Application of Integrals Test

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Application of Integrals Test - 22

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Weekly Quiz Competition

Question 1
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Area bounded by $$\displaystyle y=2\sqrt { x } $$ and $$ x=3\sqrt { y } $$ is equal to (in sq. units)

A
12

B
8

C
10

D
6

Solution

First we have to find the intersection point.
Hence, $$2\sqrt{x}=\left (\dfrac{x}{3}\right)^{2}$$
$$\Rightarrow 18\sqrt{x}=x^{2}$$
$$\Rightarrow \sqrt{x}(18-x\sqrt{x})=0$$
$$\Rightarrow x=0$$ and $$x\sqrt{x}=18$$
$$\Rightarrow x^{3}=18^{2}$$
Now the area bounded by the curve is
$$I=\displaystyle \int_{0} ^{18^{\frac{2}{3}}} 2\sqrt{x}-\dfrac{x^{2}}{9} dx$$
$$=\left [\dfrac{4x^{\frac{3}{2}}}{3}-\dfrac{x^{3}}{27}\right]_{0} ^{18^{\frac{2}{3}}}$$
$$=\dfrac{4\times 18}{3}-\dfrac{18^{2}}{27}$$
$$=24-12$$
$$=12$$ sq units.
Question 2
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The area between the parabola $$y =x^2$$ and the line $$y = x$$ is

A
$$\dfrac{1}{6}$$ sq. units

B
$$\dfrac{1}{3}$$ sq. units

C
$$\dfrac{1}{2}$$ sq. units

D
None of these

Solution

For points of intersection of the equations $$y=x^2$$ and $$y=x$$
$$\therefore x^2-x=0 \Rightarrow x(x-1)=0$$
$$x=0$$ or $$x=1\Rightarrow Y=0$$ or $$y =1$$
Hence, the coordinates of their points of intersection are O(0, 0) and P (1, 1).
$$\therefore$$ Required area (shaded region)
$$= \displaystyle \int_0^1 (x - x^2) dx = \left [ \frac{x^2}{2} - \frac{x^3}{3} \right ]_0^1$$
$$= \displaystyle \left [ \left ( \frac{1}{2} - \frac{1}{3} \right ) - 0 \right ] = \frac{1}{6} $$ sq. units
Question 3
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Area lying in the first quadrant and bounded by the circle $$x^2 + y^2 = 4$$ and the lines $$x = 0$$ and $$x = 2$$ is

A
$$\pi$$

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

C
$$\dfrac {\pi}{3}$$

D
$$\dfrac {\pi}{4}$$

Solution

The area bounded by the circle and the lines, $$x=0$$ and $$x=2$$, in the first quadrant is represented as shaded region in the plot.
$$\therefore$$ Area $$OAB=\int_0^2y dx$$
$$=\displaystyle \int_0^2\sqrt {4-x^2}dx$$
$$=\left [\dfrac {x}{2}\sqrt {4-x^2}+\dfrac {4}{2}\sin^{-1}\dfrac {x}{2}\right ]_0^2$$
$$=2\left (\dfrac {\pi}{2}\right )=\pi $$sq. units
Thus, the correct answer is A.
Question 4
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The area bounded by $${y}^{2}=4x$$ and $${x}^{2}=4y$$ is

A
$$\cfrac { 20 }{ 3 } $$ sq. units

B
$$\cfrac { 16 }{ 3 } $$ sq. units

C
$$\cfrac { 14 }{ 3 } $$ sq. units

D
$$\cfrac { 10 }{ 3 } $$ sq. units

Solution

The intersection point of a curves $${y}^{2}=4x$$ and $${x}^{2}=4y$$ are $$O(0,0)$$ and $$A(4,4)$$
$$\therefore$$ Required area $$\displaystyle \int _{ 0 }^{ 4 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx } $$
$$=\displaystyle \int _{ 0 }^{ 4 }{ \left( \sqrt { 4x } -\cfrac { { x }^{ 2 } }{ 4 } \right) dx } $$
$$={ \left[ \cfrac { 2{ x }^{ 3/2 } }{ 3/2 } -\cfrac { { x }^{ 3 } }{ 12 } \right] }_{ 0 }^{ 4 }\quad $$
$$=\cfrac { 4 }{ 3 } \left[ { 4 }^{ 3/2 }-0 \right] -\cfrac { 1 }{ 12 } \left[ { 4 }^{ 3 }-0 \right] $$
$$=\cfrac { 32 }{ 3 } -\cfrac { 16 }{ 3 } =\cfrac { 16 }{ 3 } $$sq. units
Question 5
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The area of the region bounded by the y-axis, $$y = \cos x$$ and $$y = \sin x, 0\leq x \leq \dfrac {\pi}{2}$$ is

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

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

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

D
$$\sqrt {2}$$

Solution

Refer image 1.
Finding point of Intersection $$B$$
Solving
$$y=\cos x$$ and $$y=\sin x$$
$$\cos x=\sin x$$
Refer image 2.
At $$x=\dfrac{\pi}{4}$$, both are equal
Also,
$$y=\cos x=\cos \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$$
So, $$B=\left(\dfrac{\pi}{4},\dfrac{1}{\sqrt{2}}\right)$$
Refer image 3.
Refer image 4.
Area $$ABCO$$
Area ABCO $$\displaystyle \int^{\pi/4}_0 ydx$$
Here, $$y=\cos x$$
Thus,
Area ABCO= $$\displaystyle \int^{\pi/4}_0 \cos x dx$$
$$=[\sin x]^{\pi/4}_0$$
$$=[\sin\dfrac{\pi}{4}-\sin 0]$$
$$=\dfrac{1}{\sqrt{2}}-0$$
$$=\dfrac{1}{\sqrt{2}}$$

Refer image 5.
Area $$BCO$$
Area BCO $$\displaystyle \int^{\pi/4}_0 ydx$$
Here, $$y=\sin x$$
Thus,
Area BCO = $$\displaystyle \int^{\pi/4}_0 \sin x dx$$
$$=-[\cos x]^{\pi/4}_0$$
$$=-[\cos\dfrac{\pi}{4}-\cos(0)]$$
$$=-[\dfrac{1}{\sqrt{2}}-1]$$
$$=1-\dfrac{1}{\sqrt{2}}$$
Therefore
Area Required = Area ABCD - Area BCO
$$=\dfrac{1}{\sqrt{2}}-[1-\dfrac{1}{\sqrt{2}}]$$
$$=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-1$$
$$=\dfrac{2}{\sqrt{2}}-1$$
$$=\sqrt{2}-1$$
$$\therefore$$ Option $$B$$ is correct.
Question 6
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Area bounded between the curve $$x^2=y$$ and the line $$y=4x$$ is

A
$$\displaystyle\frac{32}{3}$$sq unit

B
$$\displaystyle\frac{1}{3}$$sq unit

C
$$\displaystyle\frac{8}{3}$$sq unit

D
$$\displaystyle\frac{16}{3}$$sq unit

Solution

Given curves are $$x^2=y$$ and $$y=4x$$
Intersection points are $$(0, 0)$$ and $$(4, 16)$$
$$\therefore$$ Required area $$=\int^4_0(4x-x^2)dx$$
$$=\left[\displaystyle\frac{4x^2}{2}-\frac{x^3}{3}\right]^4_0$$
$$=\displaystyle\left[32-\frac{64}{3}\right]$$
$$=\displaystyle\frac{32}{3}$$sq unit
Question 7
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The area enclosed between the curve $$\displaystyle y=1+{ x }^{ 2 }$$, the y-axis and the straight line $$\displaystyle y=5$$ is given by

A
$$\displaystyle \frac { 14 }{ 3 } $$ sq unit

B
$$\displaystyle \frac { 7 }{ 3 } $$ sq unit

C
$$\displaystyle 5$$ sq unit

D
$$\displaystyle \frac { 16 }{ 3 } $$ sq unit

Solution

Given curve is $$\displaystyle y=1+{ x }^{ 2 }$$ and line $$\displaystyle y=5$$
$$\displaystyle \therefore $$ Required area $$\displaystyle =\int _{ 1 }^{ 5 }{ x } dy=\int _{ 1 }^{ 5 }{ \sqrt { y-1 } dx } $$
$$\displaystyle ={ \left[ \frac { { \left( y-1 \right) }^{ { 3 }/{ 2 } } }{ { 3 }/{ 2 } } \right] }_{ 1 }^{ 5 }=\frac { 2 }{ 3 } \left[ { \left( 4 \right) }^{ { 3 }/{ 2 } }-0 \right] $$
$$\displaystyle =\frac { 16 }{ 3 } $$ sq unit.
Question 8
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The area bounded by the parabolas $$y=4x^2,\,y=\dfrac{x^2}{9}$$ and line $$y=2$$ is

A
$$\dfrac{5\sqrt{2}}{3}$$ sq units

B
$$\dfrac{10\sqrt{2}}{3}$$ sq units

C
$$\dfrac{15\sqrt{2}}{3}$$ sq units

D
$$\dfrac{20\sqrt{20}}{3}$$ sq units

Solution

$$y=4x^2\;\;\;\dots(i)$$
$$y=\dfrac{x^2}{4}\;\;\;\dots(ii)$$
$$\therefore\;A=\int_0^2\begin{bmatrix}\dfrac{\sqrt{y}}{2}-3\sqrt{y}\end{bmatrix}dy$$
$$=\begin{pmatrix}\dfrac{1}{2}-3\end{pmatrix}\int_0^2\sqrt{y}dy$$
$$=\begin{pmatrix}\dfrac{-5}{2}\end{pmatrix}.\dfrac{2}{3}\begin{bmatrix}y^{{3}/{2}}\end{bmatrix}_0^2$$
$$=-\dfrac{5}{3}(2\sqrt{2}-0)=\begin{vmatrix}\dfrac{-10\sqrt{2}}{3}\end{vmatrix}$$
$$=\dfrac{10\sqrt{2}}{3}\Rightarrow\text{Area of bounded Area}=2A=\dfrac{20\sqrt{20}}{3}$$
Question 9
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The area bounded by the curve $$y=x|x|$$, $$x$$-axis and the ordinates $$x=-1$$ and $$x=1$$ is given by

A
$$0$$

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

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

D
$$\dfrac {4}{3}$$

Solution

$$y=x^2$$ if $$x > 0$$ and $$y=-x^2$$ if $$x < 0$$
Required area $$=\displaystyle \int_{-1}^{1}y dx$$

$$=\displaystyle \int_{-1}^{1}x|x|dx$$

$$=\displaystyle \int_{-1}^{0} x^2 dx+\int_{0}^{1}x^2dx$$

$$=\left [\dfrac {x^3}{3}\right ]_{-1}^{0}+\left [\dfrac {x^3}{3}\right ]_{0}^{1}$$

$$=-\left (-\dfrac {1}{3}\right )+\dfrac {1}{3}$$

$$=\dfrac {2}{3}$$ sq. units

Thus, the correct answer is C.
Question 10
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Area bounded by the curves $$y=x^3$$, the $$x$$-axis and the ordinates $$x=-2$$ and $$x=1$$ is

A
$$-9$$

B
$$-\dfrac {15}{4}$$

C
$$\dfrac {15}{4}$$

D
$$\dfrac {17}{4}$$

Solution

Required area $$=-\displaystyle \int_{-2}^{0}x^3 dx+\int_{0}^{1}x^3 dx$$
$$=-\left [\dfrac {x^4}{4}\right ]_{-2}^{0}+\left [\dfrac {x^4}{4}\right ]_{0}^{1}$$
$$=-\left [0-\dfrac {(-2)^4}{4}\right ]+\dfrac{1}{4}$$
$$=\left (\dfrac {1}{4}+4\right )=\dfrac {17}{4}$$sq. units
Thus, the correct answer is D.