Application of Integrals Test

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Application of Integrals Test - 47

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Question 1
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Let $$S(\alpha) = \{ (x, y) : y^2 \le x, 0 \le x \le \alpha\}$$ and $$A(\alpha)$$ is area of the region $$S(\alpha)$$. If for a $$\lambda , 0 < \lambda < 4, A(\lambda) : A(4) = 2 : 5$$, then $$\lambda$$ equals

A
$$2\left(\dfrac{4}{25}\right)^{\frac{1}{3}}$$

B
$$4\left(\dfrac{4}{25}\right)^{\frac{1}{3}}$$

C
$$2\left(\dfrac{2}{5}\right)^{\frac{1}{3}}$$

D
$$4\left(\dfrac{2}{5}\right)^{\frac{1}{3}}$$

Solution

$$\displaystyle S(\alpha) =\{(x, y) : y^2 \le x, 0 \le x \le \alpha\}$$
$$\displaystyle A(\alpha)= 2 \int^{\alpha}_{0} \sqrt{x}dx = 2\alpha^{\frac{3}{2}}$$
$$A(4) = 2 \times 4^{3/2} = 16$$
$$A(\lambda) = 2\times \lambda^{3/2}$$
$$\dfrac{A(\lambda)}{A(4)} = \dfrac{2}{5} \Rightarrow \lambda = 4. \left(\dfrac{4}{25}\right)^{1/3}$$
Question 2
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If the area (in sq. units) bounded by the parabola $$y^{2} = 4\lambda x$$ and the line $$y = \lambda x, \lambda > 0$$, is $$\dfrac {1}{9}$$, then $$\lambda$$ is equal to

A
$$24$$

B
$$48$$

C
$$4\sqrt {3}$$

D
$$2\sqrt {6}$$

Solution

$$Area = \dfrac {1}{9} = \int_{0}^{\dfrac {4}{\lambda}}(\sqrt {4\lambda x} - \lambda x)dx$$
$$\Rightarrow \lambda = 24$$.
Question 3
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The area (in sq. units) of the region bounded by the curves $$y={2}^{x}$$ and $$y=\left| x+1 \right| $$, in the first quadrant is:

A
$$\cfrac { 3 }{ 2 } -\cfrac { 1 }{ \log _{ e }{ 2 } } $$

B
$$\cfrac { 1 }{ 2 } $$

C
$$\log _{ e }{ 2 } +\cfrac { 3 }{ 2 } $$

D
$$\cfrac { 3 }{ 2 } $$

Solution

Required Area
$$\int _{ 0 }^{ 1 }{ \left( \left( x+1 \right) -{ 2 }^{ x } \right) } dx={ \left( \cfrac { { x }^{ 2 } }{ 2 } +x-\cfrac { { 2 }^{ x } }{ \ln { 2 } } \right) }_{ 0 }^{ 1 }=\left( \cfrac { 1 }{ 2 } +1-\cfrac { { 2 }^{ } }{ \ln { 2 } } \right) -\left( 0+0-\cfrac { 1 }{ \ln { 2 } } \right) =\cfrac { 3 }{ 2 } -\cfrac { 1 }{ \ln { 2 } } $$
Question 4
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Area of the region bounded by $$y^2\leq 4x, x+y\leq 1, x\geq 0, y\geq 0$$ is $$a\sqrt{2}+b$$, then value of $$a-b$$ is?

A
$$4$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution

Let P be the point common to $$x+y=1$$ & $$y^2=4x$$
So $$y^2=4(1-y\Rightarrow y^2+4y-4=0$$
$$\Rightarrow y=\dfrac{-4\pm \sqrt{16+16}}{2}$$
$$\Rightarrow =-2+2\sqrt{2}$$
Hence $$P(3, -2\sqrt{2}, -2+2\sqrt{2})$$
Hence started area $$=$$ Area of region (OPN)$$+$$ Area of ($$\Delta$$OPQ)
$$=\displaystyle\int^{3-2\sqrt{2}}_02\sqrt{x}dx+\dfrac{1}{2}[-1-(3-2\sqrt{2})]^2$$
$$=\dfrac{2}{3}\cdot 2(\sqrt{2}-1)(3-2\sqrt{2})+\dfrac{1}{2}[2(\sqrt{2}-1)]^2$$
$$=\dfrac{4}{3}\left\{-7+5\sqrt{2}\right\}+2(3-2\sqrt{2})=\left(\dfrac{20}{3}-4\right)\sqrt{2}+6-\dfrac{28}{3}=\dfrac{8}{3}\sqrt{2}-\dfrac{10}{3}$$
Hence $$a=\dfrac{8}{3}, b=\dfrac{-10}{3}$$
So $$a-b=6$$.
Question 5
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If the area enclosed by the curves $${ y }^{ 2 }=4\lambda x$$ and $$y=\lambda x$$ is $$\cfrac { 1 }{ 9 } $$ square units then value of $$\lambda$$ is

A
$$24$$

B
$$37$$

C
$$48$$

D
$$38$$

Solution

$${ y }^{ 2 }=4\lambda x\quad \quad ;y=\lambda x$$
If $$\lambda> 0$$ then
Hence $$\int _{ 0 }^{ 4/\lambda }{ \left( 2\sqrt { \lambda } \sqrt { x } -\lambda x \right) } dx=\cfrac { 1 }{ 9 } $$
$${ \left( \cfrac { 2\sqrt { \lambda } { x }^{ 3/2 } }{ 3/2 } -\cfrac { \lambda { x }^{ 2 } }{ 2 } \right) }^{ \cfrac { 4 }{ \lambda } }=\cfrac { 1 }{ 9 } \Rightarrow \cfrac { 4 }{ 3 } \sqrt { \lambda } \cfrac { 8 }{ { \lambda }^{ 3/2 } } -\lambda \cfrac { 8 }{ { \lambda }^{ 2 } } =\cfrac { 1 }{ 9 } \Rightarrow \cfrac { 32 }{ 3\lambda } =\cfrac { 1 }{ 9 } \Rightarrow \lambda =24$$
Question 6
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The area bounded by the line $$y=x$$, x-axis and ordinates $$x=-1$$ and $$x=2$$ is?

A
$$\dfrac{3}{2}$$

B
$$\dfrac{5}{2}$$

C
$$2$$

D
$$3$$
Question 7
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The area bounded by curve $$y=\sin { 2x } \left( x=0\quad to\quad x=\pi \right) $$ and X-axis is ______

A
$$4$$

B
$$2$$

C
$$1$$

D
$$\cfrac { 3 }{ 2 } $$

Solution

As the crest and trugh are same we can write

$$A=2\int _{ 0 }^{ \pi /2 }{ \sin { 2x } } dx$$

$$=2{ \left[ \cfrac { -\cos { 2x } }{ 2 } \right] }_{ 0 }^{ \pi /2 }$$

$$=(1+1)$$

$$ A=2$$
Question 8
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Area of the region bounded by the curve $$y = \cos x$$ between $$x = 0$$ and $$x = \pi$$ is

A
$$1$$ sq. units

B
$$4$$ sq. units

C
$$2$$ sq. units

D
$$3$$ sq. units

Solution

Required area enclosed by the curve $$y = \cos x, x = 0$$ and $$x = \pi$$
$$A = \int_{0}^{\pi/2} \cos x dx + \left |\int_{\pi/2}^{\pi} \cos x dx\right |$$
$$=[sinx]_0^{\pi/2}+[sinx]_{\pi/2}^{\pi}$$
$$= \left [\sin \dfrac {\pi}{2} - \sin 0\right ] + \left |\sin \dfrac {\pi}{2} - \sin \pi \right |$$
$$= 1 + 1 = 2\ sq\ units$$.
Question 9
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The area bounded by the curves $$y = -x^2 + 3$$ and $$y = 0$$

A
$$\sqrt{3} + 1$$

B
$$\sqrt{3}$$

C
$$4\sqrt{3}$$

D
$$5\sqrt{3}$$

Solution

$$\textbf{Step -1: Finding the points of intersection.}$$
$$\text{The two curves are }y=-x^2+3\text{ and }y=0$$
$$\text{Substituting }y=0\text{ in }y=-x^2+3$$
$$\Rightarrow 0=-x^2+3$$
$$\Rightarrow x^2=3$$
$$\Rightarrow x=\pm\sqrt3$$
$$\text{The points of intersection are }(-\sqrt3,0)\text{ and }(\sqrt3,0)$$
$$\textbf{Step -2: Finding the area between the curves.}$$
$$\text{Area between the curves}=\int_{-\sqrt3}^{\sqrt3}(-x^2+3)dx$$
$$=2\times\int_0^{\sqrt3}(-x^2+3)dx$$
$$=2\times[\dfrac{-x^3}{3}+3x]_0^{\sqrt3}$$
$$= 2[\dfrac{-3\sqrt3}{3}+3\sqrt3]$$
$$=4\sqrt3\text{ sq. units}$$
$$\textbf{Hence , the area bounded by the curves }\mathbf{y=-x^2+3}\textbf{ and }\mathbf{y=0}\textbf{ is }\mathbf{4\sqrt3}\textbf{ sq. units.}$$
Question 10
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The area bounded by $$y = \sin^2 x , x = \dfrac{\pi}{2} $$ and $$x = \pi$$ is

A
$$\dfrac{\pi}{2}$$

B
$$\dfrac{pi}{4}$$

C
$$\dfrac{\pi}{8}$$

D
$$\dfrac{\pi}{16}$$

E
$$2\pi$$

Solution

Required area $$= \displaystyle \int_{\frac{\pi}{2}}^{\pi} \sin^2 x dx$$

$$= \displaystyle \int^{\pi}_{\tfrac{\pi}{2}} \left[\dfrac{1 - \cos 2x}{2}\right] dx$$

$$= \dfrac{1}{2} \displaystyle \int^{\pi}_{\tfrac{\pi}{2}} (1 - \cos 2x) dx$$

$$= \dfrac{1}{2} \left[x - \dfrac{\sin 2x}{2} \right]_{\tfrac{\pi}{2}}^{\pi}$$

$$= \dfrac{1}{2} \left[(\pi - 0) - \left(\dfrac{\pi}{2} - 0\right) \right]$$

$$= \dfrac{1}{2} \left[\dfrac{\pi}{2} \right] = \dfrac{\pi}{4}$$