Find $\int_{-1}^{2}(5x^3+7x^2-9x+4)dx$ by Riemann sums

Question

I want to determine this integral $\int_{-1}^{2}(5x^3+7x^2-9x+4)dx$ by Riemann sums. Clearly we can take

$$\Delta x=\frac{2-(-1)}{n}$$ And $$x_i=-1+\frac{i3}{n}$$

Then

$$\int_{-1}^{2}(5x^3+7x^2- 9x+4)dx\\=\lim_{n\to\infty}\sum^n_{i=1}\left(5(-1+\frac{i3}{n})^3+7(-1+\frac{i3}{n})^2-9(-1+\frac{i3}{n})+4\right)\frac{3}{n}$$

Clearly the expression can be simplified a bit more, however after that I don't know what else I can do to get to the value of the integral.

Answer 1

You will need to use the formulas \begin{align} \sum_{i=1}^n 1 &= n \\ \sum_{i=1}^n i &= \frac{n(n+1)}{2} \\ \sum_{i=1}^n i^2 &= \frac{n(n+1)(2n+1)}{6} \\ \sum_{i=1}^n i^3 &= \frac{n^2(n+1)^2}{4} \\ \end{align}