Let $\mathrm{I}=\int_0^2\left(\mathrm{x}^2+3\right) \mathrm{dx}$
Here, $a=0, b=2$ and
$$
\begin{aligned}
&\mathrm{h}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}}=\frac{2-0}{\mathrm{n}} \\
&\Rightarrow \mathrm{h}=\frac{2}{\mathrm{n}} \Rightarrow \mathrm{nh}=2 \Rightarrow \mathrm{f}(\mathrm{x})=\left(\mathrm{x}^2+3\right)
\end{aligned}
$$
Now,
$$
\int_0^2\left(x^2+3\right) d x=\lim _{h \rightarrow 0} h[f(0)+f(0+h)+f(0+2 h)
$$
$\because \quad \mathrm{f}(0)=3$
$\begin{aligned} \Rightarrow & f(0+h)=h^2+3 \\ & f(0+2 h) \\ \quad=4 h^2+3=2^2 h^2+3 \end{aligned}$
$\quad f[0+(n-1) h]=\left(n^2-2 n+1\right) h+3$
$=(n-1)^2 h+3$
$\quad\left[\because \Sigma n^2=\frac{n(n+1)(2 n+1)}{6}\right]$
$=\lim _{h \rightarrow 0} h\left[3 n+h^2\left(\frac{\left(n^2-n\right)(2 n-1)}{6}\right)\right]$
$=\lim _{h \rightarrow 0} h\left[3 n+\frac{h^2}{6}\left(2 n^3-n^2-2 n^2+n\right)\right]$
$=\lim _{h \rightarrow 0}\left[3 n h+\frac{2 n^3 h^3-3 n^2 h^2 \cdot h+n h \cdot h^2}{6}\right]$


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