Application of Integrals Test

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Application of Integrals Test - 61

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Weekly Quiz Competition

Question 1
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The area enclosed by the curve $$y=\sqrt{(4-x^2)}, y\geq \sqrt{2}\sin\left(\dfrac{x\pi}{2\sqrt{2}}\right)$$ and x-axis is divided by y-axis in the ratio.

A
$$\dfrac{\pi^2-8}{\pi^2+8}$$

B
$$\dfrac{\pi^2-4}{\pi^2+4}$$

C
$$\dfrac{\pi -4}{\pi +4}$$

D
$$\dfrac{2\pi^2}{\pi^2+2\pi -8}$$

Solution

$$y=\sqrt{4-x^{2}} \ldots$$ (given) $$\cdots(i)$$
$$\Rightarrow y^{2}-4 x^{2}$$
$$\Rightarrow x^{2}+y^{2}=4$$
$$y \geq \sqrt{2} \sin \left(\dfrac{\pi x}{2 \sqrt{2}}\right)$$ (given) ... (ii)
To find the period,
$$=\frac{2 \not \pi}{\not \pi} \times 2 \sqrt{2}$$
$$=4 \sqrt{2} \Rightarrow$$ period.
$$=\dfrac{4 \sqrt{2}}{2}=2 \sqrt{2}$$
$$\quad=2 \times 1.732$$
$$\quad[3.4]$$
To find the intersection point using (i) and (ii)
$$x=\sqrt{2}$$
Area of circle $$=\pi r^{2}=\pi \times y=4 \pi$$
Area of circle $$=\pi=A_{1}$$.. (iii)
from (-2,0) to (0,0)
$$A_{2}=\int_{0}^{\sqrt{2}} \sqrt{4-x^{2}}-\sqrt{2} \sin \left(\dfrac{\pi}{2 \sqrt{2}} x\right) d x$$
$$\Rightarrow \int_{0}^{\sqrt{2}} \sqrt{4-x^{2}} d x-\sqrt{2} \int_{0}^{\sqrt{2}} \sin \left[\dfrac{\pi x}{2 \sqrt{2}}\right] d x$$
$$\left[\begin{array}{ll}\therefore & \sqrt{a^{2}-x^{2}} & d x=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}\end{array}\right]$$
$$=\left.\left[\dfrac{x}{2} \sqrt{4-x^{2}}+\dfrac{4}{2} \sin ^{-1} \dfrac{x}{2}\right]_{0}^{\sqrt{2}}-\dfrac{\sqrt{2}}{\left(\dfrac{\pi}{2 \sqrt{2}}\right)}[-\operatorname{cos}\left(\dfrac{\pi x}{2 \sqrt{2}}\right)\right]_{0}^{\sqrt{2}}$$
$$=\left[\dfrac{2}{2} \pi \sqrt{2}+\dfrac{4}{2} \sin ^{-1} \left[\dfrac{1}{\sqrt{2}}\right]\right]+\dfrac{2 \times 2}{\pi}\left|\cos \left(\dfrac{\pi x}{2 \sqrt{2}}\right)\right|_{0}$$
$$=\left[1+\dfrac{4}{2} \sin ^{-1}\left[\dfrac{1}{\sqrt{2}}\right]\right]+\dfrac{4}{\pi}\left[\cos \left(\dfrac{\pi}{2\sqrt 2} \times \sqrt2 )-\operatorname{cos} 0\right]\right.$$
$$\Rightarrow[1+\pi / 2-4 / \pi]$$
$$\Rightarrow \dfrac{2 \pi+\pi^{2}-8}{2 \pi}$$.. (iv)
Using equation (iii) and (iv).
$$\dfrac{A_{1}}{A_{2}}=\dfrac{\pi \times 2 \pi}{2 \pi+\pi^{2}-8}=\dfrac{2 \pi^{2}}{2 \pi+\pi^{2}-8}$$ Answer (D)
Question 2
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The area of the region enclosed between by the $${x^2} + {y^2} = 16$$ and the parabola $${y^2} = 6x$$.

A
$$\dfrac{2}{3} (\sqrt 3 + 4\pi)$$ sq. units

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

C
$$\dfrac{2}{3} (\sqrt 3 + 8\pi)$$ sq. units

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

Solution

Point of intersection of the parabola and the circle is obtained by solving the equations:

$${x}^{2}+{y}^{2}=16$$ and $${y}^{2}=6x$$

$$\Rightarrow\,{x}^{2}+6x-16=0$$

$$\Rightarrow\,{x}^{2}+8x-2x-16=0$$

$$\Rightarrow\,x\left(x+8\right)-2\left(x+8\right)=0$$

$$\Rightarrow\,\left(x-2\right)\left(x+8\right)=0$$

$$\Rightarrow\,x=2,\,x=-8$$

$$\therefore\,x=2$$ is the only possible solution(from the fig.)
$$\therefore\,$$ when $$x=2,\,y=\pm\sqrt{6\times 2}=\pm\,2\sqrt{3}$$

$$\therefore\,B\left(2,2\sqrt{3}\right)$$ and $${B}^{\prime}\left(2,-2\sqrt{3}\right)$$ are the points of intersection of parabola and the circle.

Required area$$=area\,of\,OBA{B}^{\prime}O=2\,area\,of \,OBAO$$
$$=2\left[area\,of\,OBDO+area\,of\,DBAD\right]$$

$$=2\left[\displaystyle\int_{0}^{2}{\sqrt{6x}dx}+\displaystyle\int_{2}^{4}{\sqrt{16-{x}^{2}}dx}\right]$$

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

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

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

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

$$=2\left[\dfrac{2\sqrt{3}}{3}+8\times\dfrac{2\pi}{6}\right]$$

$$=\dfrac{4\sqrt{3}}{3}+\dfrac{16\pi}{3}$$
Question 3
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In the square ABCD, the "shaded" region is the intersection of two circular regions centered at B and D respectively. If AB= 10, then what is the area of the shaded region?

A
$$25(\pi-2)$$

B
$$50(\pi-2)$$

C
$$25\pi$$

D
$$50\pi$$

E
$$ 40\pi (5-\sqrt{2})$$

Solution
Question 4
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Area bounded by $$|x-1| \le 2$$ and $$x^{2}-y^{2}=1$$, is

A
$$6 \sqrt{2}+\dfrac{1}{2}$$ In $$|3+2\sqrt{2}|$$

B
$$6 \sqrt{2}+\dfrac{1}{2}$$ In $$|3-2\sqrt{2}|$$

C
$$6 \sqrt{2}-$$ In $$|3+2\sqrt{2}|$$

D
$$none\ of\ these$$

Solution

Given :
$$|x-1|\leq 2$$ and $$x^2-y^2=1$$
We have to find area bounded by both curve.
As we know
$$|x|\leq a$$ means $$-a \leq x \leq a$$
So $$|x-1|\leq 2$$
$$-2\leq x-1\leq 2$$
$$-2\leq x-1\leq 2$$
$$-2+1\leq x\leq 2+1$$
$$\boxed{-1 \leq x \leq 3}$$
Area of curve $$|x-1|\leq 2$$
Area $$=2 \int_{3}^{1} \sqrt {x^2-1}dx$$
$$\because \int \sqrt {x^2-a^2}dx=\dfrac{x}{2}=\dfrac{a^2}{2}ln(x+\sqrt{x^2-a^2})$$
$$=2\left[\dfrac{x}{2}\sqrt{x^2-1}-\dfrac{1}{2}in (x+\sqrt{x^2-1})\right]^3_1$$
$$2\left[\dfrac{3}{2}\sqrt8-\dfrac{1}{2}in (3+\sqrt 8)\right]-2\left[\dfrac{1}{2}\sqrt 0-\dfrac{1}{2}in (1+0)\right]$$
$$6\sqrt 2 in (3+2\sqrt 2)-0$$
$$A=6\sqrt 2-in (3+2\sqrt 2)$$
Question 5
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Directions For Questions

Three circles passes through centre of each other, each having radius $$\sqrt{3}$$ unit. They are inscribed in a rectangle of length $$4\sqrt{3}$$ unit and breadth $$2\sqrt{3}$$ unit.

...view full instructions

Find area curved by three circles

A
$$(5\pi-3\sqrt{3})units^{2}$$

B
$$(5\pi+4\sqrt{3})units^{2}$$

C
$$(5\pi+3\sqrt{3})units^{2}$$

D
$$(5\pi+3\sqrt{2})units^{2}$$

Solution

Simple area concept
Question 6
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Directions For Questions

A continuous function $$f\left(x\right)$$ satisfying $${x}^{4}-4{x}^{2}\le f\left(x\right)\le 2{x}^{2}-{x}^{3}$$ for all $$x\in\left[0,2\right]$$ such that the area bounded by $$y=f\left(x\right),y={x}^{4}-4{x}^{2}$$ the $$y-$$axis and the line $$x=t\left(0\le t\le 2\right)$$ is $$k$$ times the area bounded by $$y=f\left(x\right),y=2{x}^{2}-{x}^{3},y-$$axis and the line $$x=t, \left(0\le t\le 2\right).$$ Answer the following questions:

...view full instructions

If $$k=2$$ then $$f\left(x\right)$$ attains point of inflection at

A
$$0$$

B
$$\sqrt{2}$$

C
$$-\sqrt{2}$$

D
None of these

Solution
Question 7
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The triangle formed by the tangent to the parabola $$y^2=4x$$ at the point whose abscissa lies in the interval $$\left[a^2, 4a^2\right]$$, the ordinate and the x-axis, has the greatest area equal to?

A
$$12a^3$$

B
$$8a^3$$

C
$$16a^3$$

D
$$20a^3$$
Question 8
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Consider the two curves
$${ C }_{ 1 } :{ y }^{ 2 }=4x $$
$$ { C }_{ 2 } : { x }^{ 2 }+ { y }^{ 2 } - 6x + 1 = 0$$
Then, the area of region between these curves?

A
$$\dfrac{20}{3}-2\pi$$

B
$$\dfrac{10}{3}-2\pi$$

C
$$\dfrac{20}{3}-\pi$$

D
$$\dfrac{10}{3}-\pi$$
Question 9
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Area bounded by the curves $$y=\log _{ e }{ x } \quad$$ and $$y={ \left( \log _{ e }{ x } \right) }^{ 2 }$$ is ?

A
$$e-2$$

B
$$3-e$$

C
$$e$$

D
$$e-1$$
Question 10
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The area enclosed between the curve $$y=x^3$$ and $$y=\sqrt{x}$$ is

A
$$\dfrac{5}{3}$$

B
$$\dfrac{5}{4}$$

C
$$\dfrac{5}{12}$$

D
None of these

Solution

Given two curves are
$$y=x^3$$
and $$y=\sqrt x$$
Both curves intersect at point $$(1,1)$$
Now equating the curves we get,
$$x^3=\sqrt x;$$ when $$x=1$$
$$\therefore$$ The curve represented by $$y=\sqrt x$$ is upward
to the curve represented by $$y=x^3$$
$$\therefore $$ Area enclosed by the curves= Area of the shaded region
$$\Rightarrow $$ Area enclosed by the curves $$=\int_{1}^{0}(\sqrt x-x^3)dx$$
$$=\\int_{1}^{0}x\dfrac{1}{2}dx-\int_{2}^{0}x^3dx$$
$$=\left[\dfrac{x^{3/2}}{\dfrac{3}{2}}\right]^1_0-\left[\dfrac{x^4}{4}\right]^1_0$$
$$=\left[\dfrac{(1)^{3/2}}{\dfrac{3}{2}}-0\right]-\left[\dfrac{(1)^4}{4}-0$\right]$$
$$=\dfrac{1}{\dfrac{3}{2}}-\dfrac{1}{4}$$
$$=\dfrac{2}{3}-\dfrac{1}{4}$$
$$=\dfrac{8-3}{12}$$
$$\therefore$$ Area enclosed by the curves $$=\dfrac{5}{12}$$ sq. units