Sumas de Riemann f(x) =2x^2-3x en el intervalo [0,4]

Veamos...

$$\begin{align}&\Delta x = \frac{4}{n}\\&x_i = \frac{4i}n\\&f(x_i) = 2(\frac{4i}n)^2-3 \cdot \frac{4i}n = \frac{32i^2}{n^2}-\frac{12i}n\\&S = \sum_{i=1}^nf(x_i) \Delta x = \sum_{i=1}^n (\frac{32i^2}{n^2}-\frac{12i}n)\cdot \frac{4}n=\\&\sum_{i=1}^n \frac{128i^2}{n^3}-\frac{48i}{n^2}=\\&\sum_{i=1}^n \frac{128i^2}{n^3} - \sum_{i=1}^n \frac{48i}{n^2}=\\&\frac{128}{n^3} \sum_{i=1}^n i^2 -  \frac{48i^2}{n^2} \sum_{i=1}^n i=\\&\frac{128}{n^3} (\frac{n(n+1)(2n+1)}{6}) -  \frac{48i^2}{n^2} (\frac{n(n+1)}{2})=...\\&\end{align}$$

Creo que a partir de ahí ya podrás seguirlo...

Salu2

Gustavo Omar Fellay es de mucha ayuda