Application of Integrals Test

Result

Application of Integrals Test - 37

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The angle between the curves $$y=sin x$$ and $$ y=cos x$$ is :

A
$$tan^{-1}(2\sqrt{2})$$

B
$$tan^{-1}(3\sqrt{2})$$

C
$$tan^{-1}(3\sqrt{3})$$

D
$$tan^{-1}(5\sqrt{2})$$

Solution
Question 2
1 / -0

The area bounded by parabola $$y^{2}=x$$, straight line $$y=4$$ and $$y-axis$$ is-

A
$$\dfrac{16}{3}$$

B
$$7\sqrt{2}$$

C
$$\dfrac{32}{3}$$

D
$$\dfrac{64}{3}$$

Solution

R.E.F image
Taking a rectangle of width
dy and length as x-coordinate
of parabola, Area would be,
$$ A = ^{4}\int _{0}xdy = ^{4}\int _{0}y^{2}dy =\left [\dfrac{y^{3}}{3}\right]_{4}^{0} $$
$$ = \dfrac{64}{3}\Rightarrow (D) $$
Question 3
1 / -0

The area bounded by $$\dfrac { | x | } { a } + \dfrac { | y | } { b } = 1$$ where $$a > 0$$ and $$b > 0$$ is

A
$$\dfrac { 1 } { 2 } a b$$

B
$$ab$$

C
$$2ab$$

D
$$4ab$$

Solution

$$\dfrac{|x|}{a}+\dfrac{|y|}{b}=1$$
as $$a > o,b>o$$
We should consider only 1st quadrant
are of $$\triangle =\dfrac{1}{2}\times a\times b$$
$$=\dfrac{1}{2}ab$$.
Question 4
1 / -0

Area bounded by $$y=x^2$$and line $$y=x$$

A
2/3

B
1/6

C
1/3

D
1/4

Solution
Question 5
1 / -0

The area bounded by the curve $$y=(x+1)^2,y=(x-1)^2$$ and the line $$y=0$$ is

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

B
$$\dfrac{2}{3}$$

C
$$\dfrac{1}{4}$$

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

Solution

R.E.F image
$$ \rightarrow y = (x+1)^{2} $$ is obtained
by shifting origin to (-1,0)
in $$ x^{2} = y, $$ for $$ y = (x-1)^{2}$$
Similarly (1,0)
As graph is symmetric
about y-axis, area A
would be, $$ A = 2 \int_{0}^{1} (x-1)^{2}dx $$
$$ = 2\int_{0}^{1} x^{2}-2x+1dx = 2\left[x\dfrac{3}{3}-x^{2}+x\right]_{0}^{1} $$
$$ = 2\left(\dfrac{1}{3}-1+1\right) = \dfrac{2}{3}\Rightarrow (B) $$
Question 6
1 / -0

The area (in $$sq\ unit$$) of the region $$\left\{(x,y):y^{2}\ge 2x\ and\ x^{2}+y^{2} \le 4x,x \ge 0,y\ \ge 0\right\}$$ is:-

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

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

C
$$\pi-\dfrac {8}{3}$$

D
$$\pi-\dfrac {4\sqrt {2}}{3}$$

Solution
Question 7
1 / -0

Area common to the cutve $$y=\sqrt {9-x^{2}}$$ and $$x^{2}-y^{2}=6x$$ is:

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

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

C
$$3\left(\pi +\dfrac {\sqrt {3}}{4}\right)$$

D
$$None\ of\ these$$

Solution
Question 8
1 / -0

The area enclosed by the curves $$y=x^{2},y=x^{3},x=0$$ and $$x=p$$, where $$p > 1$$, is $$\dfrac{1}{6}$$. then $$p$$ equals

A
$$8/3$$

B
$$16/3$$

C
$$4/3$$

D
$$2$$
Question 9
1 / -0

If the point $$(\lambda, \lambda +1)$$ lies inside the region bounded by the curve $$x=\sqrt{25-y^2}$$ and y-axis, then $$\lambda$$ belongs to the interval.

A
$$(-1, 3)$$

B
$$(-4, 3)$$

C
$$(-\infty, -4)\cup (3, \infty)$$

D
None of these

Solution
Question 10
1 / -0

Area of the region bounded by the curve $$y=e^x$$ and lines $$x=0$$ and $$y=e$$ is?

A
$$e-1$$

B
$$1$$

C
$$2$$

D
$$e^2-1$$

Solution