Application of Integrals Test

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Application of Integrals Test - 58

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Weekly Quiz Competition

Question 1
1 / -0

The area (in square units) of the region bounded by the curves $$x=y^2$$ and $$x=3-2y^2$$ is

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

B
2

C
3

D
4

Solution

Required area $$\displaystyle =2\int_0^1(3-2y^2-y^2)dy=6\int_0^1(1-y^2)dy=6\left(y-\dfrac{y^3}{3}\right)_0^1=4$$
Since given curves are symmetrical about x-axis
Question 2
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The area (in square units) bounded by the curves $$x\, =\, -2y^2$$ and $$x\, =\, 1-3y^2$$ is

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

B
$$1$$

C
$$\displaystyle \frac{4}{3}$$

D
$$\displaystyle \frac{5}{3}$$

Solution

$$x=-2y^2$$

$$x=1-3y^2$$

$$-2y^2=1-3y^2\Rightarrow y^2=1\Rightarrow y=\pm 1$$

Area between the curves $$= A_1-A_2$$

$$\displaystyle =\int^1_{-1}(1-3y^2+2y^2)dy$$

$$=\int^1_{-1}(1-y^2)dy$$

$$=[y-\dfrac{y^3}{3}]^1_{-1}=\dfrac{4}{3}$$
Question 3
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The area of the region bounded by the curves $$x^{2} + y^{2} = 8$$ and $$y^{2} = 2x$$ is

A
$$2\pi + \dfrac {1}{3}$$

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

C
$$2\pi + \dfrac {4}{3}$$

D
$$\pi + \dfrac {4}{3}$$

Solution

Given curves,
$$x^{2} + y^{2} = 8$$ .....(i)
and $$y^{2} = 2x$$ .....(ii)

On solving Eqs. (i) and (ii), we get
$$x^{2} + 2x - 8 = 0$$
$$x^{2} + 4x - 2x - 8 = 0$$
$$x (x - 4) - 2 (x + 4) = 0$$
$$(x- 2) (x +4) = 0$$
$$\therefore x = 2$$ and $$y = \pm 2$$

$$\therefore$$ Required area
$$= 2[\text {Area of OAP} + \text {Area of PAB}]$$

$$= 2\left [\displaystyle\int_{0}^{2} \sqrt {2x} dx + \int_{2}^{2\sqrt {2}} \sqrt {8 - x^{2}} dx \right ]$$

$$= 2\left [\sqrt {2} \left (x^{3/2} \cdot \dfrac {2}{3}\right )^{2}_{0} + \left (\dfrac {x}{2} \sqrt {8 - x^{2}} + \dfrac {8}{2} \sin^{-1} \dfrac {x}{2\sqrt {2}} \right )_{2}^{2\sqrt {2}} \right ]$$

$$= 2\left [\dfrac {2\sqrt {2}}{3} (2^{3/2}) + 4\times \dfrac {\pi}{2} - 2 - 4\times \dfrac {\pi}{4}\right ]$$

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

$$= 2\left [\dfrac {8}{3} - 2 + \pi \right ] + 2\left (\dfrac {2}{3} + \pi \right ) = 2\pi + \dfrac {4}{3}$$
Question 4
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Area common to the curves $$5x^2 = 0$$ and $$ 2x^2 + 9 = 0$$ is equal to

A
$$12 \sqrt 3 $$

B
$$ 6 \sqrt3$$

C
$$36$$

D
$$18$$

Solution

Given curves
$$y=5x^2$$
$$\Rightarrow x^2=\dfrac{y}{5}$$ ....(1)
which is a parabola opening upward having vertex at (0,0)

Other curve is $$2x^2+9=y$$ .....(2)
$$\Rightarrow x^2=\dfrac{y-9}{2}$$
which is a parabola having vertex at (0,9).

Now, solving eqn (1) and (2), we get
$$5x^2=2x^2+9$$
$$\Rightarrow 3x^2=9$$
$$\Rightarrow x=\pm\sqrt{3}$$
$$\Rightarrow y=15$$
Point of intersection of the curves is $$(-\sqrt{3},15)$$ and $$(\sqrt{3},15)$$

Required area $$=2(ar OAB)$$
$$=2 \int ^\sqrt 3 _0 (y_1-y_2)dx$$
$$= 2 \int ^\sqrt 3 _0 \{ (2x^2 + 9) - 5 x^2\} dx$$
$$ 2 \int ^\sqrt 3 _ 0 (9-3x^2) dx$$
$$ = 2 \left( 9x - 3 \displaystyle \frac{x^3}{3}\right) ^\sqrt3 _0$$
$$ = 2(9 \sqrt 3 - 3 \sqrt 3 ) = 12 \sqrt 3 $$ sq. units
Question 5
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The area of the region, bounded by the curves $$y = \sin^{-1} x + x (1 - x)$$ and $$y = \sin^{-1} x - x (1 - x)$$ in the first quadrant, is

A
$$1$$

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

C
$$\dfrac {1}{3}$$

D
$$\dfrac {1}{4}$$

Solution

$$\sin^{-1} x$$ is defined, if $$-1 \leq x \leq 1$$
In first quadrant $$0\leq x\leq 1$$ and $$x(1 - x) \geq 0$$
$$\therefore y = \sin^{-1} x + x (1 - x)$$ ...... (i)
Lies above $$y = \sin^{-1} x - x (1 - x)$$ ...... (ii)
On solving, we get
$$2x (1 - x) = 0$$
$$\Rightarrow x = 0, 1$$
$$\therefore$$ Required area $$=\displaystyle \int_{0}^{1} (y_{1} - y_{2})dx$$
$$=\displaystyle \int_{0}^{1} [\left \{\sin^{-1} x + x(1 - x)\right \} - \left \{\sin^{-1} x - x(1 - x) \right \}]dx$$
$$= 2\displaystyle \int_{0}^{1} (x - x^{2}) dx$$
$$= 2\left [\dfrac {x^{2}}{2} - \dfrac {x^{3}}{3}\right ]_{0}^{1} = 2\left (\dfrac {1}{2} - \dfrac {1}{3}\right ) = \dfrac {1}{3}$$
Question 6
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If the area bounded by the curves $$y=a{ x }^{ 2 }$$ and $$x=a{ y }^{ 2 }$$, $$(a>0)$$ is $$1$$ sq.units, then the value of $$a$$ is

A
$$\cfrac { 2 }{ 3 } $$

B
$$\cfrac { 1 }{ \sqrt 3 } $$

C
$$1 $$

D
$$4$$

Solution

Points of intersection of $$y=ax^2$$ and $$x=ay^2$$ are $$(0,0)$$ and $$\left (\dfrac {1}{a},\dfrac {1}{a}\right)$$.
Hence, $$\displaystyle \int _0 ^{\tfrac1a} \left (\sqrt {\dfrac {x}{a}}-ax^2\right)dx=1$$
$$\Rightarrow \displaystyle\cfrac{2x^{\tfrac32}}{3\sqrt a} \big|_0^{\tfrac1a} - \cfrac{ax^3}3\bigr|_0^{\tfrac1a} =1$$
$$\Rightarrow \cfrac{2}{3a^2} - \cfrac1{3a^2} =1$$
$$\Rightarrow \cfrac{1}{3a^2} =1$$
$$\Rightarrow a=\dfrac {1}{\sqrt3}$$ ....As $$(a>0)$$
Question 7
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Area bounded by curve $$y=x^2$$ and $$y=2-x^2$$ is ?

A
$$\dfrac{8}{3}$$ sq units

B
$$\dfrac{3}{8}$$ sq units

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

D
None of these

Solution

$$y=x^2$$ and $$y=2-x^2$$

now both he curve intersect each other ,
from both the equation
$$ x^2=2-x^2$$
$$ x=+1,-1$$
now area bounded by both curve is
$$A =\displaystyle \int_{-1}^{1}(2-x^2)-\displaystyle \int_{-1}{1}(x^2)$$
because both are even function hence

$$A =2[\displaystyle \int_{0}^{1}(2-x^2)-\displaystyle \int_{0}{1}(x^2)$$]

$$A=2[(2x-\dfrac{x^3}{3})-(\dfrac{x^3}{3})]$$

now on putting upper and lower value of limit ,we will get
$$A=\dfrac{8}{3}$$
Question 8
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The area bounded by the parabolas $$y^2 = 4a(x + a)$$ and $$y^2 = - 4a (x - a)$$ is

A
$$\dfrac{16}{3} a^2$$ sq units

B
$$\dfrac{8}{3} $$ sq units

C
$$\dfrac{4}{3} a^2$$ sq units

D
None of these

Solution

$$y^{2}=4a\left ( x+a \right )$$

$$y=\pm \sqrt{4ax+4a^{2}}$$

As the area bounded by this curve is in -ve $$x$$ quadrant . Therfore
$$y=-\sqrt{-4ax+4a^{2}}$$

Area bounded by this curve will be $$\int_{-a}^{0}\sqrt{4ax+4a^{2}dx} $$
Area bounded by this curve will be $$ \dfrac {8}{3} a^{2} $$

Similarly , The area bounded by $$y^{2}=-4a\left ( x-a \right )$$ curve will be
$$\int_{0}^{a}\sqrt{-4ax+4a^{2}dx}$$
Area = $$ \dfrac {8}{3} a^{2} $$

thus , total bounded area will be $$ \dfrac {16 }{3} a^{2} $$
Question 9
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The area of the region bounded by the curves $$y = 2^{x}, y = 2x - x^{2}$$ and $$x = 2$$ is

A
$$\dfrac {3}{\log 2} - \dfrac {4}{3}$$

B
$$\dfrac {3}{\log 2} - \dfrac {4}{9}$$

C
$$\dfrac {3}{2} - \dfrac {\log 2}{9}$$

D
None of these

Solution

Given curves are $$y = 2^{x}, y = 2x - x^{2}$$ and $$x = 2$$
$$\therefore$$ Required area $$= \int_{0}^{2}[2^{x} - (2x - x^{2})]dx$$
$$= \int_{0}^{2} (2^{x} -2x + x^{2})dx$$
$$= \left [\dfrac {2^{x}}{\log 2} - x^{2} + \dfrac {x^{3}}{3}\right ]_{0}^{2}$$
$$= \dfrac {4}{\log 2} - 4 + \dfrac {8}{3} - \dfrac {1}{\log 2}$$
$$= \dfrac {3}{\log 2} - \dfrac {4}{3}$$.
Question 10
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The area of the portion of the circle $${ x }^{ 2 }+{ y }^{ 2 }=64$$ which is exterior to the parabola $${ y }^{ 2 }=12x$$, is

A
$$\left( 8\pi -\sqrt { 3 } \right) $$ sq units

B
$$\dfrac { 16 }{ 3 } \left( 8-\sqrt { 3 } \right) $$ sq units

C
$$\dfrac { 16 }{ 3 } \left( 8\pi -\sqrt { 3 } \right) $$ sq units

D
None of the above

Solution

Required shaded area $$=$$

$$=Area\, of \, circle -2\left[ \displaystyle\int _{ 0 }^{ 4 }{ 2\sqrt { 3 } \sqrt { x } dx } +\displaystyle\int _{ 4 }^{ 8 }{ \sqrt { 64-{ x }^{ 2 } } dx } \right] $$

$$=64\pi -2\left[ { \left( 2\sqrt { 3 } { x }^{ { 3 }/{ 2 } }\times \dfrac { 2 }{ 3 } \right) }_{ 0 }^{ 4 }+{ \left( \dfrac { x }{ 2 } \sqrt { 64-{ x }^{ 2 } } +\dfrac { 64 }{ 2 } \sin ^{ -1 }{ \dfrac { x }{ 8 } } \right) }_{ 4 }^{ 8 } \right] $$

$$=64\pi -\dfrac { 64 }{ \sqrt { 3 } } -32\pi +16\sqrt { 3 } +\dfrac { 32\pi }{ 3 }$$

$$=\dfrac { 128\pi }{ 3 } -\dfrac { 16\sqrt { 3 } }{ 3 } $$

$$=\dfrac { 16 }{ 3 } \left( 8\pi -\sqrt { 3 } \right) $$ sq units.