Application of Integrals Test

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Application of Integrals Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by the curve $$y = 2x - {x^2}$$ and the line $$y = x$$ is

A
$$\dfrac{1}{2}$$

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

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

D
$$\dfrac{1}{6}$$

Solution
Question 2
1 / -0

Area bounded by the curve $$y = \sin ^ { - 1 } x , y - a x i s$$ and $$y = \cos ^ { - 1 } x$$ is equal to

A
$$( 2 + \sqrt { 2 } )$$ sq. unit

B
$$( 2 - \sqrt { 2 } )$$ sq. unit

C
$$( 1 + \sqrt { 2 } )$$ sq. unit

D
$$( \sqrt { 2 } - 1 )$$ sq. unit

Solution
Question 3
1 / -0

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

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

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

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

D
$$1$$

Solution
Question 4
1 / -0

Find the area enclosed between $$y=1+\left| x+1 \right| ,y=4$$.

A
$$9$$ sq. units

B
$$25$$ sq. units

C
$$18$$ sq. units

D
$$21$$ sq. units

Solution

Given $$y=1+\left| x+1 \right| $$
$$=\begin{cases} -x,\quad \quad x<-1 \\ x+2,\quad \quad x>-1 \end{cases}$$
Area $$=24-\left[ \int _{ -4 }^{ -1 }{ -x } dx+\int _{ -1 }^{ 2 }{ \left( x+2 \right) dx } \right] $$
$$==24-\left[ -\cfrac { 1 }{ 2 } (1-16)+{ \left[ \cfrac { { x }^{ 2 } }{ 2 } +2x \right] }_{ -1 }^{ 2 } \right] =24-\left[ \cfrac { 15 }{ 2 } +2+4-\cfrac { 1 }{ 2 } +2 \right] $$
$$=24-\left( 8+7 \right) =24-15=9$$
Question 5
1 / -0

Area bounded by the curves $$y=x\sin x\ and x-axis$$ between $$x=0\ and x=2\pi$$ is

A
$$2\pi$$

B
$$3\pi$$

C
$$4\pi$$

D
None of these

Solution
Question 6
1 / -0

The area of the figure bounded by the parabola $$(y-2)^2=x-1$$, the tangent to it at the point with the ordinate $$x=3$$, and the x-axis is

A
$$7$$ sq. units

B
$$6$$ sq. units

C
$$9$$ sq. units

D
None of these

Solution

Given parabola is $$(y - 2)^2 = x - 1$$
$$\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{2(y-2)}$$

When $$y = 3, x=2$$
$$\therefore \dfrac{dy}{dx} = \dfrac{1}{2}$$.
Tangent at $$(2,3)$$ is $$y - 3 = \dfrac{1}{2}(x-2) \Rightarrow x-2y + 4 = 0$$

$$\therefore$$ required area
$$= \displaystyle \int _0^3 ((y-2)^2 + 1) dy - \int_0^3 (2y-4) dy$$

$$ = \left| \dfrac{(y - 2)^3}{3} + y\right|_0^3 - |y^2 - 4y|_0^3$$

$$= \dfrac{1}{3} + 3 + \dfrac{8}{3} - (9 - 12 ) = 9$$ sq. units.
Question 7
1 / -0

The area bounded by the curve $$y=\cos x$$ and $$y=\sin x$$ between the ordinates $$x=0$$ and $$x=\dfrac {3\pi}{2}$$ is

A
$$4\sqrt {2}+2$$

B
$$4\sqrt {2}-1$$

C
$$4\sqrt {2}+1$$

D
$$4\sqrt {2}-2$$

Solution

$$\displaystyle \int_0^{3\pi/2}|\sin x-\cos x|dx----(1)$$
$$\displaystyle \int_0^{\pi / 4}(\cos x-\sin x)dx +\displaystyle \int_{\pi /4}^{5\pi /4}(\sin x-\cos x)dx$$
$$\rightarrow \ \displaystyle \int_{5\pi /4}^{3\pi/2}(\cos x-\sin x)dx [\sin x+\cos x]_0^{\pi /4}$$
$$\rightarrow \ [\cos x-\sin x]_{\pi /4}^{5\pi /4}\rightarrow {\sin x+\cos x}_{5\pi /4}^{3\pi /4}$$
$$=\sqrt {2}-1+\sqrt {2}+\sqrt {2}+\sqrt {2}-1$$
$$4\sqrt {2}-2$$
Question 8
1 / -0

Let $$y=g(x)$$ be the inverse of a bijective mapping $$f:R\rightarrow R\quad f(x)=3{ x }^{ 3 }+2x$$. The area bounded by graph of $$g(x)$$, the axis and the ordinate at $$x=5$$ is

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

B
$$\displaystyle\frac { 7 }{ 4 } $$

C
$$\displaystyle\frac { 9 }{ 4 } $$

D
$$\displaystyle\frac { 13 }{ 4 } $$

Solution

Required Area $$=\displaystyle \int_{0}^{s}{f^{-1}(x)dx}$$
$$I=\displaystyle \int_{f^{-1}(0)}^{f^{-1}(s)}{f^{-1}(f(t)).f'(t)dt}$$
put $$x=f(t)$$
$$f(0)=0$$ $$\displaystyle \int_{f^{-1}(0)}^{f^{-1}(s)}{(f'(t)dt}$$
$$f^{-1}(0)=0$$ $$\displaystyle \int_{0}^{1}{t(9t^2+2)dt}$$
$$f^{-1}(s)=1$$ $$\displaystyle \int_{0}^{1}{t(9t^3+2t)dt}$$
$$={ \left. \dfrac { 9+4 }{ 4 } +\dfrac { 2t^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }$$
$$=\dfrac{3}{4}+1=\dfrac{13}{4}\ Square\ Unit$$
Question 9
1 / -0

The area bounded by the curves $$|x|+|y|\ge1$$ and $$x^2+y^2\le1$$ is

A
2 sq unit

B
$$\pi$$ sq unit

C
$$(\pi-2)$$ sq unit

D
None of these

Solution

Required area $$\pi \times 1^2-4\times \dfrac {1}{2}\times 1\times 1$$
$$= \pi -2$$
$$\therefore (\pi -2)sq \ unit$$
Question 10
1 / -0

The area of the figure bounded by the parabola $$x = - 2 y ^ { 2 } \text { and } x = 1 - 3 y ^ { 2 } $$ is ?

A
$$1 / 6$$

B
$$2/3$$

C
$$3/2$$

D
$$4/3$$

Solution

$$x = - 2{y^2} \to \left( i \right)\,\,\& \,\,x = 1 - 3{y^2} \to \left( {ii} \right)$$
Equating equation (i) & (ii)
$$\begin{array}{l} -2{ y^{ 2 } }=1-3{ y^{ 2 } } \\ -2{ y^{ 2 } }+3{ y^{ 2 } }=1 \\ { y^{ 2 } }=1 \\ y\pm 1 \end{array}$$
So, area of region bounded by
$$\begin{array}{l} A\Rightarrow 2\int _{ 0 }^{ 1 }{ \left[ { \left( { 1-3{ y^{ 2 } } } \right) -\left( { -2{ y^{ 2 } } } \right) } \right] dy } \\ =2\int _{ 0 }^{ 1 }{ \left( { 1-{ y^{ 2 } } } \right) dy } \\ =2\left[ { y-\frac { { { y^{ 3 } } } }{ 3 } } \right] _{ 0 }^{ 1 }\Rightarrow 2\left[ { \left( { 1-\frac { 1 }{ 3 } } \right) -\left( { 0-\frac { 0 }{ 3 } } \right) } \right] \\ =2\left[ { \frac { 1 }{ 1 } -\frac { 1 }{ 3 } } \right] \Rightarrow 2\left[ { \frac { 2 }{ 3 } } \right] =\frac { 4 }{ 3 } \, \, sq.\, \, units \end{array}$$