Find out area of the region described by

${(x, y): y^2 \leqslant 2 x, y \geqslant 4 x-1}$

Solution:-

$y^2 \leqslant 2 x .$

$y^2=2 x$

$\ 2 x=-2$

$y^2=0$

$0>-2$

$y \geqslant 4 x-1$

$y=4 x-1$

$y \geqslant 4 x-1$

$4 x-1=-1$

y=1

1>-1

$4 x-\left.1\right|_ {x=0}=-1 \left.y\right|_{-2}=-2$

-2<-1

$y^2=2 x, \quad y=4 x-1$

$y^2 \leqslant 2 x, \quad y \geqslant 4 x-1$

$\begin{aligned} & y^2=2 x, y=4 x-1, \\ & y=4 \frac{y^2}{2}-1\end{aligned}$

$\ 2 y^2-y-1=0, \\ y=\frac{1 \pm \sqrt{1+8}}{4}=\frac{1 \pm 3}{4}=1,-\frac{1}{2}$

$\int_{-1 / 2}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) dy$

$=\frac{1}{4}(\frac{y^2}{2}+y)|_ {-1 / 2} ^1-\frac{1}{2}$ $\frac{y^3}{3}|_{-1 / 2} ^1$

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

$=\frac{1}{4}\left(\frac{1}{2}+1-\frac{1}{8}+\frac{1}{2}\right)-\frac{1}{6}\left(1+\frac{1}{8}\right)$

$=\frac{1}{4}\left(\frac{15}{8}\right)-\frac{9}{48} \\ =\frac{15}{32}-\frac{3}{16}=\frac{9}{32}$

$\quad y=\sin x+\cos x,\\ y=|\cos x-\sin x|$

Find out area enclosed by above two curves and the lines $x=0$ and $x=\frac{\pi}{2}$.

solution:

$y = \sin x + \cos x$

$\quad y(0)=1, y\left(\frac{\pi}{2}\right)=1$

$y^{\prime}=\cos x-\sin x>0,\left[0,\frac{\pi}{4}\right)$

$y^{\prime}<0 \quad\left(\frac{\pi}{4}, \frac{\pi}{2}\right]$

$y^{\prime}\left(\frac{\pi}{4}\right)=0,y\left(\frac{\pi}{4}\right)=\sqrt{2}$

$y=\cos x-\sin x,\quad\left[0, \frac{\pi}{4}\right]$

$y^{\prime}=-\sin x-\cos x < 0,[0, \frac{\pi}{4}]$

$\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow y=-(\cos x-\sin x)$

$y^{\prime}=\sin x+\cos x>0,\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$

$y(0)=1, y\left(\frac{\pi}{4}\right)=0$

$y\left(\frac{\pi}{2}\right)=1$

$A=\int_0^{\pi / 4} {[\sin x+\cos x}{-(\cos x-\sin x) dx}$

$+\int_{\pi / 4}^{\pi / 2}[\sin x+\cos x-(-\cos x + \sin x) d x]$

$=2 \int_0^{\pi / 4} \sin x d x+2 \int_{\pi / 4}^- {\pi / 2} \cos x d x$

$=\left.2(-\cos x)\right|_ 0 ^{\pi / 4}+2(\sin x)_{\pi / 4}^{\pi / 2}$

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

Required Area $=4\left(1-\frac{1}{\sqrt{2}}\right)$Required Area $=2A$

$y=x-b x^2, \quad b>0$

$y=\frac{x^2}{b}, \quad b>0$

$\frac{x^2}{b}=x-b x^2$

$\frac{x^2}{b}+b x^2=x \quad x=0$

$x^2 (\frac{1+b^2}{b}) =x$ $x=\frac{b}{1+b^2}$

Area Required

$=\int_0^{b / 1+b^2}\left[x-b x^2-\frac{x^2}{b}\right] d x$

$\text { Required \ Area }=\frac{x^2}{2}-\frac{b}{3} x^3-\left.\frac{x^3}{3 b}\right|_0 ^{b/1+b^2}$

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

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

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

= $\frac{3-2}{6} \frac{b^2}{\left(1+b^2\right)^2}=\frac{b^2}{6\left(1+b^2\right)^2}$

$A=\frac{1}{6} .\frac{b^2}{\left(1+b^2\right)^2}, \quad b>0$

for which value of $b$ Area is maximum.

$\begin{aligned} \frac{d A}{d b} & =\frac{1}{6} \frac{\left(1+b^2\right)^2 2 b-b^2 2\left(1+b^2\right) 2 b}{\left(1+b^2\right)^4} \\ & =\frac{1}{6} \frac{1+b^2}{\left(1+b^2\right)^4}\left[2 b+2 b^3-4 b^3\right] \\ & =\frac{1}{6} \frac{1}{\left(1+b^2\right)^3}\left(2 b-2 b^3\right) \\ & =\frac{1}{3} b \frac{\left(1-b^2\right)}{\left(1+b^2\right)^3} \end{aligned}$

$\frac{d A}{d b}=\frac{1}{3} b \frac{\left(1-b^2\right)}{\left(1+b^2\right)^3}$

$\frac{d A}{d b}=0, b=0, b=-1, b=+1, \quad b>0$

So only possible value of " $b$ " $=1$

$b > 1, \frac{d A}{d b} < 0$

$0 < b < 1,\frac{d A}{d b} > 0$

at $b=1$ Area is maximum

$\frac{d A}{d b}=\frac{1}{3} b\left(\frac{(1-b^2)}{\left(1+b^2)\right.}\right) 3$

$I=\int_2^4 \frac{\log x^2 d x}{\log x^2+\log \left(36-12x+x^2\right)}$

solution:

$=\int_2^4 \frac{2 \log x d x}{2 \log x+\log (6-x)^2}$

$I =\int_2^4 \frac{2 \log x d x}{2\log x+2 \log (6-x)}\quad \cdots \cdots (1)$

$\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$

$I=\int_2^4 \frac{\log (6-x) d x}{\log (6-x)+\log (6-(6-x))}$

$I=\int_2^4 \frac{\log (6-x) d x}{\log (6-x)+\log x} \quad \cdots \cdots (2)$

$\begin{aligned} & \int_{-\pi / 2}^{\pi / 2}\left[x^2+\log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x d x \\ & =\int_{-\pi / 2}^{\pi / 2} x^2 \cos x d x+\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x d x \\ & =2 \int_0^{\pi / 2} x^2 \cos x d x+0\end{aligned}$

$h(x)=\log \left(\frac{\pi-x}{\pi+x}\right) \cos x \quad$

$h(-x)=\operatorname{log}\left(\frac{\pi+x}{\pi-x}\right) \cos (-x)$

$=-\log \left(\frac{\pi-x}{\pi+x}\right) \cos x$

$=-h(x) \int_{-\pi / 2}^{\pi / 2} h(x) d x=0$

$I =2 \int_0^{\pi / 2} x^2 \cos x d x$

$=2\left[\left.x^2(\sin x)\right|_0 ^{\pi / 2}-\int_0^{\pi / 2} 2 x \sin x d x\right]$

$=2\left[\frac{\pi^2}{4}-0-\left.2 x(-\cos x)\right|_0 ^{\pi / 2}+\int_0^{\pi / 2} 2(-\cos x) d x\right.$

$=2\left[\frac{\pi^2}{4}-\left.2 \sin x\right|_0 ^{\pi / 2}\right] =2\left[\frac{\pi^2}{4}-2\right]=\frac{\pi^2}{2}-4$

$\int_{\sqrt{\log (2)}} ^{\sqrt{\log (3)}} \frac{x \sin \left(x^2\right)}{\sin x^2+\sin \left[\log 6-x^2\right]}$

$x^2=t, \quad 2 x d x=d t$

$=\int_{t=\log 2}^{t=\log 3} \frac{\sin t\left(\frac{1}{2} d t\right)}{\sin t+\sin [\log 6-t]}$

$I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t d t}{\sin t+\sin [\log 6-t]}$

$\because \int_a^b f(x) dx = \int _a^b f(a+b-x) dx$

$\begin{aligned} & I= \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+ \sin t} d t \\ & 2I = \frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t + \sin (\log 6-t)}{\sin (\log 6-t)+ \sin t} d t \\ & 2 I=\frac{1}{2} \int_{\log 2}^{\log 3} d t=\frac{1}{2}[\log 3-\log 2] \\ & 2 I=\frac{1}{2} \log (\frac{3}{2}), \\ & I=\frac{1}{4} \log (\frac{3}{2}) .\end{aligned}$