$f(x)$ continuous on $[a, b], f(x)>0$

$m(b-a) \leqslant \int_a^b f(x) d x \leqslant {M(b-a)}$

$m \leqslant f(a) \leqslant M \quad\forall x[a, b]$

$2 e^{-2}<\int_0^2 e^{-x} d x<2$

$m(b-a) \leqslant \int_a^b f(x) d x \leqslant {M(b-a)}$

${\text { continuous on }[a, b] }$

Evaluate

$\int_{-1 / 2}^{1 / 2}[[x]+\log(\frac{1+x}{1-x})] d x$

$=\int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log(\frac{1+x}{1-x}) d x$

$\log (\frac{1+x}{1-x})=h(x)$

$h(-x)=\log(\frac{1-x}{1+x})$ $=-\log (\frac{1+x}{1-x})$ $=-h(x)$

$\therefore \int_{-1 / 2}^{1 / 2} \log(\frac{1+x}{1-x}) d x = 0$

$\therefore \int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log(\frac{1+x}{1-x}) d x =\int_{-1 / 2}^{1 / 2}[x] d x$

$=\int_{-1 / 2}^{0} (-1) d x$ $+\int_{0}^{1 / 2} (0) d x$ $= -x|_{-1 / 2} ^0=-\frac{1}{2}$

Evaluate

$I=\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x, \quad a>0$

$x=-t \quad d x=-d t \quad I=\int_{-\pi}^{-\pi} \frac{-\cos ^2 (-t) d t}{1+a^{-t}}$

$I=\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x \quad ......(1) =\int_ {-\pi}^\pi \frac{\cos ^2 t d t}{1+{a}^{-t}} =\int_{-\pi}^\pi \frac{a^x\cos ^2 x dx}{1+a^x} \quad ......(2)$

$\text { (1) +(2) }$

$2 I=\int_{-\pi}^\pi \frac{(1+a^{x}) \cos ^2 x}{(1+a^{x})} d x$ $2 I=\int_{-\pi}^\pi \cos ^2 x d x =2 \int_0^\pi \frac{(1+\cos 2 x)}{2} d x =\pi \Rightarrow I=\frac{\pi}{2}$

Compute

$I=\int_ {\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}} =\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan (\frac{\pi}{2}-x)}}$

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

$I=\int_ {\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}} =\int_{\pi / 6}^{\pi / 3} \frac{1+\sqrt{\tan x}-1}{1+\sqrt{\tan x}}) d x$

$I=\int_{\pi / 6}^{\pi / 3} 1 d x-\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x$

$2 I=\frac{\pi}{3}-\frac{\pi}{6} \Rightarrow I = \frac{\pi}{12}$

Evaluale $\int_0^{\pi / 4} \log (1+\tan x) d x$ $\int_0^a f(x) d x=\int_0^a f(a-x) d x$

solution:

$=\int_0^{\pi / 4} \log (1 + tan (\frac{\pi}{4}-x)) d x$

$=\int_0^{\pi / 4} \log (1+\frac{\tan \frac{\pi}{4}- tan x}{1+\tan \frac{\pi}{4} \tan x}) d x$

$=\int_0^{\pi / 4} \log (1+\frac{1-\tan x}{1+\tan x}) d x$ $=\int_0^{\pi / 4} \log (\frac{2}{1+\tan x}) d x$

$I=\int_0^{\pi / 4} \log (\frac{2}{1+\tan x}) d x \quad .....(2)$

$I=\int_0^{\pi / 4} \log (1+\tan x) d x \quad .....(1)$

$\text { (1) }+(2)$

$2 I=\int_0^{\pi / 4}[\log (\frac{2}{1+\tan x})+ log(1+\tan x)] d x$

$2 I=\int_0^{\pi / 4} \log [\frac{2}{1+\tan x} \times (1+\tan x)] d x$

$2 I=\int_0^{\pi / 4} \log 2 d x =\log 2 \ \frac{\pi}{4}$

$I=\left(\frac{\pi}{8}\right) \log 2$

$I=\int_{-\pi}^\pi \frac{2 x(1+\operatorname{sin} x)}{(1+\cos ^2 x)} d x$ Compute $I$.

solution:

$I =\int_{-\pi}^{\pi} \frac{2 x d x}{1+\cos ^2 x}+\int_{-\pi}^{\pi} \frac{2 x \sin x}{1+\cos^2 x} d x$

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

$f(-x)=\frac{-2 x}{1+\cos ^2(-x)}$ $=\frac{-2 x}{1+\cos ^2 x} = -f(x)$

$\int_{-a}^a f(x) d x=0$

$[f(x) \text { is odd }]$

$I =\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$

$g(x)=\frac{2x \sin x}{1+\cos ^2x}$

$g(-x)=\frac{2(-x) \sin (-x)}{1+\cos ^2(-x)} = \frac{2 x \sin x}{1+\cos ^2 x}$

$\quad =g(x)$

$\int_{-a}^a g(x) d x=2 \int_0^a g(x) d x$

$\int_0^a f(x) d x =\int _0^a f(a-x) d x$

$I =2 \int_0^{\pi} \frac{2 x \sin x}{1+\cos ^2 x} \quad ...(1) =2 \int_0^\pi \frac{2(\pi-x) \sin (\pi-x) d x}{1+(\cos ^2(\pi-x))}$

$I =4 \int_0^\pi \frac{(\pi- x) \sin x d x}{1+\cos ^2 x}\quad ...(2)$

$2 I=4 \int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x$

$\text { Let } \cos x=t$

$- \sin x d x=d t$

$I=2 \pi \int_1^{-1} \frac{-d t}{1+t^2}$

$=2 \pi \int_{-1}^1 \frac{d t}{1+t^2}$

$I=2 \pi \tan ^{-1} t|_{-1} ^1$

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

$I=2 \pi \frac{\pi}{2}=\pi^2$

Evaluate:

$I=\int_0^\pi \frac{x d x}{1+\cos \alpha \sin x}, 0<\alpha<\pi$

solution:

$I=\int_0^\pi \frac{x d x}{1+\cos \alpha \sin x} \quad ...(1)$

$\int_0^a f(x) d x =\int _0^a f(a-x) d x$

$I=\int_0^\pi \frac{(\pi-x) d x} {1+\cos \alpha \sin (\pi-x)} \quad ...(2)$

$2 I=\int_0^\pi \frac{\pi d x}{1+\cos \alpha \sin x}=\pi \int_0^\pi \frac{d x}{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+ \cos \alpha \ 2\sin \frac{x}{2} \cos \frac{x}{2}}$

$2 I=\pi \int_0^\pi \frac{\sec ^2 \frac{x}{2} d x}{\tan ^2 \frac{x}{2}+1+2 \tan \frac{x}{2} \cos \alpha}$

$\tan \frac{x}{2}=t$ $\sec ^2 \frac{x}{2} d x = 2 d t$ $\tan 0 = 0$ $\tan \frac{\pi}{2} = {\infty}$

$I=\frac{\pi}{2} \int_0^{\infty} \frac{2 d t}{t^2+1+2 t \cos \alpha+\cos ^2 \alpha-\cos ^2 \alpha}$ $=\pi \int_0^{\infty} \frac{d t}{(t+\cos \alpha)^2+\sin ^2 \alpha}$ $=\pi \frac{1}{\sin \alpha} \tan ^{-1}(\frac{t+\cos \alpha}{\sin \alpha}) |_0^{\infty}$

$I =\frac{\pi}{\sin \alpha}[\tan ^{-1} \infty -\tan ^{-1} \cot \alpha] , 0<\alpha<\pi$ $=\frac{\pi}{\sin \alpha}[\frac{\pi}{2}-(\frac{\pi}{2}-\alpha)]$

$I =\frac{\pi \alpha}{\sin \alpha}, 0<\alpha<\pi$

$I=\int_0^{\pi / 2} \frac{x \sin x \cos x d x}{\cos ^4 x+\sin ^4 x} \quad ...(1)$

Solution:

$\int_0^a f(x) d x$

$=\int_0^a f(a-x) d x$

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

$I=\int_0^{\pi / 2} \frac{(\frac{\pi}{2}-x) \cos x \sin x d x}{\sin ^4 x+\cos ^4 x} \quad ...(2)$

$(1)+(2)$

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

$I =\frac{\pi}{4} \int_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$ $=\frac{\pi}{4} \int_0^{\pi / 2} \frac{\frac{\sin x \cos x}{\cos ^4 x}}{{1+\tan ^4 x}} dx$

$I =\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan x \sec ^2 x d x}{1+\tan ^4 x}$ $\tan ^2 x=t$

$2 \tan x \sec ^2 x {d x}=d t$ $\tan 0=0$ $\tan \frac{\pi}{2}=\infty$

$I=\frac{\pi}{4} \int_{t=0}^{t=\infty} \frac{\frac{d t}{2}}{1+t^2}$

$I=\frac{\pi}{8} \int_0^{\infty} \frac{d t}{1+t^2} =\left.\frac{\pi}{8} \tan ^{-1} t\right|_0 ^{\infty}$

$I=\frac{\pi}{8}[\tan ^{-1} \infty-\tan ^{-1} 0]$ $\quad I=\frac{\pi^2}{16}$

$\int_0^{\pi / 4} \frac{(\sin x+\cos x)}{9+16 \sin 2 x} dx$

$=\int_0^{\pi / 4} \frac{\sin x+\cos x}{25-16+16 \sin 2 x} d x$

$=\int_0^{\pi / 4} \frac{(\sin x+\cos x) d x}{25-16(1-\sin 2 x)}$

$=\int_0^{\pi / 4} \frac{(\sin x+\cos x) d x}{25-16(\sin ^2 x+\cos ^2 x-2 \sin x \cos x)}$

$=\int_0^{\pi / 4} \frac{(\sin x+\cos x) d x}{25-16(\sin x-\cos x)^2}$

$\text { Let } \sin x-\cos x=t$ $(\cos x+\sin x) d x=d t$

$x = \frac{\pi}{4} , t = 0$ $x = 0 , t = -1$

$=\int_{-1}^0 \frac{d t}{25-16 t^2}$ $=\frac{1}{16} \int_{-1}^0 \frac{d t}{(\frac{5}{4})^2-t^2}$

$\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log |\frac{a+x}{a-x}|$

$I =\frac{1}{16} \frac{1}{2 \times \frac{5}{4}}(\log |\frac{\frac{5}{4}+t}{\frac{5}{4}-t}|)|_{-1} ^0$

$=\frac{1}{40}[\log 1-\log |\frac{1 / 4}{3 / 4}|]$

$=\frac{1}{40}(-1) \log (\frac{1}{9})$

$=\frac{1}{40} \log 9 =\frac{1}{40} \log 3^2=\frac{2}{40} \log 3$

$=\frac{1}{20} \log 3$

Sketch this region bounded between $y=x^2$ and $y=\frac{2}{1+x^2}$. Find out its area.

Solution:

$y=\frac{2}{1+x^2} ; \quad y^{\prime}=\frac{-2 \quad 2 x}{(1+x^2)^2}=\frac{-4 x }{(1+x^2)^2}$

$\forall x>0, y^{\prime}<0$

$\forall x<0, y^{\prime}>0$

$y(0)=\frac{2}{1+0}=2$ $y^{\prime}(0)=0$

$x^2=\frac{2}{1+x^2}$

$y=\frac{2}{1+y}$

$y^2+y-2=0$

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

$y=1=x^2 \\ \Rightarrow x= \pm 1$

$\text { Required Area } =\int_{-1}^1 \underbrace{[\frac{2}{1+x^2}-x^2] d x} _{\text { elementary area }}$

$=2 \int_{-1}^1 \frac{d x}{1+x^2}-\int_{-1}^1 x^2 d x$

$=2 \tan ^{-1} x|_{-1} ^1$

$-\frac{x^3}{3}|_{-1} ^1$

$=2(\frac{\pi}{4}-(-\frac{\pi}{4})-\frac{1}{3}(1-(-1))=\pi-\frac{2}{3}$

Sketch the curves and identify the region bounded by $x=\frac{1}{2}, x=2, y=\log x$ and $y=2^x$.

$\int_{\frac{1}{2}}^2\left[2^x-\log x\right] d x$

$=\biggl[\frac{2^x}{\log 2}-(x \log x-x)\biggl] |_{\frac{1}{2}} ^2$

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

$=\frac{4-\sqrt{2}}{\log 2}+\frac{3}{2}-\frac{5}{2} \log 2$