Let $$I=\int_0^2\left(x^2+3\right) d x$$

$$\begin{array}{ll} \text { Here, } & a=0, b=2 \text { and } h=\frac{b-a}{n}=\frac{2-0}{n} \\ \Rightarrow & h=\frac{2}{n} \Rightarrow n h=2 \Rightarrow f(x)=\left(x^2+3\right) \end{array}$$

Now, $\quad \int_0^2\left(x^2+3\right) d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)+\ldots+f\{0+(n-1) h\}]\quad\text{.... (i)}$

$\because\quad f(0)=3$

$$\begin{aligned} \Rightarrow \quad f(0+h)=h^2+3, f(0+2 h) & =4 h^2+3=2^2 h^2+3 \\ f[0+(n-1) h] & =\left(n^2-2 n+1\right) h+3=(n-1)^2 h+3 \end{aligned}$$

From Eq. (i),

$$\int_0^2\left(x^2+3\right) d x=\lim _{h \rightarrow 0} h\left[3+h^2+3+2^2 h^2+3+3^2 h^2+3+\ldots+(n-1)^2 h^2+3\right]$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} h\left[3 n+h^2\left\{1^2+2^2+\ldots+(n-1)^2\right\}\right] \\ & =\lim _{h \rightarrow 0} h\left[3 n+h^2\left(\frac{(n-1)(2 n-2+1)(n-1+)}{6}\right)\right]\left[\because \Sigma n^2=\frac{n(n+1)(2 n+1)}{6}\right] \\ & =\lim _{h \rightarrow 0} h\left[3 n+h^2\left(\frac{\left(n^2-n\right)(2 n-1)}{6}\right)\right] \end{aligned}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} h\left[3 n+\frac{h^2}{6}\left(2 n^3-n^2-2 n^2+n\right)\right] \\ & =\lim _{h \rightarrow 0}\left[3 n h+\frac{2 n^3 h^3-3 n^2 h^2 \cdot h+n h \cdot h^2}{6}\right] \\ & =\lim _{h \rightarrow 0}\left[3 \cdot 2+\frac{2 \cdot 8-3 \cdot 2^2 \cdot h+2 \cdot h^2}{6}\right]=\lim _{h \rightarrow 0}\left[6+\frac{16-12 h+2 h^2}{6}\right] \\ & =6+\frac{16}{6}=6+\frac{8}{3}=\frac{26}{3} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Let }\quad & I=\int_0^2 e^x d x \\ \text{Here,}\quad & a=0 \text { and } b=2 \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \therefore & h=\frac{b-a}{n} \\ \Rightarrow & n h=2 \text { and } f(x)=e^x \end{array}$$

Now, $$\int_0^2 e^x d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)+\ldots+f\{0+(n-1) h\}]$$

$\therefore \quad I=\lim _{h \rightarrow 0} h\left[1+e^h+e^{2 h}+\ldots+e^{(n-1) h}\right]$

$$ \begin{aligned} & =\lim _{h \rightarrow 0} h\left[\frac{1 \cdot\left(e^h\right)^n-1}{e^h-1}\right]=\lim _{h \rightarrow 0} h\left(\frac{e^{n h}-1}{e^h-1}\right) \\ & =\lim _{h \rightarrow 0} h\left(\frac{e^2-1}{e^h-1}\right) \\ & =e^2 \lim _{h \rightarrow 0} \frac{h}{e^h-1}-\lim _{h \rightarrow 0} \frac{h}{e^h-1} \quad \left[\because \lim _{h \rightarrow 0} \frac{h}{\mathrm{e}^h-1}=1\right]\\ & =e^2-1=e^2-1 \end{aligned}$$

Let $I=\int_0^1 \frac{d x}{e^x+e^{-x}}=\int_0^1 \frac{e^x}{1+e^{2 x}} d x$

Put $$e^x=t$$

$\Rightarrow \quad e^x d x=d t$

$\therefore \quad I=\int_1^e \frac{d t}{1+t^2}=\left[\tan ^{-1} t\right]_1^e$

$$\begin{aligned} & =\tan ^{-1} e-\tan ^{-1} 1 \\ & =\tan ^{-1} e-\frac{\pi}{4} \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int_0^{\pi / 2} \frac{\tan x d x}{1+m^2 \tan ^2 x} d x \\ & =\int_0^{\pi / 2} \frac{\frac{\sin x}{\cos x}}{1+m^2 \cdot \frac{\sin ^2 x}{\cos ^2 x}} d x \\ & =\int_0^{\pi / 2} \frac{\frac{\sin x}{\cos x}}{\frac{\cos ^2 x+m^2 \sin ^2 x}{\cos ^2 x}} d x \\ & =\int_0^{\pi / 2} \frac{\sin x \cos x d x}{1-\sin ^2 x+m^2 \sin ^2 x} d x \\ & =\int_0^{\pi / 2} \frac{\sin x \cos x}{1-\sin ^2 x\left(1-m^2\right)} d x \end{aligned}$$

$$\begin{array}{lr} \text { Put } & \sin ^2 x=t \\ \Rightarrow & 2 \sin x \cos x d x=d t \end{array}$$

$$\begin{aligned} \therefore\quad I & =\frac{1}{2} \int_0^1 \frac{d t}{1-t\left(1-m^2\right)} \\ & =\frac{1}{2}\left[-\log \left|1-t\left(1-m^2\right)\right| \cdot \frac{1}{1-m^2}\right]_0^1 \\ & =\frac{1}{2}\left[-\log \left|1-1+m^2\right| \cdot \frac{1}{1+m^2}+\log |1| \cdot \frac{1}{1-m^2}\right] \\ & =\frac{1}{2}\left[-\log \left|m^2\right| \cdot \frac{1}{1-m^2}\right]=\frac{2}{2} \cdot \frac{\log m}{\left(m^2-1\right)} \\ & =\log \frac{m}{m^2-1} \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int_1^2 \frac{d x}{\sqrt{(x-1)(2-x)}}=\int_1^2 \frac{d x}{\sqrt{2 x-x^2-2+x}} \\ & =\int_1^2 \frac{d x}{\sqrt{-\left(x^2-3 x+2\right)}} \end{aligned}$$

$$\begin{aligned} & =\int_1^2 \frac{d x}{\sqrt{-\left[x^2-2 \cdot \frac{3}{2} x+\left(\frac{3}{2}\right)^2+2-\frac{9}{4}\right]}} \\ & =\int_1^2 \frac{d x}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2\right\}}} \\ & =\int_1^2 \frac{d x}{\sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{3}{2}\right)^2}}=\left[\sin ^{-1}\left(\frac{x-\frac{3}{2}}{\frac{1}{2}}\right)\right]_1^2 \end{aligned}$$

$$\begin{aligned} & =\left[\sin ^{-1}(2 x-3)\right]_1^2=\sin ^{-1} 1-\sin ^{-1}(-1) \\ & =\frac{\pi}{2}+\frac{\pi}{2} \quad\left[\because \sin \frac{\pi}{2}=1 \text { and } \sin (-\theta)=-\sin \theta\right] \\ & =\pi \end{aligned}$$