Application of Integrals Test

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Application of Integrals Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

If the area enclosed by the parabolas $$\displaystyle\ y= a-x^{2}$$ and $$\displaystyle\ y=x^{2}$$ is $$18\sqrt{2}$$ sq.units. Find the value of 'a'

A
1

B
2

C
5

D
9

Solution

The intersection point is $$(\sqrt{\dfrac{a}{2}},\dfrac{a}{2}),(-\sqrt{\dfrac{a}{2}},\dfrac{a}{2})$$
$$\int y_{1}-y_{2}.dx$$
$$\int a-2x^{2}.dx$$
$$=[ax-\dfrac{2x^{3}}{3}]_{-\sqrt{\frac{a}{2}}} ^{\sqrt{\frac{a}{2}}}$$

$$=2[ax-\dfrac{2x^{3}}{3}]_{0} ^{\sqrt{\frac{a}{2}}}$$

$$2[\dfrac{a\sqrt{a}}{\sqrt{2}}-\dfrac{2a\sqrt{a}}{6\sqrt{2}}]$$

$$=\sqrt{2}[a\sqrt{a}-\dfrac{a\sqrt{a}}{3}]$$

$$=\dfrac{2\sqrt{2}a\sqrt{a}}{3}=18\sqrt{2}$$

$$2a\sqrt{a}=54$$
$$a\sqrt{a}=27=3^{3}$$
Squaring both sides, we get
$$a^{3}=3^{6}$$
$$a=3^{2}$$
Hence
$$a=9$$.
Question 2
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The ratio in which the area bounded by the curves $$y^2=4x$$ and $$x^2=4y$$ is divided by the line $$x = 1$$ is

A
64 : 49

B
15 : 34

C
15 : 49

D
None o fthese

Solution

Area = $$\int _{ 0 }^{ 4 }{ \sqrt { 4x } } -\frac { { x }^{ 2 } }{ 4 } $$
= $$2\times \frac { 2 }{ 3 } \times 8-\frac { (4)^{ 3 } }{ 12 } $$
= $$\frac { 32 }{ 3 } -\frac { 4 }{ 3 } \times 4$$
= $$\frac { 16 }{ 3 } $$
$${ A }_{ 1 }=\int _{ 0 }^{ 1 }{ \sqrt { 4x } -\frac { x^{ 2 } }{ 4 } } $$
= $$2\times \frac { 2 }{ 3 } \times -\frac { 1 }{ 12 } $$
= $$\frac { 4 }{ 3 } \times \frac { 1 }{ 12 } =\frac { 15 }{ 12 } $$
$${ A }_{ 2 }=\frac { 16 }{ 3 } -\frac { 15 }{ 12 } $$
= $$\frac { 64-15 }{ 12 } $$
= $$\frac { 49 }{ 12 } $$
$$\frac { { A }_{ 1 } }{ { A }_{ 2 } } =\frac { 15 }{ 49 } $$
Question 3
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The area of the region bounded by$$y=\mid x-1\mid $$ and $$ y=1$$ is

A
$$1$$

B
$$2$$

C
$$\dfrac{1}{2}$$

D
None of these

Solution

$$ y=\left|x-1\right|$$
for $$x\gt1, y=x-1$$
for $$ x\lt1, y=-x+1$$
Here area can be calculated as the area of the triangle.
Base of triangle,$$b=2$$
Height of triangle, $$h=1$$
Area of triangle,$$ A=\dfrac{1}{2}\times b\times h$$
$$A=\dfrac{1}{2}\times 2\times 1$$
Area, $$A=1cm^2$$
Question 4
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Semicircles are drawn outside by taking every side of regular hexagon as a diameter. The perimeter of hexagon is 60 cm. Find the area of complete figure formed as such.($$\pi$$ = 3.14) ($$\sqrt3$$ = 1.73)

A
495 c$$m^2$$

B
259.5c$$m^2$$

C
235.5c$$m^2$$

D
695.5c$$m^2$$

Solution

The area of a hexagon=$$\quad \dfrac { 3\sqrt { 3 } }{ 2 } { side }^{ 2 }.\quad $$
There are 6 semicircles=3 circles.
radius of each circle=r=$$\quad \dfrac { 1 }{ 2 } \times $$side of the hexagon.
ar.complete figure=ar.hexagon+3(ar.circle).
Here ,
perimeter of the regular hexagon=60cm.
So one side=$$\quad \dfrac { 60 }{ 6 } cm=10cm.\\ \therefore \quad ar.hexagon\\ =\dfrac { 3\sqrt { 3 } }{ 2 } { side }^{ 2 }\\ =\dfrac { 3\sqrt { 3 } }{ 2 } { 10 }^{ 2 }{ cm }^{ 2 }\\ =\dfrac { 3\times 1.73\times 100 }{ 2 } { cm }^{ 2 }\\ =259.5{ cm }^{ 2 }.\quad $$
Again radius of each circle=r=$$\quad \dfrac { 1 }{ 2 } \times $$side of the hexagon= $$\dfrac{10}{2}cm=5cm.$$
$$\quad ar.(3circles)=3\times \pi \times { r }^{ 2 }\\ =3\times 3.14\times 5\times 5{ cm }^{ 2 }\\ =235.5{ cm }^{ 2 }.\quad $$
So ar.complete figure=ar.hexagon+3(ar.circle).
$$=(259.5+235.5)$$ $$\quad { cm }^{ 2 }\quad $$
=495$$\quad { cm }^{ 2 }$$
Ans- Option A
Question 5
1 / -0

The area of the region bounded by the curves $$ \displaystyle y=\sqrt{x} $$ and $$ \displaystyle y=\sqrt{4-3x} $$ and $$ \displaystyle y=0 $$ is

A
$$\dfrac{4}{9}$$

B
$$\dfrac{16}{9}$$

C
$$\dfrac{8}{9}$$

D
None of these

Solution

From figure required area $$\displaystyle =\int _{ 0 }^{ 1 }{ \sqrt { x } dx } +\int _{ 1 }^{ \frac { 4 }{ 3 } }{ \sqrt { 4-3x } dx } $$
$$\displaystyle ={ \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 1 }+{ \left[ -\frac { 1 }{ 3 } \frac { { \left( 4-3x \right) }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 1 }^{ \frac { 4 }{ 3 } }=\frac { 8 }{ 9 } $$
Question 6
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The area bounded by the circle $${ x }^{ 2 }+{ y }^{ 2 }=8$$, the parabola $${ x }^{ 2 }=2y$$ and the line $$y=x$$ in $$y\ge 0$$ is

A
$$\displaystyle \frac { 2 }{ 3 } +2\pi $$

B
$$\displaystyle \frac { 2 }{ 3 } -2\pi $$

C
$$\displaystyle \frac { 2 }{ 3 } +\pi $$

D
$$\displaystyle \frac { 2 }{ 3 } -\pi $$

Solution

The required area
$$\displaystyle =\int _{ -2 }^{ 2 }{ \sqrt { 8-{ x }^{ 2 } } dx } -\int _{ -2 }^{ 0 }{ \frac { 1 }{ 2 } { x }^{ 2 }dx } -\int _{ 0 }^{ 2 }{ xdx } $$

$$\displaystyle =2\int _{ 0 }^{ 2 }{ \sqrt { 8-{ x }^{ 2 } } dx } -{ \left[ \frac { { x }^{ 3 } }{ 6 } \right] }_{ -2 }^{ 0 }-{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 2 }$$

$$\displaystyle =2{ \left[ \frac { x }{ 2 } \sqrt { 8-{ x }^{ 2 } } +\frac { 8 }{ 2 } \sin ^{ -1 }{ \frac { x }{ 2\sqrt { 2 } } } \right] }_{ 0 }^{ 2 }-\frac { 4 }{ 3 } -2$$

$$\displaystyle =2\left[ 2+4.\frac { \pi }{ 4 } \right] -\frac { 10 }{ 3 } =\frac { 2 }{ 3 } +2\pi $$
Question 7
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The area bounded by the curve $$ \displaystyle y=\sin x $$ and $$ \displaystyle y=\cos x,\forall 0\leq x\leq \pi /2 $$ is

A
$$ \displaystyle 2\left( \sqrt { 2 } -1 \right) $$

B
$$ \displaystyle 2\sqrt{2} \left ( \sqrt{3} -1\right ) $$

C
$$ \displaystyle 2\left ( \sqrt{2} +1\right ) $$

D
$$ \displaystyle \sqrt{3}-1 $$

Solution
Question 8
1 / -0

Area bounded by the curves $$ \displaystyle y=xe^{x} $$ and $$ \displaystyle y=xe^{-x} $$ and the line $$ \displaystyle \left| x \right| =1$$ is

A
$$ \displaystyle 1$$

B
$$ \displaystyle \dfrac 4 e$$

C
$$e$$

D
$$ \displaystyle -1 $$
Question 9
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The area common to the curves $$\displaystyle y^{2}=x$$ and $$x^{2}=y$$is

A
$$1$$

B
$$\displaystyle \frac{2}{3}$$

C
$$\displaystyle \frac{1}{3}$$

D
None of these

Solution

The given curves intersect at
$$y={ x }^{ 2 }={ y }^{ 4}$$ :$$y=0$$ or $$y=1$$
$$x={ y }^{ 2 }={ x }^{ 4 }$$ : $$x=0$$ or $$x=1$$
$$\displaystyle =\int _{ x=0 }^{ x=1 }{ \left( { y }_{ 1 }-{ y }_{ 2 } \right) } dx=\int _{ x=0 }^{ 1 }{ \left( \sqrt { x } -{ x }^{ 2 } \right) } dx$$
$$\displaystyle =\left[ \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ \dfrac { 3 }{ 2 } } -\dfrac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ 1 }=\dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } =\dfrac { 1 }{ 3 } $$
Question 10
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Area of the region bounded by the curves $$y=x-1$$ and $$y=3-|x|$$ is

A
$$3$$

B
$$4$$

C
$$6$$

D
$$2$$

Solution

Consider the curves $$ y = |x-1| $$ & $$ y = 3-|x| $$
when $$ x< 0 $$
$$ 3-(-x) = (-x+1) $$
$$ \Rightarrow 3+x = -x+1 $$
$$ \Rightarrow 2x = -2 $$
$$ x = -1 $$
Put $$ x = -1 $$ in $$ y = |x-1| $$ we have
$$ y = 2 $$
When $$ x>0 $$
$$ 3-x = x- 1 $$
$$ \Rightarrow 3+1 = 2x $$
$$ x = 2 $$
put $$ x = 2 $$ in $$ y = |x-1| $$ we have
$$ y = 1 $$
Thus pt. of intersection is $$ (-1,2),(2,1) $$
$$ \therefore $$ required area $$\displaystyle = \int_{-1}^{0}(3+x+x-1)dx+\int_{0}^{1}(3-x+x-1)dx $$
$$\displaystyle +\int_{1}^{2}3-x-x+1dx $$
$$\displaystyle \Rightarrow \int_{-1}^{0}(2x+2)dx+\int_{0}^{1}2dx+\int_{1}^{2}(4-2x)dx $$
$$\displaystyle \Rightarrow \left ( \frac{2x^{2}}{2} \right )^{1}+(2x)_{-1}^{0}+(2x)_{0}^{1}+(4x)_{1}^{2}-\left ( \frac{2x^{2}}{2} \right )_{1}^{2} $$
$$ \Rightarrow -1+2(-(-1))+2+4(2-1)-(2^{2}-1^{2}) $$
$$ \Rightarrow -1+2+2+4-3 = \boxed{4} $$