Mathematics M.C.Q (1 Marks)

Area bounded by curve $y=\frac{1}{x}, y=0, x=1, x=10$ is

$\int \limits_1^{10} \frac{1}{x} d x$$=[\log _e x]_1^{10}$$=\log _e 10$

$A =\log _e 10=2.303$

$B =1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots+\frac{1}{9}$

$C =\frac{1}{2}+\frac{1}{3}+\frac{1}{4} \ldots+\frac{1}{10}$

Clearly, $B > C$

Also, $\quad B > A$

$\left[\because \log _e(1+n) < 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}\right]$

$B > A > C$

$2 A > B + C$

$A-C > B-A$

 $A=\frac{1}{2} \times 4 \times 8+\int_4^6\left(6 x-x^2\right) d x $

$A=\frac{76}{3} $

$12 A=304$

$ =\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13} $

$ \therefore \alpha+\beta=171$

$ A=\frac{41}{3} $

$ 12 A=41 \times 4=164$

$ \text { Area }=\left|\frac{4 x^2}{2}-\frac{x^3}{3}-\frac{(x-4)^3}{9}\right|_1^4 $

$ =\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right| $

$ \Rightarrow(27-21)=6$

$ y(x+4)=16 \ldots $ (1)

$x+y=6 $

$ x+y=6$    $....(2)$

on solving, $(1) \& (2)$

we get $x=4, x=-2$

 Area = $ \int_{-2}^4\left((6-x)-\left(\frac{16}{x+4}\right)\right) d x$

$ =30-32 \ln 2$

$ \text { Shaded area }=\int_{-\frac{1}{2}}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y $

$ =\left(\frac{1}{4}\left(\frac{y^2}{2}+y\right)-\frac{y^3}{6}\right)_{-\frac{1}{2}}^1=\frac{9}{32}$

$Image$

$ \int_0^6(g(x)-f(x)) d x $

$ \int_0^6\left(\left(7 x-x^2\right)-\left(x^2-5 x\right)\right) d x $

$ \int_0^6\left(12 x-2 x^2\right) d x=\left[12 \frac{x^2}{2}-\frac{2 x^3}{3}\right]_0^6 $

$ \Rightarrow 6(6)^2-\frac{2}{3}(6)^3 $

$ =216-144=72 \text { unit }^2$

$A=\int_{-2}^0-x^2-2 x=\frac{4}{3}$

   $=9$

$(y-2)^2=x-1 \text { and } x-2 y+4=0$

$x=2(y-2)$

$\frac{x^2}{4}=x-1$

$x^2-4 x+4=0$
$(x-2)^2=0 $
$x=2
$Exclose area (w.r.t. y-axis) $=\int_0^3 x \mathrm{dy}-$ Area of $\Delta$.

$ =\int_0^3\left((y-2)^2+1\right) d y-\frac{1}{2} \times 1 \times 2 $

$ =\int_0^3\left(y^2-4 y+5\right) d y-1$

$ =\left[\frac{y^3}{3}-2 y^2+5 y\right]_0^3-1 $

$ =9-18+15-1=5$

$ \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0$

$\text { Case-I : y>0 }$

$\frac{x(x-1)(x-2)}{(x-3)(x-4)}>0$

$x \in(0,1) \cup(2,3)$

 $\text { Case - II : y<0 }$

$\frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \in(1,2) \cup(3,4)$

$ \text { Area }=2 \int_0^4 \sqrt{x} d x $

$ =2 \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^4=\frac{32}{3}$

$ 2 y^2=k x $

$ \text { Point of intersection } \rightarrow $

$ \mathrm{ky}^2=2\left(\mathrm{y}-\frac{2 \mathrm{y}^2}{\mathrm{k}}\right) $

$ \mathrm{y}=0 \quad \mathrm{ky}=2\left(1-\frac{2 \mathrm{y}}{\mathrm{k}}\right) $

$ \mathrm{ky}+\frac{4 \mathrm{y}}{\mathrm{k}}=2 $

$ y=\frac{2}{k+\frac{4}{k}}=\frac{2 k}{k^2+4} $

$ A=\int_0^{\frac{2 k}{k^2+4}}\left(\left(y-\frac{k y^2}{2}\right)-\left(\frac{2 y^2}{k}\right)\right) \cdot d y $

$ =\frac{\mathrm{y}^2}{2}-\left.\left(\frac{\mathrm{k}}{2}+\frac{2}{\mathrm{k}}\right) \cdot \frac{\mathrm{y}^3}{3}\right|_0 ^{\frac{2 \mathrm{k}}{\mathrm{k}^2+4}} $

$ =\left(\frac{2 \mathrm{k}}{\mathrm{k}^2+4}\right)^2\left[\frac{1}{2}-\frac{\mathrm{k}^2+4}{2 \mathrm{k}} \times \frac{1}{3} \times \frac{2 \mathrm{k}}{\mathrm{k}^2+4}\right] $

$ =\frac{1}{6} \times 4 \times\left(\frac{1}{\mathrm{k}+\frac{4}{\mathrm{k}}}\right)^2$

$A \cdot M \geq G \cdot M \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2$

$\mathrm{k}+\frac{4}{\mathrm{k}} \geq 4$

Area is maximum when $\mathrm{k}=\frac{4}{\mathrm{k}}$

$\mathrm{k}=2,-2$

$ A=\left[x+\frac{3 x^2}{2}-\frac{2 x^3}{3}-\ln x\right]_{\frac{1}{2}}^{\frac{1+\sqrt{5}}{2}} $

$ A=\frac{1+\sqrt{5}}{2}+\frac{3}{2}\left(\frac{1+\sqrt{5}}{2}\right)^2-\frac{2}{3}\left(\frac{1+\sqrt{5}}{2}\right)^3-\ln \left(\frac{1+\sqrt{5}}{2}\right) $

$ -\frac{1}{2}-\frac{3}{2}\left(\frac{1}{4}\right)+\frac{2}{3}\left(\frac{1}{8}\right)+\ln \left(\frac{1}{2}\right) $

$ A=\frac{1}{2}+\frac{\sqrt{5}}{2}+\frac{3}{8}+\frac{3}{4} \sqrt{5}+\frac{15}{8}-\frac{4}{3}-\frac{2}{3} \sqrt{5} $

$ -\frac{1}{2}-\frac{3}{8}+\frac{1}{12}-\ln (1+\sqrt{5}) $

$ =\sqrt{5}\left(\frac{1}{2}+\frac{3}{4}-\frac{2}{3}\right)+\frac{15}{8}-\frac{4}{3}+\frac{1}{12}-\ln (1+\sqrt{5}) $

$ =\frac{14}{24} \sqrt{5}+\frac{15}{24}-\ln (1+\sqrt{5})$

$Image$

$ \mathrm{A}=\int_0^3 3 \mathrm{x}-(3 \mathrm{x}-\mathrm{x} \sqrt{\mathrm{x}}) \mathrm{dx}+\int_3^9\left(\frac{27-3 \mathrm{x}}{2}-(3 \mathrm{x}-\mathrm{x} \sqrt{\mathrm{x}})\right) \mathrm{dx} $

$ \mathrm{A}=\int_0^3 \mathrm{x}^{3 / 2} \mathrm{dx}+\int_3^9 \frac{27}{2}-\frac{9 \mathrm{x}}{2}+\mathrm{x}^{3 / 2} \mathrm{dx} $

$ \mathrm{A}=\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_0^3+\frac{27}{2}[\mathrm{x}]_3^9-\frac{9}{2}\left[\frac{\mathrm{x}^2}{2}\right]_3^9+\left[\frac{2 \mathrm{x}^{5 / 2}}{5}\right]_3^9 $

$ \mathrm{~A}=\frac{2}{5}\left(3^{5 / 2}\right)+\frac{27}{2}(6)-\frac{9}{4}(72)+\frac{2}{5}\left(9^{5 / 2}-3^{5 / 2}\right) $

$ \mathrm{A}=\frac{2}{5}\left(3^{5 / 2}\right)+81-162+\frac{2}{5} \times 3^5-\frac{2}{5} \times 3^{5 / 2} $

$ \mathrm{~A}=\frac{486}{5}-81=\frac{81}{5} $

$ 10 \mathrm{~A}=162 $

$ A n s .=4$

$ {\left[\ln x+\frac{a}{x}\right]_1^2} $

$ \ell \operatorname{nn} 2+\frac{a}{2}-a=\log _e 2-\frac{1}{7} $

$ \frac{-a}{2}=-\frac{1}{7} $

$ a=\frac{2}{7} $

$ 7 a=2 $

$ 7 a-3=-1$

$ \int_0^{\pi / 4} \sin x=(\cos x)_{\pi / 4}^0=1-\frac{1}{\sqrt{2}} $

$Image$

$ \int_{-\pi}^{-3 \pi / 4}(\sin x-\cos x)=(-\cos x-\sin x)_{-\pi}^{-3 \pi / 4} $

$ =(\cos x+\sin x)_{-3 \pi / 4}^{-\pi} $

$ =(-1+0)-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right) $

$ =-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} $

$ \int_{\pi / 4}^{\pi / 2} \cos x d x=(\sin x)_{\pi / 4}^{\pi / 2}=1-\frac{1}{\sqrt{2}} $

$ A=4 $

$ A^2=16$

$ \operatorname{ar}(\text { para from } 0 \text { to } 2) $

$ =\int_0^2 \sqrt{8-\mathrm{x}^2} \mathrm{dx}-\int_0^2 \sqrt{2 \mathrm{x}} \mathrm{dx} $

$ =\left[\frac{\mathrm{x}}{2} \sqrt{8-\mathrm{x}^2}+\frac{8}{2} \sin ^{-1} \frac{\mathrm{x}}{2 \sqrt{2}}\right]_0^2-\sqrt{2}\left[\frac{\mathrm{x} \sqrt{\mathrm{x}}}{3 / 2}\right]_0^2 $

$ =\frac{2}{2} \sqrt{8-4}+\frac{8}{2} \sin ^{-1} \frac{2}{2 \sqrt{2}}-\frac{2 \sqrt{2}}{3}(2 \sqrt{2}-0) $

$ \Rightarrow 2+4 \cdot \frac{\pi}{4}-\frac{8}{3}=\pi-\frac{2}{3}$

$ x^2+y^2=5$

$\therefore$ Area of shaded region as shown in the figure will be

$Image$

$ \mathrm{A}_1=\int_0^1 \sqrt{4 \mathrm{x}} \mathrm{dx}+\int_1^{\sqrt{5}} \sqrt{5-\mathrm{x}^2} \mathrm{dx} $

$ =\frac{4}{3} \cdot\left[\mathrm{x}^{\frac{3}{2}}\right]_0^1+\left[\frac{\mathrm{x}}{2} \sqrt{5-\mathrm{x}^2}+\frac{5}{2} \sin ^{-1} \frac{\mathrm{x}}{\sqrt{5}}\right]_1^{\sqrt{5}} $

$ =\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) $

$ \therefore \text { Required Area }=2 \mathrm{~A}_1 $

$ =\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1}\left(\frac{1}{\sqrt{5}}\right) $

$ =\frac{2}{3}+5\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{\sqrt{5}}\right) $

$ =\frac{2}{3}+5 \cos ^{-1} \frac{1}{\sqrt{5}} $

$ =\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)$

$Image$

$ \frac{x^2}{18}+\frac{3 x^2}{18}=1 \Rightarrow 4 x^2=18 \Rightarrow x^2=\frac{9}{2} $

$ \int_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \frac{\sqrt{18-x^2}}{\sqrt{3}} d x $

$ =\frac{1}{\sqrt{3}}\left(\frac{x \sqrt{18-x^2}}{2}+\frac{18}{2} \sin ^{-1} \frac{x}{3 \sqrt{2}}\right)_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} $

$ =\frac{1}{\sqrt{3}}\left(9 \times \frac{\pi}{2}-\frac{3}{2 \sqrt{2}} \times \frac{3 \sqrt{3}}{\sqrt{2}}-9 \times \frac{\pi}{6}\right)$

Required Area

$ =\frac{1}{2} \times \frac{9}{2}+\left(\frac{18 \pi}{6}-\frac{9 \sqrt{3}}{4}\right) \frac{1}{\sqrt{3}} $

$ =\sqrt{3} \pi$

$y^2=-4(x-1)$

$A=\int \limits_{-4}^2\left(\frac{4-y^2}{4}-\frac{y-2}{2}\right) d y=9$

i.e. $y=3 x+2$

Point of intersection with curve $(2,8)$

So Area $=\int \limits_{-1}^2\left((3 x+2)-x^3\right) d x=\frac{27}{4}$

$\left|\int \limits_{-\pi}^{\frac{-3 \pi}{4}} \sin x d x\right|+\left|\int \limits_{\frac{-3 \pi}{4}}^{\frac{-\pi}{2}} \cos x d x\right|+\int \limits_{\frac{-\pi}{2}}^{\frac{\pi}{4}} \cos x d x+\int \limits_{\frac{\pi}{4}}^\pi \sin x d x$

$=4$

$\Rightarrow y^2-2 y+1=-x+1$

$(y-1)^2=-(x-1)$

$y=-x$

Points of intersection

$x^2+2 x=-x$

$x^2+3 x=0$

$x=0,-3$

$A=\int \limits_0^3\left(-y^2+2 y+y\right) d y$

$=\frac{3 y^2}{2}-\left.\frac{y^3}{3}\right|_0 ^3=\frac{9}{2}$

$8 A=36$

and $y=x^2+6$ is $\pm 2$

$\text { Area }=2 \int \limits_0^2\left(x^2+6-\frac{5 x^2}{2}\right) d x=\int \limits_0^{\frac{2 \alpha}{5}}\left(\alpha x-\frac{5 x^2}{2}\right) d x$

$\Rightarrow \int \limits_0^{\frac{2 \alpha}{5}}\left(\alpha x-\frac{5 x^2}{2}\right) d x=16$

$\Rightarrow \alpha^3=600$

$16(x+2)^2-64-(y-2)^2+4+44=0$

$16(x+2)^2-(y-2)^2=16$

$\frac{(x+2)^2}{1}-\frac{(y-2)^2}{16}=1$

$A=\int \limits_{-2}^{\sqrt{6}}\left(2-\left(x^2-4\right)\right) d x$

$A=\int \limits_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}$

$A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)$

$A=\frac{12 \sqrt{6}}{3}+\frac{28}{3}$

$A=4 \sqrt{6}+\frac{28}{3}$

$\text { shaded portion }=\text { circular }( OABC )$

$-\operatorname{Ar}(\triangle OAB )$

$=\frac{\pi(4)}{4}-\frac{1}{2}(2)(1)$

$A =(\pi-1)$

$\text { Area } B=\operatorname{Ar}(\triangle AOB )+\text { Area of arc of circle }( ABC )$

$=\frac{1}{2}(1)(2)+\frac{\pi(2)^2}{4}=\pi+1$

$\frac{ A }{ B }=\frac{\pi-1}{\pi+1}$

$\Delta=\frac{8}{3}(3 \sqrt{3}-1)+21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)-6 \sqrt{3}$

$\frac{1}{2}\left(\Delta-21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)\right)=\frac{2 \sqrt{3}-\frac{8}{3}}{2}$

$=\sqrt{3}-\frac{4}{3}$

Intersection point of $\cos x-\sin x=\sin x$

$\Rightarrow \quad \tan x =\frac{1}{2}$

Let $\psi=\tan ^{-1} \frac{1}{2}$

So, $\tan \psi=\frac{1}{2}, \sin \psi=\frac{1}{\sqrt{5}}, \cos \psi=\frac{2}{\sqrt{5}}$

$Area =\int \limits_\psi^{\pi / 2}(\sin x-|\cos x-\sin x|) d x$

$=\int \limits_\psi^{\pi / 4}(\sin x-(\cos x-\sin x)) d x +\int \limits_{\pi / 4}^{\pi / 2}(\sin x-(\sin x-\cos x)) d x$

$=\int \limits_\psi^{\pi / 4}(2 \sin x-\cos x) d x+\int_{\pi / 4}^{\pi / 2} \cos x d x$

$=[-2 \cos x-\sin x]_\psi^{\pi / 4}+[\sin x]_{\pi / 4}^{\pi / 2}$

$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2 \cos \psi+\sin \psi+\left(1-\frac{1}{\sqrt{2}}\right)$

$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\left(\frac{2}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}\right)+1-\frac{1}{\sqrt{2}}$

$=\sqrt{5}-2 \sqrt{2}+1$

$y^2=8 x$

Solving it

$x ^2=8 x$

$x=0,8$

$y=0,8$

$x=2$ will intersect occur at

$y^2=16 \Rightarrow \quad y=\pm 4$

Area of shaded

$=\int \limits_2^8(\sqrt{8 x}-x) d x=\int_2^8(2 \sqrt{2} \sqrt{x}-x) d x$

$=\left[2 \sqrt{2} \cdot \frac{x^{3 / 2}}{3 / 2}-\frac{x^2}{2}\right]_0^8$

$=\left(\frac{4 \sqrt{2}}{3} \cdot 2^{9 / 2}-32\right)-\left(\frac{4 \sqrt{2}}{3} \cdot 2^{9 / 2}-2\right)$

$=\frac{128}{3}-32-\frac{16}{3}+2=\frac{112-90}{3}=\frac{22}{3}= A$

$\therefore 3 A =22$

$D = p ^2-5 p = p ( p -5)$

$\therefore q =9$

$0 \leq y \leq( x -9)^2$

Area $=\int \limits_0^9(x-9)^2 dx =243$

$=2\left[2 x^2-x^3-x\right]_{1 / 3}^{1 / 2}$

$\therefore A=\frac{5}{108} \Rightarrow 540\,A=\frac{5}{108} \times 540=25$

$g(x)=\left[\begin{array}{ll}x^2 & x \geq 0 \\ x & x < 0\end{array}\right.$

$f o g(x)=f[g(x)]=\left[\begin{array}{cc}g(x) & g(x) \geq 0 \\ 0 & g(x) < 0\end{array}\right.$

$f o g(x)=\left[\begin{array}{cc}x^2 & x \geq 0 \\ 0 & x < 0\end{array}\right.$

$2 y-x=15$

$A=\int \limits_0^3\left(\frac{x+15}{2}-x^2\right) d x+\frac{1}{2} \times \frac{15}{2} \times 15$

$\frac{x^2}{4}+\frac{15 x}{2}-\left.\frac{x^3}{3}\right|_0 ^3+\frac{225}{4}$

$=\frac{9}{4}+\frac{45}{2}-9+\frac{225}{4}=\frac{99-36+225}{4}$

$=\frac{288}{4}=72$

Both curve are symmetric about $x =\frac{1}{2}$ Hence

$A=2 \int \limits_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}\left(\left(x-x^2\right)-(1-2 x)\right) d x$

$A=2 \int \limits_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}\left(-x^2+3 x-1\right) d x=2\left(\frac{-x^3}{3}+\frac{3}{2} x^2-x\right)_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}$

On solving $6 A +11=5 \sqrt{5}$

$(6 A +11)^2=125$

$\Rightarrow \quad A =\frac{ x ^3}{3}-\left.\frac{3 x ^2}{2}\right|_{-1} ^0+\frac{3 x ^2}{2}-\left.\frac{ x ^3}{3}\right|_0 ^2$

$\Rightarrow \quad A =\frac{11}{6}+\frac{10}{3}=\frac{31}{6}$

$\therefore \quad 12\,A =62$

$=\left(\frac{x^3}{3}\right)_1^2+8(\ell \operatorname{nx})_2^8-(x)_1^8$

$=\frac{7}{3}+8(2 \ell n 2)-7$

$=16 \ell n 2-\frac{14}{3}$

$\therefore \text { Required area }=\frac{1}{2}(1+3) \times 2=4$

$x^2=8-x^2$

$x^2=4$

$x= \pm 2$

$2\left(1.7+\int \limits_1^2\left(8-2 x^2\right) d x\right)-2 \int \limits_0^1\left(x^2\right) d x$

$=2\left[7+\left(8 x-\frac{2 x^3}{3}\right)_1^2\right]-2\left(\frac{x^3}{3}\right)_0^1$

$=2\left[7+\left(16-\frac{16}{3}\right)-\left(8-\frac{2}{3}\right)\right]-2\left(\frac{1}{3}\right)$

$=2\left[7+\frac{32}{3}-\frac{22}{3}\right]=2\left[7+\frac{10}{3}\right] \frac{-2}{3}$

$=\frac{60}{3}=20$

$=\left[\frac{ x ^3}{3}+\frac{3 x }{4}\right]_0^{\frac{1}{2}}+1$

$A =\frac{1}{24}+\frac{3}{8}+1$

$12 A =\frac{1}{2}+\frac{36}{8}+12$

$=\frac{1}{2}+\frac{9}{2}+12$

$=5+12$

$=17$

$A=\int \limits_1^{\sqrt{2}}\left(x-\left(2-x^2\right)\right) d x+\int \limits_{\sqrt{2}}^2\left(x-\left(x^2-2\right)\right) d x$

$=\left(1-2 \sqrt{2}+\frac{2 \sqrt{2}}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+\left(2-\frac{8}{3}+4\right)-\left(1-\frac{2 \sqrt{2}}{3}+2 \sqrt{2}\right)$

$=-4 \sqrt{2}+\frac{4 \sqrt{2}}{3}+\frac{7}{6}+\frac{10}{3}=\frac{-8 \sqrt{2}}{3}+\frac{9}{2}$

$6 A =-16 \sqrt{2}+27$

$\therefore 6 A +16 \sqrt{2}=27$

Solving circle and parabola simultaneously :

$2 y+y^2-4 y+4=4$

$y^2-2 y=0$

$y=0,2$

Put $y =2$ in $x ^2=2 y \rightarrow x = \pm 2$

$\Rightarrow(2,2)$ and $(-2,2)$

$=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi$

$=2\left[\int \limits_0^2 \frac{ x ^2}{2} dx -(4-\pi)\right]$

$=2\left[\left.\frac{ x ^3}{6}\right|_0 ^2-4+\pi\right]$

$=2\left[\frac{4}{3}+\pi-4\right]$

$=2\left[\pi-\frac{8}{3}\right]$

$=2 \pi-\frac{16}{6}$

Tangent at $(1,3)$

$y =4 x -1$

$A =\int \limits_0^1\left(2 x ^2+1\right) dx \text {-area of }(\Delta QOT )-\text { area of }$

$(\Delta PQR )+\text { area of }(\Delta QRS )$

$A =\left(\frac{2}{3}+1\right)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{60}$

$2 y^2=3(3-y)$

$2 y^2+3 y-9=0$

$2 y^2-3 y+6 y-9=0$

$(2 y-3)(y+2)=0 ; y=3 / 2$

$\text { Area }\left[\int_0^{\frac{3}{2}}\left(x_R-x_2\right) d y\right)-A_1$

$=\int \limits_0^{\frac{3}{2}}\left((3-y)-\frac{2 y^2}{3}\right) d y-\frac{\pi}{8}(2)$

$A=\left(3 y-\frac{y^2}{2}-\frac{2 y^3}{9}\right)_0^{\frac{3}{2}}-\frac{\pi}{4}$

$4 A+\pi=4\left[\frac{9}{2}-\frac{9}{8}-\frac{3}{4}\right]=\frac{21}{2}=10.50$

$\therefore 4(4 A+\pi)=42$

Hence required area $=\frac{1}{2} \times 2 \times 2-\int_0^2 \frac{x^2}{2} d x-\left(\frac{\pi}{4}-\frac{1}{2}\right)$

$=\frac{7}{6}-\frac{\pi}{4} \Rightarrow n=5$

$y=\sqrt{2} x.........(2)$

$y^{2}=2 x^{2}$

$\Rightarrow 8 x=2 x^{2}$

$\Rightarrow x=0 \,\,and \,\,4$

Area $:=\int\limits_{1}^{4}\, 2 \sqrt{2} \sqrt{x}-\sqrt{2} x \,d x$

$=2 \sqrt{2}\left(\frac{x^{\frac{3}{2}}}{3 / 2}\right)_{1}^{4}-\sqrt{2}\left(\frac{x^{2}}{2}\right)_{1}^{4}$

$=\frac{4 \sqrt{2}}{3}(8-1)-\frac{\sqrt{2}}{3}(16-1)$

$=\frac{28 \sqrt{2}}{3}-\frac{15 \sqrt{2}}{2}=\frac{11 \sqrt{2}}{6}$

$=2 \int \limits_{0}^{1} \frac{1-y^{2}}{4} d y=\frac{1}{2}\left|y-\frac{y^{3}}{3}\right|_{0}^{1}=\frac{1}{3}$

on subtracting, we get $(a-b)^{2}=a^{2}-2 a b+b^{2}=12$