Application of Integrals Test

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Application of Integrals Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

The area of (in sq. units ) of the region described by $$ A={(x,y): x^2+y^2 \leq 1\ and\ y^2 \leq 1-x }$$

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

B
$$\dfrac{\pi}{2}-\dfrac{4}{3}$$

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

D
$$\dfrac{\pi}{2}+\dfrac{2}{3}$$

Solution
Question 2
1 / -0

The area of the region bounded by the curves $$y=|x-1|$$ and $$y=3-|x|$$ is?

A
$$2$$ sq. units

B
$$3$$ sq. units

C
$$4$$ sq. units

D
$$6$$ sq. units

Solution
Question 3
1 / -0

The area enclosed between the curves $${y^2} = x$$ and $$y = |x|\;is$$ -

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

B
$$1$$

C
$$\dfrac {1}{6}$$

D
$$\dfrac {1}{3}$$

Solution

POint of intersection P
$$y^2=x\\x^2=x\\x=1\quad and\quad 0$$
Area enclosed $$=\int_0^1{\sqrt x-x}\\ \Rightarrow\cfrac{2}{3}x^{2/3}-\cfrac{x^2}{2}]^1_0\\ \Rightarrow\cfrac{2}{3}-\cfrac{1}{2}\\ \Rightarrow\cfrac{4-3}{6}=\cfrac{1}{6}$$
Question 4
1 / -0

The area bounded by $$x^2=4ay$$ and $$y=2a$$ is?

A
$$\dfrac{16\sqrt{2}a^2}{3}$$

B
$$\dfrac{16a^2}{3}$$

C
$$\dfrac{8a^2}{3}$$

D
$$\dfrac{8\sqrt{2}a^2}{3}$$

Solution

REF.Image
$$x^{2}=4ay, y=2a$$

$$A= A_{1}+A_{2}$$

$$=2A_{2}$$

$$= 2\int_{0}^{2\sqrt{2}a}(2a-\dfrac{x^{2}}{4a})dx$$

$$= 2[2ax-\dfrac{x^{3}}{4a\times 3}]_{0}^{2\sqrt{2}a}$$

$$= 2[4\sqrt{2}a^{2}-\dfrac{4a^{2}\sqrt{2}}{3}]$$

$$= \dfrac{16}{3}\sqrt{2}a^{2}$$
Question 5
1 / -0

Area of the region containing all points $$(x, y)$$ satisfying $$0\le y\le \sqrt{4-x^{2}}, y \le x^{2}+x+1$$ and $$y=\left[\sin^{2}\dfrac{\pi}{4}+\cos\dfrac{x}{4}\right]$$ is equal to ( where [.] denotes the greatest integer function ).

A
$$\dfrac{4\pi-1}{6}+\sqrt{3}$$

B
$$\dfrac{4\pi+1}{6}+\sqrt{3}$$

C
$$\dfrac{2\pi-1}{6}+\sqrt{3}$$

D
$$\dfrac{2\pi+1}{6}+\sqrt{3}$$
Question 6
1 / -0

If area bounded by to curves $$y^2 = 4ax$$ and y=mx is $$\dfrac{a^2}{3}$$, then the value of m is

A
$$2$$

B
$$-1$$

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

D
none of these

Solution

Given,

$$y^2=4ax,y=mx$$ curves intersect at $$\left ( 4\dfrac{a}{m^2},\dfrac{a}{m} \right )$$

Area enclosed is,

$$A=\displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx$$

$$\therefore \displaystyle \int_{0}^{4\frac{a}{m^2}}(\sqrt{4ax}-mx)dx=\dfrac{a^2}{3}$$ .... given

$$\left[2\sqrt a \ \dfrac{x^{\tfrac 32 }}{\tfrac 32 }- m \dfrac {x^2} 2 \right]^{4\frac{a}{m^2}}_0=\dfrac{a^2}{3}$$

$$\dfrac {4\sqrt a} 3 {({4\frac{a}{m^2}})^{\tfrac 3 2}}-\dfrac m 2 {(4\frac{a}{m^2})}^2-0-0=\dfrac{a^2}{3}$$

$$\dfrac{32 a ^2} {3m^3}-\dfrac{8 a ^2} {m^3}=\dfrac{a^2}{3}$$

$$\dfrac{8}{3}\dfrac{a^2}{m^3}=\dfrac{a^2}{3}$$

$$m^3=8$$

$$\therefore m=2$$
Question 7
1 / -0

The are boundede by the curve $$y=x^{2},y=-x$$ and $$y^{2}=4x-3$$ is $$k$$, them the value of $$9k$$ is

A
$$2$$

B
$$3$$

C
$$0$$

D
$$4$$

Solution
Question 8
1 / -0

If $${\int}_{0}^{1}\left(4x^{3}=f(x)\right)f(x)dx=\dfrac{4}{7}$$, then the area of region bounded by $$y=f(x),x-$$ axis and the line $$x=$$ and $$x=2$$ is

A
$$\dfrac{11}{2}$$

B
$$\dfrac{13}{2}$$

C
$$\dfrac{15}{2}$$

D
$$\dfrac{17}{2}$$
Question 9
1 / -0

The area bounded by curves $$ 3 x^2 + 5 y= 32$$ and $$ y = |x-2| $$ is

A
25

B
33/2

C
17/2

D
33

Solution
Question 10
1 / -0

The area of the plane region bounded by the curves $$x+{ 2y }^{ 2 }=0$$ and $$x+{ 3y }^{ 2 }=1$$ is equal to

A
$$\cfrac { 5 }{ 3 } sq.unit$$

B
$$\cfrac { 1 }{ 3 } sq.unit$$

C
$$\cfrac { 2 }{ 3 } sq.unit$$

D
$$\cfrac { 4 }{ 3 } sq.unit$$

Solution

The given curves are $$x+2y^2=0$$ & $$x+3y^2=1$$
$$1$$st curve
$$\Rightarrow x^2+2y^2=0$$
$$\Rightarrow x^2=-2y^2$$
$$\therefore y^2=-x/2$$ (parabola)
$$2nd$$ curve
$$\Rightarrow x+3y^2=1$$
$$\Rightarrow 3y^2=1-x$$
$$y^2=(1-x)/3$$ (parabola)
So, area between parabolas is required
Solving
$$-\dfrac{x}{2}=\dfrac{(1-x)}{3}$$
$$\Rightarrow -3x=2-2x$$
$$\Rightarrow -3x+2x=2$$
$$\Rightarrow -x=2$$
$$=x=-2$$
$$\therefore y=1, -1$$
$$(-2, -1)$$ & $$(-2, 1)$$ are points of intersection.
Required area
$$=2|\displaystyle\int^1_0(-2y^2-1+3y^2)dy|$$
$$=2|\displaystyle\int ^1_0(y^2-1)dy|=2\left[y-\dfrac{y^3}{3}\right]^1_0$$
$$=2|\left(\dfrac{1}{3}-1\right)|=2\times 2/3=4/3$$ square units.