Application of Integrals Test

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Application of Integrals Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

Area bounded by $$\displaystyle y=2x-{ x }^{ 2 }$$ & $$\displaystyle (x-1{ ) }^{ 2 }+{ y }^{ 2 }=1$$ in first quadrant, is:

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

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

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

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

Solution

Area of circle is $$\pi r^2$$
But here only half circle is in $$1^{st}$$ quadrant
Then, $$A^*=\dfrac{\pi r^2}{2}$$ .......$$( r=1)$$

$$\Rightarrow A^{*}=\dfrac{\pi}{2} ......(1)$$

Now area bounded by parabola $$y=2x-x^2$$ is

$$A^@=\displaystyle \int_{0}^{2}(2x-x^2)$$

$$A=\dfrac{4}{3} ......(2)$$

Now area bounded by both curves is
$$A=A^*-A^@ $$

$$\Rightarrow A=\dfrac{\pi}{2}-\dfrac{4}{3}$$
Question 2
1 / -0

In what ratio does the x-axis divide the area of the region bounded by the parabolas $$\displaystyle y=4x-x^{2}$$ and $$\displaystyle y=x^{2}-x$$?

A
4 : 121

B
4 : 144

C
4 : 169

D
4 : 100

Solution
Question 3
1 / -0

For what value of 'a' is the area of the figure bounded by $$\displaystyle y=\frac{1}{x}, y=\frac{1}{2x-1}$$ $$x = 2$$ & $$x = a$$ equal to $$\displaystyle ln\frac{4}{\sqrt{5}}$$?

A
$$\displaystyle a=4\:$$

B
$$\displaystyle a=8\:$$

C
$$\displaystyle a=4\: or \frac{2}{5}\left ( 6-\sqrt{21} \right )$$

D
none of these

Solution

The required area is
$$=\int_{2} ^{a} \dfrac{1}{x}-\dfrac{1}{2x-1}.dx$$

$$=[lnx-\dfrac{ln(2x-1)}{2}]_{2} ^{a}$$

$$=[ln\dfrac{x}{\sqrt{2x-1}}]_{2} ^{a}$$

$$=-ln\dfrac{(2)}{\sqrt{3}}+ln\dfrac{a}{\sqrt{2a-1}}$$

$$=ln\dfrac{4}{\sqrt{5}}$$

Hence
$$ln(\dfrac{a}{\sqrt{2a-1}})=ln\dfrac{4}{\sqrt{5}}+ln\dfrac{(2)}{\sqrt{3}}$$

$$\Rightarrow ln(\dfrac{a}{\sqrt{2a-1}})=ln\dfrac{8}{\sqrt{15}}$$
Hence
$$2a-1=15$$
Hence
$$a=8$$.
Therefore a=8 is one solution.
Now
$$\dfrac{a^{2}}{2a-1}=\dfrac{64}{15}$$

$$\Rightarrow 15a^{2}=128a-64$$

$$\Rightarrow 15a^{2}-128a+64=0$$

Hence $$a=8$$ and $$a=\dfrac{8}{\sqrt{15}}$$
Question 4
1 / -0

If the area enclosed by the parabolas $$\displaystyle y=a-x^{2}$$ and $$\displaystyle y=x^{2}$$ is $$\displaystyle 18\sqrt {2}$$ sq. units Find the value of 'a'

A
$$a = -9$$

B
$$a= 6$$

C
$$a =9$$

D
$$a=-6$$

Solution

For the point of intersection
$$x^{2}=a-x^{2}$$
$$2x^{2}=a$$
$$x=\pm\sqrt{\dfrac{a}{2}}$$
Hence the required area will be
$$=\int_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}} x^{2}-(a-x^{2}).dx$$

$$=\int_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}} 2x^{2}-a.dx$$

$$=2\int_{0} ^{\sqrt{\frac{a}{2}}} 2x^{2}-a.dx$$

$$=2[\dfrac{2x^{3}}{3}-ax]_{0} ^{\sqrt{\frac{a}{2}}}$$

$$=2[\dfrac{2.a\sqrt{a}}{6.\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}]$$

$$=2[\dfrac{a\sqrt{a}}{3\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}]$$

$$=18\sqrt{2}$$

$$|\dfrac{a\sqrt{a}}{3\sqrt{2}}-\dfrac{a\sqrt{a}}{\sqrt{2}}|=9\sqrt{2}$$

$$\dfrac{2a\sqrt{a}}{3\sqrt{2}}=9\sqrt{2}$$

$$a\sqrt{a}=27$$

$$a^{\frac{3}{2}}=3^{3}$$

$$a=3^{2}$$
Hence $$a=9$$.
Question 5
1 / -0

Find the area enclosed between the curves $$\displaystyle y=\log_{e}\left ( x+e \right ), x=\log_{e}\left ( 1/y \right )$$ & the x-axis

A
1 sq. units

B
2 sq. units

C
3 sq. units

D
4 sq. units

Solution

Area enclosed
$$ = \int _{1-e} ^0 ln(x+e) dx + \int _{0} ^{\infty} e^{-x} dx$$
$$ = (xln(x+e) -x)| _{1-e} ^0 -e^{-x} | _{0} ^{\infty}$$
$$ =2$$
Question 6
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Let $$f(x)$$ be a continuous function given by $$\displaystyle f\left ( x \right )=2x$$ for $$\displaystyle \left | x \right |\leq 1$$ for $$\displaystyle f\left ( x \right )=x^{2}+ax+b$$ for $$\displaystyle \left | x \right |> 1$$. Find the area of the region in the third quadrant bounded by the curves $$\displaystyle x=-2y^{2}$$ and $$y = f(x)$$ lying on the left of the line $$8x + 1 = 0$$

A
$$\dfrac{235}{192} ; a = 2 ; b = - 1$$

B
$$\dfrac{235}{192} ; a = -1 ; b = 2$$

C
$$\dfrac{257}{192} ; a = -1 ; b = 2$$

D
$$\dfrac{257}{192} ; a = 2 ; b = - 1$$

Solution
Question 7
1 / -0

Let $$\displaystyle C_{1}$$ & $$\displaystyle C_{2}$$ be two curves passing through the origin as shown in the figure A curve C is said to "bisect the area" the region between $$\displaystyle C_{1}$$ & $$\displaystyle C_{1}$$ if for each point P of C the two shaded regions A & B shown in the figure have equal areas Determine the upper curve $$\displaystyle C_{2}$$ given that the bisecting curve C has the equation $$\displaystyle y=x^{2}$$ & that the lower curve $$\displaystyle C_{1}$$ has the equation $$\displaystyle y=x^{2}/2$$

A
$$\displaystyle \left ( 16/9 \right )x^{2}$$

B
$$\displaystyle \left ( 25/9 \right )x^{2}$$

C
$$\displaystyle \left ( 25/16 \right )x^{2}$$

D
$$\displaystyle \left ( 9/25 \right )x^{2}$$

Solution

According to question
$$\displaystyle \int_{0}^{a^{2}}\left ( -f^{-1}\left ( y \right )+\sqrt{y} \right )\:\:dy=\int_{0}^{a}\left ( x^{2}-\frac{x^{2}}{2} \right )dx$$
$$\displaystyle \Rightarrow \left [ f^{-1}(a^{2})-a \right ]2a=-\frac{a^{2}}{2}\Rightarrow f^{-1}(a^{2})=\frac{3a}{4}\Rightarrow f\left ( \frac{3a}{4} \right )-a^{2}$$
or $$\displaystyle f(x)=\frac {16}{9}x^{2}$$
Question 8
1 / -0

Find the area bounded by $$y = x + sinx$$ and its inverse between $$x = 0$$ and $$x = \displaystyle 2\pi$$

A
$$2$$

B
$$4$$

C
$$6$$

D
$$8$$

Solution

graph of $$x+\sin x $$ and its inverse is shown in the figure which is symmetric with $$y=x$$
let $$f(x)=x+\sin x$$
from the figure it is clear that $$A=\int _{ 0 }^{ 2\pi }{ \left( f\left( x \right) -f^{ -1 }\left( x \right) \right) dx } =4\int _{ 0 }^{ \pi }{ \left( f\left( x \right) -x \right) dx } =4\int _{ 0 }^{ \pi }{ \left( x+\sin { x } -x \right) dx } =8$$
Question 9
1 / -0

The smaller area enclosed by $$y=f(x)$$, where $$f(x)$$ is polynomial of least degree satisfying $$\displaystyle{ \left[ \lim _{ x\rightarrow 0 }{ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } } \right] }^{ \tfrac { 1 }{ x } }=e$$ and the circle $$x^2+y^2=2$$ above the $$x-$$axis is

A
$$\displaystyle\frac { \pi }{ 2 } +\frac { 3 }{ 5 } $$

B
$$\displaystyle\frac { \pi }{ 2 } -\frac { 3 }{ 5 } $$

C
$$\dfrac { \pi }{ 2 } -\dfrac { 6 }{ 5 } $$

D
None of these

Solution

Since $$\displaystyle\lim _{ x\rightarrow 0 }{ { \left[ 1+\frac { f\left( x \right) }{ { x }^{ 3 } } \right] }^{ \tfrac { 1 }{ x } } } $$ exists, so $$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { f\left( x \right) }{ { x }^{ 3 } } } =0$$

$$\therefore f(x)={ a }_{ 4 }{ x }^{ 4 }+{ a }_{ 5 }{ x }^{ 5 }+...+{ a }_{ n }{ x }^{ n },a_n\neq 0,n\ge 4$$

Since, $$f(x)$$ is of least degree $$\Rightarrow f(x)={ a }_{ 4 }{ x }^{ 4 }$$
The graph of $$y=x^4$$ and $$x^2+y^2=2$$ are shown in the figure
$$\therefore$$ The required area $$\displaystyle=2\int _{ 0 }^{ 1 }{ \left( \sqrt { 2-{ x }^{ 2 } } -{ x }^{ 4 } \right) } dx=\frac { \pi }{ 2 } +\frac { 3 }{ 5 } $$
Question 10
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The area of the region described by $$\left \{(x, y)/ x^{2} + y^{2} \leq 1\ and\ y^{2} \leq 1 - x\right \}$$ is

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

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

C
$$\dfrac {\pi}{2} + \dfrac {4}{3}$$

D
$$\dfrac {\pi}{2} - \dfrac {4}{3}$$

Solution