please help me integrate this $$\int_1^3 (e^{2-3x}+x^2+1)dx$$ as a limit of a sum
I know that there is an identity used to solve this but how to apply that identity. Here is the identity $$\int^b_af(x)dx=\lim_{h \to 0}\:\:h[f(a)+f(a+h)+f(a+2h+...+f[a+(n-1)h])]$$
$$I=\int_{1}^{3} (e^{2-3x}+x^2+1) dx =e^2 \int_{1}^{3} e^{-3x} dx+\int_{1}^{3} (x^2+1) dx$$ $$\implies I= \lim_{n \to \infty} \frac{1}{n}\left(e^2 \sum_{k=n}^{3n} e^{-3k/n}+\sum_ {k=n}^{3n} \frac{k^2}{n^2}+\sum_{k=n}^{3n}1\right)$$
$$\implies \lim_{n \to \infty}\left[ \frac{e^2}{n}\left( \frac{e^{-9-3/n}-1}{e^{-3/n}-1}-\frac{e^{-3-3/n}-1}{e^{-3/n-1}-1}\right)+\frac{1}{n^3} \left( \frac{3n(3n+1)(6n+1)}{6}-\frac{n(n+1)(2n+1)}{6}\right)+\frac{3n-n}{n}\right]$$ let $h=1/n$ $$\implies I=(e^{-7}-e^{-1})\lim_{h\to 0} \frac{h}{e^{-3h}-1}+\lim_{n \to \infty} \frac{1}{n^3}\frac{ 52n^3}{6}+2.$$ $$\implies \frac{(e^{-1}-e^{-7})}{3}+\frac{32}{3}$$