}\right)^2sq\:units.$$
Question 10
1 / -0

The area of the region bounded by the curves $$\mathrm{y}=\sqrt{x}$$ and $$y=\sqrt{4-3x}$$ and $$\mathrm{y}=0$$ is:

A
4/9

B
16/9

C
8/9

D
9/2

Solution

$$\mathrm{y}=\sqrt{x}$$ ; $$y=\sqrt{4-3x}$$
$$\overset{1}{\underset{0}{\int}}\sqrt{x}dx+\overset{4/3}{\underset{1}{\int}}\sqrt{4-3x}dx$$
$$\dfrac{2}{3}1+\dfrac{2}{9}(4-3x)^{3/2}|_{1}^{4/3}$$
$$\dfrac{2}{3}-\dfrac{2}{9}(0-1)$$
$$=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}sq\:units$$