Application of Integrals Test

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Application of Integrals Test - 60

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Weekly Quiz Competition

Question 1
1 / -0

What is the area of the region bounded by the parabola $${ y }^{ 2 }=6(x-1)$$ and $${ y }^{ 2 }=3x$$

A
$$\cfrac { \sqrt { 6 } }{ 3 } $$

B
$$\cfrac { 2\sqrt { 6 } }{ 3 } $$

C
$$\cfrac { 4\sqrt { 6 } }{ 3 } $$

D
$$\cfrac { 5\sqrt { 6 } }{ 3 } $$

Solution

Solving $$y^2=6(x-1)$$ and $$y^2=3x$$ we get

$$6(x-1)=3x$$

$$\Rightarrow x=2$$

Hence $$y=\pm \sqrt{6}$$

$$y^2=6(x-1)\Rightarrow x=1+\dfrac{y^2}{6}$$ and

$$y^2=3x \Rightarrow x=\dfrac{y^2}{3}$$

Area $$=\int_{-\sqrt{6}}^{\sqrt{6}}\left(1+\dfrac{y^2}{6}-\dfrac{y^2}{3}\right)dy$$

$$=2\int_{0}^{\sqrt{6}}\left(1-\dfrac{y^2}{6}\right)dy$$

$$=2\left[y-\dfrac{y^3}{18}\right]_{0}^{\sqrt{6}}$$

$$=2 \times \dfrac{2\sqrt{6}}{3}=\dfrac{4\sqrt{6}}{3}$$

Area $$=\dfrac{4\sqrt{6}}{3}$$
Question 2
1 / -0

Area bounded by the curves $$\displaystyle y = \left[ \frac{x^2}{64} + 2 \right]$$ ([.] denotes the greatest integer function) $$y = x - 1$$ and $$x = 0$$ above the x-axis is

A
2

B
3

C
4

D
none of these
Question 3
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The area bounded by the curves $$x= a \cos^3t, y= a \sin^3 t$$ is

A
$$\dfrac{3\pi a^2}{8}$$

B
$$\dfrac{3\pi a^2}{16}$$

C
$$\dfrac{3\pi a^2}{32}$$

D
None of the above

Solution

$$x=a\cos^{3}t$$   , $$y=a\sin^{2}t$$
$$x^{\dfrac{2}{3}}+y^{\dfrac{2}{3}}=a^{\dfrac{2}{3}}$$
$$A=\int_{0}^{2\pi}{x}dy$$
$$A=a^{2}\int_{0}^{2\pi}{\cos^{3}t\times3\sin^{2}t\times\cos t}dt$$
$$=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt$$      $$\rightarrow(1)$$
similarily   $$A=\int_{0}^{2\pi}{y}dx$$
$$=3a^{2}\int_{0}^{2\pi}{\sin^{2}t\times\cos^{2}t}dt$$        $$\rightarrow(2)$$
Adding (1) and (2)
$$2A=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt
$$A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{\sin^{2}2}dt$$    $$\rightarrow(3)$$
Similarily $$A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{\cos^{2}t}dt$$       $$\rightarrow(4)$$
By putting $$t=t+\dfrac{\pi}{4}$$ in (3)
Adding (3) and (4)
$$2A=\dfrac{3a^{2}}{8}\int_{0}^{2\pi}{dt}$$
$$A=\dfrac{3a^{2}}{8}\pi$$
Question 4
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$$Let\quad f(x)=2-\left| x-1 \right| and\quad g(x)={ \left( x-1 \right) }^{ 2 },\quad then\quad $$

A
area bounded by $$f(x)$$ and $$g(x)$$ is $$\cfrac { 7 }{ 6 } $$

B
area bounded by $$f(x)$$ and $$g(x)$$ is $$\cfrac { 7 }{ 3 } $$

C
area bounded by $$f(x)$$$$g(x)$$ and $$x-$$ axis is $$\cfrac { 5 }{ 3 } $$

D
area bounded by $$f(x)$$$$g(x)$$ and $$x-$$ axis is $$\cfrac { 5 }{ 6 } $$

Solution

Area of parabola w.r.t x axis $$=g(X)$$
$$\int_{0}^{2}{(x-1)^{2}dx}$$
$$\Rightarrow$$$$\left[\dfrac{(x-1)^{3}}{3}\right]_{0}^{2}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{2}{3}$$
Area of $$f(x)$$
$$=2\times1+\dfrac{1}{2}\times2\times1$$
$$\Rightarrow$$ $$2+1=3$$
Area bounded by $$f(x)$$ and $$g(x)$$
$$=3-\dfrac{2}{3}=\dfrac{7}{3}$$
Question 5
1 / -0

The area bounded by $$y=\sin ^{ -1 }{ x } , y=\cos ^{ -1 }{ x }$$ and the $$x-axis$$, is given by

A
$$\sqrt { 2 }$$

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

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

D
$$2\sqrt { 2 }$$
Question 6
1 / -0

The area bounded by the curves $$y={ \left( x-1 \right) }^{ 2 },y={ \left( x+1 \right) }^{ 2 }$$ and $$y=\dfrac { 1 }{ 4 }$$ is

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

B
$$\dfrac { 2 }{ 3 } sq\ unit$$

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

D
$$\dfrac { 1 }{ 5 } sq\ unit$$
Question 7
1 / -0

Area common to the circle $$x^{2}+y^{2}=64$$ and the parabola $$y^{2}=4x$$ is

A
$$\dfrac{16}{3}(4\pi + \sqrt{3})$$

B
$$\dfrac{16}{3}(8\pi + \sqrt{3})$$

C
$$\dfrac{16}{3}(4\pi - \sqrt{3})$$

D
$$none\ of\ these$$
Question 8
1 / -0

If $$f\left(x\right)=$$max$$\left\{\sin{x},\cos{x},\dfrac{1}{2}\right\}$$, then the area of the region bounded by the curves $$y=f\left(x\right),x-$$axis $$y-$$axis and $$x=2\pi$$ is

A
$$\left(\dfrac{5\pi}{12}+3\right)$$.sq.unit

B
$$\left(\dfrac{5\pi}{12}+\sqrt{2}\right)$$.sq.unit

C
$$\left(\dfrac{5\pi}{12}+\sqrt{3}\right)$$.sq.unit

D
$$\left(\dfrac{5\pi}{12}+\sqrt{2}+\sqrt{3}\right)$$.sq.unit

Solution

$$\because f\left(x\right)=$$max$$\left\{\sin{x},\cos{x},\dfrac{1}{2}\right\}$$
Interval value of $$f\left(x\right)$$
For $$0\le x<\dfrac{\pi}{4}, \cos{x}$$
For $$\dfrac{\pi}{4}\le x<\dfrac{5\pi}{6}, \sin{x}$$
For $$\dfrac{5\pi}{6}\le x<\dfrac{5\pi}{3}, \dfrac{1}{2}$$
For $$\dfrac{5\pi}{3}\le x<2\pi, \cos{x}$$
Hence, required area
$$=\int_{0}^{\frac{\pi}{4}}{\cos{x}dx}+\int_{\frac{\pi}{4}}^{\frac{5\pi}{6}}{\sin{x}dx}+\int_{\frac{5\pi}{6}}^{\frac{5\pi}{3}}{\dfrac{1}{2}dx}+\int_{\frac{5\pi}{3}}^{2\pi}{\cos{x}dx}$$
$$=\left[\sin{x}\right]_{0}^{\frac{\pi}{4}}-\left[\cos{x}\right]_{\frac{\pi}{4}}^{\frac{5\pi}{6}}+\dfrac{1}{2}\left[x\right]_{\frac{5\pi}{6}}^{\frac{5\pi}{3}}+\left[\sin{x}\right]_{\frac{5\pi}{3}}^{2\pi}$$
$$=\left(\dfrac{1}{\sqrt{2}}-0\right)-\left(-\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\right)+\dfrac{1}{2}\left(\dfrac{5\pi}{3}-\dfrac{5\pi}{6}\right)+\left(0+\dfrac{\sqrt{3}}{2}\right)$$
On simplification, we get
$$=\left(\dfrac{5\pi}{12}+\sqrt{2}+\sqrt{3}\right)$$.sq.unit.
Question 9
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The parabola $$y=\dfrac{x^2}{2}$$ divides the circle $$x^2+y^2=8$$ into two parts. Find the area of both parts.

A
$$6\pi+\dfrac{4}{3}$$ , $$2\pi-\dfrac{4}{3}$$

B
$$6\pi-\dfrac{4}{3}$$ , $$2\pi+\dfrac{4}{3}$$

C
$$6\pi+\dfrac{2}{3}$$ , $$2\pi-\dfrac{2}{3}$$

D
$$6\pi-\dfrac{2}{3}$$ , $$2\pi+\dfrac{2}{3}$$

Solution

Let us first find the x-values of the points of intersection using the given equations of parabola $$y=\dfrac {x^2}{2}$$ and circle $$x^2+y^2=8$$ as follows:

$$x^{ 2 }+y^{ 2 }=8\\ \Rightarrow x^{ 2 }+\left( \dfrac { x^{ 2 } }{ 2 } \right) ^{ 2 }=8\quad \quad \quad \quad \quad \left( \because \quad y=\dfrac { x^{ 2 } }{ 2 } \right) \\ \Rightarrow x^{ 2 }+\frac { x^{ 4 } }{ 4 } =8\\ \Rightarrow 4x^{ 2 }+x^{ 4 }=32$$
$$\Rightarrow x^{ 4 }+4x^{ 2 }-32=0\\ \Rightarrow x^{ 4 }+8x^{ 2 }-4x^{ 2 }-32=0\\ \Rightarrow x^{ 2 }(x^{ 2 }+8)-4(x^{ 2 }+8)=0\\ \Rightarrow (x^{ 2 }-4)(x^{ 2 }+8)=0\\ \Rightarrow (x^{ 2 }-4)=0,\quad (x^{ 2 }+8)=0\\ \Rightarrow x^{ 2 }=4,\quad x^{ 2 }=-8\\ \Rightarrow x=\pm \sqrt { 4 } \\ \Rightarrow x=\pm 2$$

Let $$A_1$$ be the area of the region inside the circle and above the parabola and $$A_2$$ be the area of the region inside the circle and below the parabola. Then we have,

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

We know that the area of the circle is $$\pi r^2$$, therefore the area of the circle $$x^2+y^2=8$$ with radius $$r=\sqrt {8}$$ is:

$$\pi \left( \sqrt { 8 } \right) ^{ 2 }=8\pi$$

Thus, we have

$$A_{ 2 }=8\pi -\left( 2\pi +\dfrac { 4 }{ 3 } \right) =6\pi -\dfrac { 4 }{ 3 }$$

Hence, the area of parabola and circle is $$2\pi +\dfrac { 4 }{ 3 }$$ and $$6\pi -\dfrac { 4 }{ 3 }$$ respectively.
Question 10
1 / -0

The area of the region lying between the line x-y+2=0 and the curve x=$$\sqrt y $$.

A
9

B
9/2

C
10/3

D
none of these

Solution