Application of Integrals

 $Area=2{\displaystyle \underset{0}{\overset{\sqrt[]{2}}{\int }}\left(1-\left|x^2-1\right|\right)dx=2\left[{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-\left(1-x^2\right)\right)dx+{\displaystyle \underset{1}{\overset{\sqrt[]{2}}{\int }}\left(2-x^2\right)dx}}\right]}=\frac{8}{3}\left(\sqrt[]{2}-1\right)$

Since a is a odd natural number then $\left|{\displaystyle \underset{1}{\overset{3}{\int }}{y}^{a}dy}\right|=\frac{364}{3}\Rightarrow \left|{\left(\frac{y^{a+1}}{a+1}\right)}_1^3\right|=\frac{364}{3}\Rightarrow \frac{3^{a+1}}{a+1}=\frac{364}{3}$

a = 5

Required area is

= ${\displaystyle \underset{e-e^2}{\overset{0}{\int }}\mathcal{l}n\left(x+e^2\right)-1dx+{\displaystyle \underset{0}{\overset{\mathcal{l}n2}{\int }}2e^{-x}-1dx}}$

$=1+e-\mathcal{l}n2$

 $\overrightarrow{a}=2\widehat{i}+\widehat{j}+3\widehat{k}$

$\begin{array}{l}\overrightarrow{b}=3\widehat{i}+3\widehat{j}+\widehat{k}\\ \overrightarrow{c}=c_1\widehat{i}+c_2\widehat{j}+c_3\widehat{k}\end{array}$

$Coplnanar\Rightarrow \left|\begin{array}{ccc}\hfill 2\hfill & \hfill 1\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right|=0$

$\begin{array}{l}\Rightarrow -8c_1+7c_2+12c_3=0........(i)\\ \overrightarrow{a}.\overrightarrow{c}=5\Rightarrow 2c_1+c_2+3c_3=5........(ii)\\ \overrightarrow{b}.\overrightarrow{c}=0\Rightarrow 3c_1+3c_2+c_3=0........(iii)\end{array}$

Solving (i), (ii), (iii)

$C_1=\frac{10}{122},c_2=\frac{-85}{122},c_3=\frac{225}{122}$

$\therefore 122\left(c_1+c_2+c_3\right)=150$

 $\left|x-1\right|\le y\le \sqrt[]{5-x^2}$

$y=\left|x-1\right|,y^2+x^2=5,y\ge 0y=\sqrt[]{5-x^2}$

${\left(x-1\right)}^2+x^2=5\Rightarrow 2x^2-2x-4=0\Rightarrow x^2-x-2=0$

$\left(x-2\right)\left(x+1\right)=0\to x=-1,2$

$A={\displaystyle \underset{-1}{\overset{2}{\int }}\left(\sqrt[]{5-x^2}-\left|x-1\right|\right)dx}$

$Area={\displaystyle \underset{-1}{\overset{2}{\int }}\left(\sqrt[]{5-x^2}dx-\left|x-1\right|dx\right)}$

$={\displaystyle \underset{-1}{\overset{2}{\int }}\sqrt[]{5-x^2}dx+{\displaystyle \underset{-1}{\overset{1}{\int }}\left(x-1\right)dx-{\displaystyle \underset{1}{\overset{2}{\int }}\left(x-1\right)dx}}}$

$=\frac{5}{2}si{n}^{-1}\left(1\right)-\frac{1}{2}=\frac{5\pi }{4}-\frac{1}{2}$

Area of shaded region

$={\displaystyle {\int }_{-\left(\frac{1}{2}\right)}^0\frac{3}{2}\left({\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}}+x\right)dx+{\displaystyle {\int }_0^1{\left(1-x^{\frac{2}{3}}\right)}^{\frac{3}{2}}dx}}$

$x=si{n}^3\theta$

$\therefore dx=3si{n}^2\theta cos\theta d\theta$

$={\displaystyle {\int }_{-\frac{\pi }{4}}^{\frac{\pi }{2}}3si{n}^2\theta co{s}^4\theta d\theta +\left(0-\frac{1}{16}\right)}$

$=\frac{9\pi }{64}+\frac{1}{16}-\frac{1}{16}=\frac{36\pi }{256}=A$

$\therefore \frac{256A}{\pi }=36$

Given curve is $y=cosx$

$y=sinx$ for $0\le x\le \frac{\pi }{2}$

And $y-axis$

We know that $sinx=cosx$ at $x=\frac{\pi }{4}and<\frac{\pi }{4}<\frac{\pi }{2}$ i.e,  $cos\frac{\pi }{4}=sin\frac{\pi }{4}=1/\surd 2$

So the point of intersection is at $x=\frac{\pi }{4}$

The given area of the circle is $x^2+y^2=16-------(1)$ is a circle with centre (0,0) and radius, $\pi =4$ and the parabola is $y^2=6x$ -------------(2)

Solving (1) and (2) for x and y.

$\begin{array}{l}{x}^2+6x=16\\ =x^2+6x-16=0\\ =x^2+8x-2x-16=0\\ =x\left(x+8\right)-2\left(x+8\right)=0\\ =\left(x+8\right)\left(x-2\right)=0\\ =x=-8\&x=2\end{array}$

For, $x=-8,y^2=6\left(-8\right)=-48$

Which is not possible.

For, $x=2,y^2=6\left(2\right)=12$

$y=\pm \mathrm{2\surd 3}$

$areaOACBO=2*\left\{area\left(OADO\right)+area\left(ACDA\right)\right\}$

The given curve is $y=x\left|x\right|$

$=\left\{\left(x.xifx\ge 0\right)\left(x.\left(-x\right)ifx\le 0\right)\right\}=\left\{\left(x^{2}if,x\ge 0\right)\left(-x^{2}if,x\le 0\right)\right\}$

Which is in the form of a parabola nad the lines are $x=-1\&x-axis$

At $x=1>0,y=1^2=1$

At $$$x=-1<0,y=-1^2=-1$

Shaded area of the Ist quadrant

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}ydx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{3}\end{array}$

Shaded area of the IInd quadrant

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{0}{\int }}ydx}={\displaystyle \underset{-1}{\overset{0}{\int }}{x}^{2}dx}\\ ={\left[\frac{-x^3}{-3}\right]}_{-1}^0\\ =\frac{1}{3}\end{array}$

$\therefore$ Total area of the enclosed region $=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}uni{t}^2$

$\therefore$ Option (c) is correct.