Application of Integrals Test

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Application of Integrals Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The area of the circle $$x^2+y^2=16$$ exterior to the parabola $$y^2=6x$$ is

A
$$\dfrac {4}{3}(4\pi -\sqrt 3)$$

B
$$\dfrac {4}{3}(4\pi +\sqrt 3)$$

C
$$\dfrac {4}{3}(8\pi -\sqrt 3)$$

D
$$\dfrac {4}{3}(8\pi +\sqrt 3)$$

Solution

The given equations are
$$x^2+y^2=16........ (1)$$
$$y^2=6x ....... (2)$$

Area bounded by the circle and parabola
$$=2[Area (OADO)+Area (ADBA)]$$

$$=2[\displaystyle \int_{0}^{2}\sqrt {16x}dx+\int_{2}^{4}\sqrt {16-x^2}dx]$$

$$=2\left [\sqrt 6\left \{\dfrac {x^{\dfrac {3}{2}}}{\dfrac {3}{2}}\right \}_{0}^{2}\right ]+2\left [\dfrac {x}{2}\sqrt {16-x^2}+\dfrac {16}{2}\sin^{-1}\dfrac {x}{4}\right ]_{2}^{4}$$

$$=2\sqrt 6\times \dfrac {2}{3}\left [x^{\dfrac {3}{2}}\right ]_{0}^{2}+2\left [8\cdot \dfrac {\pi}{2}-\sqrt {16-4}-8 \sin^{-1}\left (\dfrac {1}{2}\right )\right ]$$

$$=\dfrac {4\sqrt 6}{3}(2\sqrt 2)+2\left [4\pi-\sqrt {12}-8\dfrac {\pi}{6}\right ]$$

$$=\dfrac {16\sqrt 3}{3}+8\pi - 4\sqrt 3-\dfrac {8}{3}\pi$$

$$=\dfrac {4}{3}[4\sqrt 3+6\pi - 2\sqrt 3-2\pi ]$$$$=\dfrac {4}{3}[\sqrt 3+4\pi]$$

$$=\dfrac {4}{3}[4\pi +\sqrt 3]$$ sq. units

Area of circle $$=\pi(r)^2$$
$$=\pi(4)^2$$
$$=16 \pi$$ sq. units

$$\therefore$$ Required area $$=16\pi -\dfrac {4}{3}[4\pi +\sqrt 3]$$
$$=\dfrac {4}{3}[4\times 3\pi -4\pi -\sqrt 3]$$

$$=\dfrac {4}{3}(8\pi -\sqrt 3)$$ sq. units

Thus, the correct answer is C.
Question 2
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The area of the region bounded by the curves $$y={ x }^{ 2 }$$ and $$x={ y }^{ 2 }$$ is

A
$$\dfrac {1}{3}$$

B
$$\dfrac {1}{2}$$

C
$$\dfrac { 1 }{ 4 }$$

D
$$3$$

Solution

Given curves are $$y={x}^{2}$$ and $$x={y}^{2}$$, which is the form of parabola.
The point of intersection, $$x={ \left( { x }^{ 2 } \right) }^{ 2 }$$
$$\Rightarrow x={ x }^{ 4 }\Rightarrow x\left( 1-{ x }^{ 3 } \right) =0$$
$$\Rightarrow x=0$$ and $$1={ x }^{ 3 }$$
$$\Rightarrow x=0$$ and $$x=1$$
When $$x=0$$, then $$y=0$$
When $$x=1$$, then $$y={ 1 }^{ 2 }=1$$
$$\therefore $$ The point of intersection is $$\left( 0,0 \right) $$ and $$\left( 1,1 \right) $$.
$$\therefore $$ Area of shaded region
$$=\displaystyle\int _{ 0 }^{ 1 }{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) dx } $$
$$=\displaystyle\int _{ 0 }^{ 1 }{ \left[ \sqrt { x } -{ x }^{ 2 } \right] dx } ={ \left[ \dfrac { { x }^{ { 3 }/{ 2 } } }{ { 3 }/{ 2 } } -\dfrac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 1 }$$
$$=\dfrac { 2 }{ 3 } { \left( 1 \right) }^{ { 3 }/{ 2 } }-\dfrac { { \left( 1 \right) }^{ 3 } }{ 3 } -0-0$$
$$=\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 } $$sq units
Question 3
1 / -0

Area of the region bounded by $$y = |x|$$ and $$y = |x| + 2$$, is

A
$$4\ sq. units$$

B
$$3\ sq. units$$

C
$$2\ sq. units$$

D
$$1\ sq. units$$

Solution

The graph is shown in image

In $$1^{st}$$ quadrant , intersection point of $$y=x$$ and $$y=-x+2$$ is $$(1,1)$$

similarly in $$2^{nd}$$ quadrant , intersection point of $$Y=X$$ and $$y=-x+2$$ is $$(-1,1)$$

now $$A=2\Bigg[\int_{0}^{1}\bigg[-x+2\bigg]-\int_{0}^{1}\bigg[x\bigg]\Bigg]$$
$$A=2\times{1}$$
$$A=2 sq.unit$$
Question 4
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The area included between the parabolas $$x^2=4y$$ and $$y^2=4x$$ is (in square units)

A
$$\dfrac{4}{3}$$

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

C
$$\dfrac{16}{3}$$

D
$$\dfrac{8}{3}$$

Solution

One intersection point of the two parabolas is the origin, while the other can be found out by substituting either equation into the other.
Thus we have $$\left (\dfrac{x^2}{4}\right)^2 = 4x$$
which yields $$x^4 = 64x$$ i.e. $$x = 0,4$$
Area included between the two curves can be found out by $$\displaystyle \int_0^4 \left (2\sqrt{x} - \dfrac{x^2}{4}\right) dx$$
$$=\left [ \dfrac{4x^{1.5}}{3} - \dfrac{x^3}{12}\right] _0^4$$
$$= \dfrac{32}{3} - \dfrac{64}{12}$$
$$= \dfrac{32}{6}$$
Question 5
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The area enclosed between the curves $$\displaystyle y={ x }^{ 3 }$$ and $$\displaystyle y=\sqrt { x } $$ is, (in square units):

A
$$\displaystyle \frac { 5 }{ 3 } $$

B
$$\displaystyle \frac { 5 }{ 4 } $$

C
$$\displaystyle \frac { 5 }{ 12 } $$

D
$$\displaystyle \frac { 12 }{ 5 } $$

Solution

Required area $$\displaystyle \int_0^1 (\sqrt x -x^3)dx=\left( \dfrac{2x^{3/2}}{3}-\dfrac{x^4}{4}\right)_0^1=\dfrac{2}{3}-\dfrac{1}{4}-0=\dfrac{5}{12}$$
Question 6
1 / -0

The area included between the parabolas $$\displaystyle { y }^{ 2 }=4x$$ and $$\displaystyle { x }^{ 2 }=4y$$ is

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

B
$$\displaystyle 8$$ sq unit

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

D
$$\displaystyle 12$$ sq unit

Solution

We know that, the area of region bounded by the parabolas $$\displaystyle { y }^{ 2 }=4ax$$ and $$\displaystyle { x }^{ 2 }=4by$$ is $$\displaystyle \frac { 16 }{ 3 } ab$$ sq.unit.
Therefore, $$\displaystyle { y }^{ 2 }=4ax$$ and $$\displaystyle { x }^{ 2 }=4y$$ is $$\displaystyle \frac { 16 }{ 3 } $$ sq.unit.
$$\displaystyle \left( \because a=1,b=1 \right) $$
Question 7
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The area of the region bounded by the graph of $$y = \sin x$$ and $$y = \cos x$$ between $$x = 0$$ and $$x = \dfrac {\pi}{4}$$ is

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

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

C
$$2\sqrt {2} - 2$$

D
$$2\sqrt {2} + 2$$

Solution

Area bounded by the curves $$y=\sin x$$ and $$y=\cos x$$ is given by
$$A=\displaystyle \int _{0} ^{\tfrac{\pi}{4}} (\cos x-\sin x) dx$$
$$=[\sin x+\cos x]_{0} ^{\tfrac{\pi}{4}}$$
$$=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-1$$
$$=\dfrac{2}{\sqrt{2}}-1$$
$$=\sqrt{2}-1$$ sq. units.
Question 8
1 / -0

Area of the region satisfying $$x \le 2, y \le |x|,x-axis$$ and $$x\ge 0$$ is:

A
$$4$$ sq unit

B
$$1$$ sq unit

C
$$2$$ sq unit

D
None of these

Solution

Required area = Area of shaded region OAB
$$=^2_oydx=^2_0xdx=\frac{x^2}{2}^2_0$$
$$=$$2 sq unit
Alternate Solution
Required area = Area of $$\Delta$$ OAB
$$=\frac{1}{2}\times 2\times 2$$
$$=$$ 2 sq unit
Question 9
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The area of the region bounded by the curves $$y = x^{3}, y = \dfrac {1}{x}, x = 2$$ is

A
$$4 - \log_{e}2$$

B
$$\dfrac {1}{4} + \log_{e}2$$

C
$$3 - \log_{e}2$$

D
$$\dfrac {15}{4} - \log_{e}2$$

Solution

First of all we draw the graph,
$$y = x^{3}, y = \dfrac {1}{x}, x = 2$$
$$\therefore$$ Required area i.e., $$(OMPNO)$$
$$=\displaystyle \int_{0}^{1} x^{3} dx + \int_{1}^{2} \dfrac {1}{x}dx$$

$$=\left [\dfrac {x^{4}}{4}\right ]_{0}^{1} + [\log_{e}x]_{1}^{2} = \dfrac {1}{4} + \log_{e} 2$$
Question 10
1 / -0

The area (in square units) bounded by the curves $$y^{2} = 4x$$ and $$x^{2} = 4y$$ is

A
$$\dfrac {64}{3}$$

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

C
$$\dfrac {8}{3}$$

D
$$\dfrac {2}{3}$$

Solution

$$y^{2} = 4x$$
$$x^{2} = 4y$$
solving two equations
$$\dfrac {y^{2}}{4} = x$$
$$\left (\dfrac {y^{2}}{4}\right )^{2} = 4y$$
$$\dfrac {y^{4}}{4\times 4} = 4y$$
$$y^{4} = 64y$$
$$y(y^{2} - 64) = 0$$
$$y = 0, y^{3} = 64$$
$$y = 4$$
$$\int_{0}^{4} \left (\dfrac {y^{2}}{4} - 2\sqrt {y}\right )dy$$
$$= \left [\dfrac {y^{3}}{4\times 3} - \dfrac {y^{\dfrac {B}{2}}}{3}\right ]_{0}^{4}$$
$$= \left |\dfrac {-16}{3}\right | = \dfrac {16}{3}$$