$$\begin{gathered} \mathrm{I} = {\int }_1^3({\mathrm{x}}^2 + 3\mathrm{x}+ {\mathrm{e}}^{\mathrm{x}}) \mathrm{dx} \\ = {\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right) \mathrm{dx} \\ \mathrm{when} \mathrm{f}\left(\mathrm{x}\right) = {\mathrm{x}}^2 + 3\mathrm{x} + {\mathrm{e}}^{\mathrm{x}}; \mathrm{a} =1, \mathrm{b} =3 \\ \mathrm{h} =\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}} = \frac{3-1}{\mathrm{n}} =\frac{2}{\mathrm{n}} \\ \mathrm{f} (\mathrm{a} +\mathrm{rh}) = \mathrm{f}( 1 +\mathrm{rh}) = (1 +\mathrm{rh}{)}^2 + 3(1 +\mathrm{rh}) +{\mathrm{e}}^{1+\mathrm{rh}} \\ = 4 + 5\mathrm{rh} +{\mathrm{r}}^2{\mathrm{h}}^2 +\mathrm{e} \mathrm{x} {\mathrm{e}}^{\mathrm{rh}} \\ = {\mathrm{r}}^2{\mathrm{h}}^2 + 5\mathrm{rh} + 4 +\mathrm{e} \mathrm{x} {\mathrm{e}}^{\mathrm{rh}} \end{gathered}$$

$$\begin{gathered} {\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right) \mathrm{dx} = \underset{\mathrm{n} \to \infty }{\lim } \sum _{\mathrm{r} =1}^{\mathrm{n}} \mathrm{hf}(\mathrm{a} +\mathrm{rh}) \\ \underset{\mathrm{n}\to \infty }{\lim } \sum _{\mathrm{r} = 1}^{\mathrm{n}}\mathrm{h} ({\mathrm{r}}^2{\mathrm{h}}^2 + 5\mathrm{rh} + 4 + \mathrm{ex} {\mathrm{e}}^{\mathrm{rh}}) \\ \underset{\mathrm{n}\to \infty }{\lim } \left(\underset{\mathrm{r} = 1}{\overset{\mathrm{n}}{{\sum }^2}}{\mathrm{r}}^2{\mathrm{h}}^2 +\sum _{\mathrm{r} = 1}^{\mathrm{n}}{\mathrm{rh}}^2 + \sum _{\mathrm{r} = 1}^{\mathrm{n}}4\mathrm{h} +\mathrm{e}\sum _{\mathrm{r} = 1}^{\mathrm{n}}{\mathrm{e}}^{\mathrm{rh}} \mathrm{x} \mathrm{h}\right) \\ \underset{\mathrm{n}\to \infty }{\lim }\left(\frac{8}{{\mathrm{n}}^3} \mathrm{x} \frac{\mathrm{n} (\mathrm{n} +1)(2\mathrm{n} +1)}{6} \mathrm{x} 5 \mathrm{x} \frac{4}{{\mathrm{n}}^2}\mathrm{x}\frac{\mathrm{n}(\mathrm{n}+1)}{2} + 4 \mathrm{x} \frac{2}{\mathrm{n}} \mathrm{x} \mathrm{n} + \left({\mathrm{e}}^{\mathrm{h}}\left(\frac{{\mathrm{e}}^{\mathrm{nh}}-1}{{\mathrm{e}}^{\mathrm{h}}-1}\right)\right).\mathrm{h}\right) \end{gathered}$$

$$\begin{gathered} {\int }_a^{b}f\left(x\right) dx = \frac{4}{3} x 1 x \underset{n\to \infty }{lim} \left(1 + \frac{1}{n}\right) x \underset{n\to \infty }{lim} \left(2 + \frac{1}{n}\right)+\underset{n\to \infty }{lim} 10 x 1 x \left(1 + \frac{1}{n}\right) + 8 + \underset{h\to 0}{lim}\frac{e^{h+1}(e^2-1)}{\left({\displaystyle \frac{e^h-1}{h}}\right)} \\ = \frac{4}{3} x 1 x 2 + 10 x 1 x 1 + 8 + \frac{e(e^2 - 1)}{1} \\ = \frac{8}{3} + 10 + 8 + e^3-e \\ = \frac{8}{3} + 18 + e^3-e \\ = \frac{62}{3} + e^3 -e \end{gathered}$$