Evaluate \(\int_2^3 x^2 d x\) as the limit of a sum

\[
\begin{aligned}
& \text { Let } I=\int_2^3 x^2 d x \\
& \qquad \quad I=\int_2^3 x^2 d x=\int_2^3 f(x) d x \text {, where } f(x)=x^2 \\
& \text { Also, } a=2, b=3 \\
& \lim _{h \rightarrow 0} h[f(2)+f(2+h)+f(2+2 h) \\
& \text { where, } n h=b-a=3-2=1 \\
& =\lim _{h \rightarrow 0} h\left[2^2+(2+h)^2+(2+2 h)^2+\ldots+(2+(n-1) h)^2\right] \\
& =\lim _{h \rightarrow 0} h\left[2^2+\left(2^2+h^2+4 h\right)+\left(2^2+2^2 h^2+8 h\right)\right. \\
& =\lim _{h \rightarrow 0} h\left[2^2 h+h^2\left(1^2+2^2+\ldots . .+(n-1)^2\right)\right. \\
& \left.=\lim _{h \rightarrow 0}\left[4 n h+\frac{h^3(n-1) n(2(n-1)+2+1)}{6}+4+(n-1)\right)\right] \\
& =\lim _{h \rightarrow 0}\left[4 n h+\frac{(n h-h)(n h)(2 n h-h)}{6}+2(n h-h)(n h)\right] \\
& =\lim _{h \rightarrow 0}\left[4 \times 1+\frac{(1-h) \cdot 1(2-h)}{6}+2(1-h) \cdot 1\right] \\
& =4+\frac{2}{6}+2=6+\frac{1}{3}=\frac{19}{3}
\end{aligned}
\]