Application of Integrals Test

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Application of Integrals Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Area bounded by $$y = x^2 $$ and $$ y = \dfrac{2}{1 + x^2}$$ is:

A
$$\pi - \dfrac{1}{3}$$

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

C
$$2 \pi - \dfrac{1}{3}$$

D
none of these

Solution

$$y=x^{2}, y=\dfrac{2}{1+x^{2}}$$
$$x^{2}=\dfrac{1}{1+x^{2}}$$, For intersection Points
$$(x^{2})^{2}+x^{2}-2=0$$
$$x^{2}=1$$
$$\therefore x=\pm 1$$
Area bounded $$=\displaystyle\int_{-1}^{1}\left(\dfrac{2}{1+x^{2}}-x^{2}\right)dx$$
$$=2\tan^{-1}(x)-\dfrac{x}{3}\left.\dfrac{}{}\right|_{-1}^{1}$$
$$=2\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)-\dfrac{2}{3}=\pi-\dfrac{2}{3}$$
Question 2
1 / -0

The area between the curves $$y= \tan x, y=2 \sin x $$ and $$x$$-axis in $$-\dfrac{\pi}{3} \leq x \leq \dfrac{\pi}{3}$$ is

A
$$2-\ln2$$

B
$$3-ln2$$

C
$$4-\ln2$$

D
None of these

Solution

Due to symmetry
$$2\int_{0}^{\dfrac{\pi}{3}}{2\sin x-\tan x}dx$$
$$=2[-2\cos x+\ln \cos x]_{0}^{\dfrac{\pi}{3}}$$
$$|2\ln|\cos x|1-4\cos x|_{0}^{\dfrac{\pi}{3}}$$
$$=2-2\ln 2$$
Question 3
1 / -0

The area of the region bounded by the curve $${a}^{4}{y}^{2}=\left(2a-x\right){x}^{5}$$ is to that of the circle whose radius is $$a$$, is given by the ratio

A
$$4:5$$

B
$$5:8$$

C
$$2:3$$

D
$$3:2$$

Solution

Given curve $${a}^{4}{y}^{2}=\left(2a-x\right){x}^{5}$$
cut off $$x-$$axis,when $$y=0$$
$$0=\left(2a-x\right){x}^{5}$$ is
$${A}_{1}=\int_{0}^{2a}{\dfrac{\sqrt{\left(2a-x\right)}{x}^{\frac{5}{2}}}{{a}^{2}}dx}$$
Put $$x=2a{\sin}^{2}{\theta}$$
$$\therefore dx=4a\sin{\theta}\cos{\theta}d\theta$$
$$\therefore {A}_{1}=\int_{0}^{{\frac{\pi}{2}}}{\dfrac{\sqrt{2a}\cos{\theta}{\left(2a\right)}^{\frac{5}{2}}{\sin}^{5}{\theta}\times 4a\sin{\theta}\cos{\theta}}{{a}^{2}}d\theta}$$
$$=32{a}^{2}\int_{0}^{\frac{\pi}{2}}{{\sin}^{6}{\theta}{\cos}^{2}{\theta}d\theta}$$
$$=32{a}^{2}\dfrac{\left(5.3.1\right)\left(1\right)}{8.6.4.2}.\dfrac{\pi}{2}$$ (by Walli's formula)
$$=\dfrac{5\pi{a}^{2}}{8}$$
Area of circle,$${A}_{2}=\pi{a}^{2}$$
$$\therefore \dfrac{{A}_{1}}{{A}_{2}}=\dfrac{5}{8}$$
$$\Rightarrow {A}_{1}:{A}_{2}=5:8$$
Question 4
1 / -0

The area of the region described by $$A=((x,y): {x}^{2}+{y}^{2}\le1)$$ and $$B=((x,y):{y}^{2}\le1-x)$$

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

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

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

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

Solution

Area $$ABCA$$ $$\dfrac{\pi}{2}r^{2}=\dfrac{\pi}{2}$$
Due to symmetry
$$ADCA=2AODA$$
$$=2\int_{0}^{1}{\sqrt{1-x}}$$
$$=\dfrac{-2[(1-x)^{{3}/{2}}]_{0}^{1}}{\dfrac{3}{2}}$$
$$=\dfrac{-4}{2}{-1}=\dfrac{4}{3}$$
Hence the total area = $$\dfrac{\pi}{2} $$+ $$ \dfrac {4}{3}$$
Question 5
1 / -0

Area bounded by the curves $$y=\left[\dfrac{{x}^{2}}{64}+2\right], y=x-1$$ and $$x=0$$ above $$x-$$axis is ($$\left[.\right]$$ denotes the greatest integer function.)

A
$$2$$.sq.unit

B
$$3$$.sq.unit

C
$$4$$.sq.unit

D
none of these

Solution

From the figure shown
$$-8<x<8\Rightarrow y=2$$
$$\therefore$$ Required Area$$=\dfrac{1}{2}\left(1+3\right)\times 2$$
$$=4$$.sq.unit.
Question 6
1 / -0

Area of the region bounded by the curves $$y\left|y\right|\pm x\left|x\right|=1$$ and $$y=\left|x\right|$$ is:

A
$$\dfrac{\pi}{8}$$ sq.unit

B
$$\dfrac{\pi}{4}$$ sq.unit

C
$$\dfrac{\pi}{2}$$ sq.unit

D
$$\pi$$ sq.unit

Solution

In First quadrant $${y}^{2}\pm{x}^{2}=1$$
and in Second quadrant $${y}^{2}\pm{x}^{2}=1$$
i.e., $${y}^{2}+{x}^{2}=1$$
and $${y}^{2}-{x}^{2}=1$$
and in Third quadrant $$-{y}^{2}-{x}^{2}=1$$
and in fourth quadrant $$-{y}^{2}\pm{x}^{2}=1$$
i.e., $${x}^{2}-{y}^{2}=1$$ $$\left(\because -{x}^{2}-{y}^{2}\neq1\right)$$
$$\therefore $$Required Area$$=$$area of $$OABCO$$
$$=\dfrac{1}{4}$$(area of circle)
$$=\dfrac{1}{4}\pi{r}^{2}$$
$$=\dfrac{1}{4}\pi{\left(1\right)}^{2}$$
$$=\dfrac{\pi}{4}$$sq.units
Question 7
1 / -0

Find the area included between the parabolas $$y^{2} = x$$ and $$x = 3 - 2y^{2}$$.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$A=(1,1)$$ , $$B=(0,\sqrt{\dfrac{3}{2}})$$ , $$C=(0,-\sqrt{\dfrac{3}{2}})$$ , $$D=(1,-1)$$
Due to symmetry we can just calculate the area in the first quadrant and multiply by 2
$$=\int_{0}^{1}{3-2y^{2}}dy -\int_{0}^{1}y^{2}dy$$
$$=[3y-\dfrac{2y^{3}}{3}]-\dfrac{y^{3}}{3}=3-\dfrac{2}{3}-\dfrac{1}{3}$$
Area of first quadrant $$=2$$
$$Total\ \text {area} = 4$$
Question 8
1 / -0

The area of the region bounded by $$1-{y}^{2}=\left|x\right|$$ and $$\left|x\right|+\left|y\right|=1$$ is

A
$$\dfrac{1}{3}$$.sq.unit

B
$$\dfrac{2}{3}$$.sq.unit

C
$$\dfrac{4}{3}$$.sq.unit

D
$$1$$.sq.unit

Solution

$$1-y^{2}=|x|$$
$$|x|+|y|=1$$
$$\begin{aligned} \text { Area } &=\int_{0}^{1} x d y \\ &=\int_{0}^{1}\left(1-y^{2}\right) d y \end{aligned}$$
$$=\left[y-\dfrac{y^{3}}{3}\right]_{0}^{1}$$
$$=1-1 / 3$$
$$=\dfrac{2}{3}$$ sq.unit (Area of one region)
$$\begin{aligned} \text { Area } &=\dfrac{2}{3}-\dfrac{1}{2} \\ &=4\left[\dfrac{2}{3}-\dfrac{1}{2}\right] \\ &=\dfrac{4}{6}=\dfrac{2}{3} \text { squnits } \end{aligned}$$
Correct option is (B).
Question 9
1 / -0

The area of the figure bounded by the curves $$y=\left|x-1\right|$$ and $$y=3-\left|x\right|$$ is

A
$$1$$ sq.unit

B
$$2$$ sq.unit

C
$$3$$ sq.unit

D
$$4$$ sq.unit

Solution

Since, $$y=\left|x-1\right|=\begin{cases}x-1;x\ge 1 \\ 1-x, x<1 \end{cases}$$ ..................$$\left(1\right)$$
and $$y=3-\left|x\right|=\begin{cases} 3-x; x\ge 0 \\ 3+x; x<0\end{cases}$$ ..................$$\left(2\right)$$
Solving eqns$$\left(1\right)$$ and $$\left(2\right)$$, we get
$$x=2$$ and $$x=-1$$
$$\therefore$$ Required Area$$=\left|\int_{-1}^{2}{\left(3\left|x\right|-\left|x-1\right|\right)}\right|$$
$$=\left|\int_{-1}^{0}{\left(2x+2\right)dx}+\int_{0}^{1}{2.dx}+\int_{1}^{2}{\left(4-2x\right)dx}\right|$$
$$=\left[{x}^{2}+2x\right]_{-1}^{0}+\left[2x\right]_{0}^{1}+\left[4x-{x}^{2}\right]_{1}^{2}$$
$$=4$$.sq.unit.
Question 10
1 / -0

The area between the curve $$y=2{x}^{4}-{x}^{2},$$ the $$x-$$axis and the ordinates of two minima of the curve is:

A
$$\dfrac{7}{120}$$ sq.unit

B
$$\dfrac{9}{120}$$ sq.unit

C
$$\dfrac{11}{120}$$ sq.unit

D
$$\dfrac{13}{120}$$ sq.unit

Solution

$$y=2{x}^{4}-{x}^{2}$$
$$\therefore \dfrac{dy}{dx}=8{x}^{3}-2x,$$ for max. or min. $$\dfrac{dy}{dx}=0$$
$$x=-\dfrac{1}{2},0,\dfrac{1}{2}$$
Then, $$\left(\dfrac{{d}^{2}y}{d{x}^{2}}\right)_{x=-\frac{1}{2}}>0, \left(\dfrac{{d}^{2}y}{d{x}^{2}}\right)_{x=0}<0$$
and $$\left(\dfrac{{d}^{2}y}{d{x}^{2}}\right)_{x=\frac{1}{2}}>0$$
$$\therefore $$Required Area$$=\left|\int_{-\frac{1}{2}}^{\frac{1}{2}}{\left(2{x}^{4}-{x}^{2}\right)dx}\right|$$
$$=\left[\dfrac{2{x}^{5}}{5}-\dfrac{{x}^{3}}{3}\right]_{-\frac{1}{2}}^{\frac{1}{2}}$$
$$=\dfrac{7}{120}$$.sq.unit.