Application of Integrals Test

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Application of Integrals Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

The area bounded by $$y= x^{2}$$ and $$y= 1-x^{2}$$ is

A
$$\dfrac{\sqrt{8}}{3}$$

B
$$\dfrac{16}{3}$$

C
$$\dfrac{32}{3}$$

D
$$\dfrac{17}{3}$$

Solution

Required are $$\displaystyle =2\left[ \int _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }{ \left( 1+{ x }^{ 2 } \right) } dx-\int _{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }{ { x }^{ 2 } } dx \right] $$

$$\displaystyle ={ 2\left[ x+\dfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }-{ 2\left[ \dfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ \dfrac { 1 }{ \sqrt { 2 } } }=\dfrac { \sqrt { 8 } }{ 3 } $$
Question 2
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The area bounded by $$ \displaystyle \begin{vmatrix}y\end{vmatrix}=1-x^{2} $$ is

A
$$8/3$$

B
$$4/3$$

C
$$16/3$$

D
None of these

Solution

Clearly given curve is symmetrical about both the axis,
Hence required area $$=\displaystyle 4.\int_0^1(1-x^2)dx =4. \left[x-\frac{x^3}{3}\right]_0^1=8/3$$
Question 3
1 / -0

The area common to the circle $$x^2+y^2=16a^2$$ and the parabola $$y^2=6ax$$ is

A
$$\displaystyle \frac { 4{ a }^{ 2 } }{ 3 } \left( 4\pi -\sqrt { 3 } \right) $$

B
$$\displaystyle \frac { 4{ a }^{ 2 } }{ 3 } \left( 8\pi -3 \right) $$

C
$$\displaystyle \frac { 4{ a }^{ 2 } }{ 3 } \left( 4\pi +\sqrt { 3 } \right) $$

D
None of these

Solution

Solving the two, we get $$\displaystyle { x }^{ 2 }+6a-16{ a }^{ 2 }=0$$
or $$\displaystyle \left( x+8a \right) \left( x-2a \right) =0$$
$$\therefore \quad x=2a$$
$$\therefore \quad y=\pm 2\sqrt { 3a } $$
The required common area $$=2[APOA]$$
$$=2\int { ydx } +2\int { ydx } $$
$$\displaystyle =2\int _{ 0 }^{ 2a }{ \sqrt { 6a } \sqrt { x } dx } +2\int _{ 2a }^{ 4a }{ \sqrt { { \left( 4a \right) }^{ 2 }-{ x }^{ 2 } } dx } $$
$$\displaystyle =2.\sqrt { 6a } \frac { 2 }{ 3 } \left[ { x }^{ 3/2 } \right] _{ 0 }^{ 2a }+2.\left[ \frac { x }{ 2 } \sqrt { { \left( 4a \right) }^{ 2 }-{ x }^{ 2 } } +\frac { 1 }{ 2 } { \left( 4a \right) }^{ 2 }\sin ^{ -1 }{ \frac { x }{ 4a } } \right] _{ 2a }^{ 4a }$$
$$\displaystyle =2.\sqrt { 6a } \frac { 2 }{ 3 } \left( 2a \right) \sqrt { 2a } +2\left[ \left( 0-a2\sqrt { 3a } \right) +8{ a }^{ 2 }\left( \sin ^{ -1 }{ 1 } -\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] $$
$$\displaystyle =2.2\sqrt { 3 } .\frac { 4 }{ 3 } { a }^{ 2 }+2\left[ -2\sqrt { 3 } { a }^{ 2 }\left( \frac { \pi }{ 2 } -\frac { \pi }{ 6 } \right) \right] $$
$$\displaystyle =\frac { 16 }{ 3 } \sqrt { 3 } { a }^{ 2 }-4\sqrt { 3 } { a }^{ 2 }+16{ a }^{ 2 }\frac { \pi }{ 3 } $$
$$\displaystyle =\frac { 4\sqrt { 3 } { a }^{ 2 } }{ 3 } +\frac { 16\pi { a }^{ 2 } }{ 3 } =\frac { 4{ a }^{ 2 } }{ 3 } \left( 4\pi +\sqrt { 3 } \right) $$
Question 4
1 / -0

The area of the region(s) enclosed by the curves $$\displaystyle y=x^{2}$$ and $$\displaystyle y=\sqrt{\left | x \right |}$$ is:

A
1/3

B
2/3

C
1/6

D
1

Solution

The curve $$y={ x }^{ 2 }$$ and $$y=\sqrt { x } $$ intersect at $$O(0,0)$$ and $$P(1,1)$$
Therefore required area
$$\displaystyle =\int _{ 0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 2 } \right) dx } ={ \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }=\frac { 1 }{ 3 } $$
Question 5
1 / -0

Let 'a' be a positive constant number. Consider two curves $$\displaystyle C_{1}:y=e^{x},C_{2}:y=e^{a-x}$$. Let S be the area of the part surrounded by $$\displaystyle C_{1}$$,$$\displaystyle C_{2}$$ and the y-axis, then

A
$$\displaystyle \lim_{a\rightarrow \infty }S=1$$

B
$$\displaystyle \lim_{a\rightarrow 0}\frac{S}{a^{2}}=\frac{1}{4}$$

C
Range of S is $$\displaystyle \left ( 0, \infty \right )$$

D
S(a) is neither odd nor even

Solution

$$\displaystyle C_{1}:y=e^{x},C_{2}:y=e^{a-x}$$
Point of intersection
$$e^x={ e }^{ \left( a-x \right) }$$
$$x=\dfrac{a}{2},y={ e }^{ \dfrac { a }{ 2 } }$$
$$S=\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { C }_{ 2 } } dx-\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { C }_{ 1 } } dx$$
$$S=\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { e }^{ a-x } } dx-\int _{ 0 }^{ \dfrac { a }{ 2 } }{ { e }^{ x } } dx={ \left[ { e }^{ a-x } \right] }_{ 0 }^{ \dfrac { a }{ 2 } }-{ \left[ { e }^{ x } \right] }_{ 0 }^{ \dfrac { a }{ 2 } }$$
$$S=-({ e }^{ \dfrac { a }{ 2 } }-{ e }^{ a })-({ e }^{ \dfrac { a }{ 2 } }-1)=({ e }^{ a }{ -2e }^{ \dfrac { a }{ 2 } }+1)={ \left( ({ e }^{ \dfrac { a }{ 2 } }-1) \right) }^{ 2 }$$
$$\lim _{ a\rightarrow 0 } \dfrac { s }{ a^{ 2 } } =\left[ \dfrac { { ({ e }^{ \dfrac { a }{ 2 } }-1) }^{ 2 } }{ \dfrac { a }{ 2 } } \right] *\dfrac { { a }^{ 2 } }{ 4{ a }^{ 2 } } =\dfrac { 1 }{ 4 } $$
Question 6
1 / -0

The area bounded by the curves $$x^2+y^2\le 8$$ and $$y^2\ge 4x$$ lying in the first quadrant is not equal to

A
$$\displaystyle 32\left( \frac { \pi }{ 8 } -\frac1 3\right) $$

B
$$\displaystyle \frac { 32 }{ 3 } \left( \frac { 3\pi }{ 8 } -1 \right) $$

C
$$\displaystyle 4\pi -\frac { 32 }{ 3 } $$

D
$$\displaystyle \frac { 1 }{ 3 } \left( 12\pi -32 \right) $$

Solution

Given $$x^2+y^2\le 8x$$
Let $$x^2+y^2=8x$$ ...(1)
$$y^2\ge 4x, y^2=4x$$ ...(2)
From (1) and (2), we have
$$x^2=4x\Rightarrow x=0,x=4$$
$$\therefore$$ Set of points $$($$ or points of intersection $$)$$ are $$(0,0),(4,4)$$.
Required area $$\displaystyle =\int _{ 0 }^{ 4 }{ \sqrt { { 4 }^{ 2 }-{ \left( x-4 \right) }^{ 2 } } } -\int _{ 0 }^{ 4 }{ 2\sqrt { x } dx } \quad \quad \quad \left( \because { x }^{ 2 }+{ y }^{ 2 }=8x \right) $$
$$\displaystyle \Rightarrow { \left( x-4 \right) }^{ 2 }+{ y }^{ 2 }={ 4 }^{ 2 }\Rightarrow { y }^{ 2 }={ 4 }^{ 2 }-{ \left( x-4 \right) }^{ 2 }$$
$$\displaystyle ={ \left[ \frac { \left( x-4 \right) }{ 2 } \sqrt { { 4 }^{ 2 }-{ \left( x-4 \right) }^{ 2 } } +\frac { 16 }{ 2 } \sin ^{ -1 }{ \left( \frac { x-4 }{ 4 } \right) } \right] }_{ 0 }^{ 4 }-2\times \frac { 2 }{ 3 } { \left[ { x }^{ 3/2 } \right] }_{ 0 }^{ 4 }$$
$$\displaystyle =8\times \frac { \pi }{ 2 } -\frac { 4 }{ 3 } \times { \left( 4 \right) }^{ 3/2 }=\left( 4\pi -\frac { 32 }{ 3 } \right) =\frac { 32 }{ 3 } \left( \frac { 3\pi }{ 8 } -1 \right) $$ sq. units.
Question 7
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Area enclosed by the curves $$\displaystyle y=\ln x;y=\ln\left | x \right |;y=\left | \ln x \right |$$ and $$\displaystyle y=\left | \ln\left | x \right | \right |$$ is equal to

A
$$2$$

B
$$4$$

C
$$8$$

D
Cannot be determined

Solution

Required area = $$2 \displaystyle \int_{0}^{1} |log |x||dx$$
$$=2[(x|log|x||)]_{0}^{1}- \displaystyle\int_{0}^{1}\left(-\dfrac{1}{2}\right)xdx$$
$$=2[(1-0)+(x)_{0}^{1}]$$
$$= 4 sq.units$$
Question 8
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The area of the figure bounded by the lines $$x= 0,\: x= \dfrac{\pi}{2},\: f\left ( x \right )= \sin x$$ and $$g\left ( x \right )= \cos x$$ is

A
$$2\left ( \sqrt{2}-1 \right )$$

B
$$ \sqrt{3}-1$$

C
$$2\left ( \sqrt{3}-1 \right )$$

D
$$2\left ( \sqrt{2}+1 \right )$$

Solution

For $$x=0\ to\ \dfrac{\pi}{4}\ \cos x>\sin x\ and\ for\ x=\dfrac{\pi}{4}\ to\ \dfrac{\pi}{2} \sin x>\cos x$$

So, Required area $$=\displaystyle \int _{ 0 }^{ \dfrac { \pi }{ 4 } }{ \left( \cos { x } -\sin { x } \right) dx } +\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ \left( -\cos { x } +\sin { x } \right) dx } $$

$$\displaystyle ={ \left[ \sin { x } +\cos { x } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }+{ \left[ -\cos { x } -\sin { x } \right] }_{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }=2\left( \sqrt { 2 } -1 \right) $$
Question 9
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The area of the figure bounded by the curves $$\displaystyle y=\ln x$$ & $$\displaystyle y=\left ( \ln x \right )^{2}$$ is

A
$$e+1$$

B
$$e-1$$

C
$$3-e$$

D
$$1$$

Solution

Given curves, $$\displaystyle y=\ln x$$ & $$\displaystyle y=\left ( \ln x \right )^{2}$$
Points of intersection of given curves are, $$(1,0)$$ and $$(e,1)$$
Hence required area is,
$$A = \displaystyle \int_{1}^{e}\left ( \ln x-(\ln x)^{2} \right )dx$$
On solving it by parts we get
$$A\displaystyle =[x\ln x-x]_{1}^{e}-\left[x(\ln x)^{2}-\int_{1}^{e} 2\ln x dx\right]$$
$$A\displaystyle =[x\ln x-x]_{1}^{e}-\left[x(\ln x)^{2}- 2x\ln x +2x\right]_{1}^{e}$$
$$A=\left[x\ln x-x-x(\ln x)^{2}+2x\ln x-2x\right]_{1}^{e}$$
$$A=\displaystyle \left|3x\left ( \ln x-1 \right )\right|_{1}^{e}-|x(\ln x)^{2}|_{1}^{e}=3-e$$
Question 10
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The area of the smaller portion between curves $$x^2 + y^2 = 8$$ and $$y^2 = 2x$$ is

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

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

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

D
$$\displaystyle \pi + \frac{4}{3}$$

Solution

Given curves
$$x^2 + y^2 = 8$$ ....(1)
Center is (0,0) and radius $$r=2\sqrt{2}$$
$$y^2 = 2x$$ ....(2)
Solving (1) and (2), we get
$$x^2+2x=8$$
$$\Rightarrow x=2,-4$$
At $$x=2 \Rightarrow y=2,-2$$
Hence, the point of intersection of the curves is $$(2,2)$$ and $$(2,-2)$$

Now, required area $$=2(area OPA)$$
$$=2[ar(OPB)+ar(PBA)]$$
$$= \displaystyle 2 \int_0^2 \sqrt{2x} dx + 2 \int_{2}^{2 \sqrt 2} \sqrt{8 - x^2} dx$$

$$=2\sqrt{2}[\dfrac{x^{3/2}}{3/2}]_{0}^2 +2 [\dfrac{x}{2}\sqrt{8 - x^2}+4\sin^{-1}{\dfrac{x}{2\sqrt{2}}}]_{2}^{2\sqrt{2}}$$

$$=\dfrac{4\sqrt{2}}{3} (2\sqrt{2})+2(0+{2}{\pi}-2-\pi)$$

Required Area $$= \displaystyle \left( 2 \pi + \frac{4}{3} \right )$$ sq.units