Application of Integrals Test

Result

Application of Integrals Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

The area of the region bounded by the curves $$y = x^2$$ and $$y = |x|$$ is

A
$$\dfrac{1}{6}$$

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

C
$$\dfrac{5}{6}$$

D
$$\dfrac{5}{3}$$

Solution
Question 2
1 / -0

Area bounded by the curves $$y = \sin x ,$$ tangent drawn to it at $$x = 0$$ and the line $$x = \frac { \pi } { 2 }$$ is equal to

A
$$\frac { \pi ^ { 2 } - 4 } { 2 }$$ sq.units

B
$$\frac { \pi ^ { 2 } - 4 } { 4 }$$ sq.units

C
$$\frac { \pi ^ { 2 } - 2 } { 4 }$$ sq.units

D
$$\frac { \pi ^ { 2 } - 2 } { 2 }$$ sq.units
Question 3
1 / -0

The area enclosed by the line y = x + 1, X- axis and the lines x = -3 and x = 3 is

A
5

B
7

C
10

D
9

Solution
Question 4
1 / -0

The area of the region
A=$$\left [ \left ( x,y \right ) :0\leq y\leq x\left | x \right |+1and -1\leq x\leq 1\right ]$$. in sq. units, is:

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

B
$$\frac{1}{3}$$

C
2

D
$$\frac{4}{3}$$

Solution
Question 5
1 / -0

As shown in the figure of an ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$$. The area of shaded region is .......
.

A
$$12\pi $$

B
$$3(\pi -2)$$

C
$$4(\pi -2)$$

D
$$12(\pi -2)$$
Question 6
1 / -0

The area bounded by the parabolas $$y = {\left( {x + 1} \right)^2}\;and\;y = {\left( {x - 1} \right)^2}$$ and the line $$y = \frac{1}{4}$$ is

A
$${\text{4sq}}{\text{.}}\;{\text{units}}$$

B
$$\frac{1}{6}\;{\text{sq}}{\text{.}}\;{\text{units}}$$

C
$$\frac{4}{3}\;{\text{sq}}{\text{.}}\;{\text{units}}$$

D
$$\frac{1}{3}\;{\text{sq}}{\text{.}}\;{\text{units}}$$
Question 7
1 / -0

The area enclosed by parabola $${ y }^{ 2 }=64x$$ and its latus-rectum is $$\lambda $$ then $$3\lambda $$ then $$3\lambda$$ =

A
2048

B
2408

C
2804

D
2084

Solution
Question 8
1 / -0

The area enclosed between the curves $$y = a x ^ { 2 }$$ and $$x = a y ^ { 2 } ( a > 0 )$$ is $$1$$ sq. unit, then the value of $$a$$ is

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

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

C
$$1$$

D
$$\dfrac { 1 } { 3 }$$
Question 9
1 / -0

The area of the region lying between the line $$x-y+2=0 $$ and the curve $$x=\sqrt{y}$$ is

A
$$9$$

B
$$\dfrac {9}{2}$$

C
$$\dfrac {10}{3}$$

D
none

Solution

Given
$$x-y+2$$ and $$x=\sqrt{y}$$
$$x^{2}=y$$
$$\therefore $$ Required area $$A=\int_{0}^{2}[x-x^{2}+2]dx$$
This represents the point of intersection of curves
so, by equating the equations we get,
$$x+2=x^{2}$$
$$\Rightarrow x^{2}-x-2=0$$
$$\Rightarrow x^{2}-2x+x-2=0$$
$$\Rightarrow x(x-2)+1(x-2)=0$$
$$\Rightarrow (x-2)(x+1)=0$$
$$\Rightarrow x=-1,2$$
Since x> 0 x=1 is not valid
hence a=2
$$A=\int_{0}^{2}[x+2-x^{2}]dx$$
$$A=\int_{0}^{2} xdx +\int_{0}^{2}2dx-\int_{0}^{2}x^{2}dx$$
$$A=[\frac{x^{2}}{2}+2x-\frac{x^{3}}{3}]_{0}^{2}$$
$$A=2+4-\frac{8}{3}=\frac{6+12-8}{3}$$
$$A=\frac{10}{3}$$
Question 10
1 / -0

the volume of a solid obtained by revolving about y-axis enclosed between the ellipse $${x}^{2}+9{y}^{2}=9$$ and the straight line $$x+3y=3$$ in the first quadrant is

A
$$3\pi$$

B
$$4\pi$$

C
$$6\pi$$

D
$$9\pi$$