Q.Evaluate ∫₀² (3x² + 2x) dx as a limit of sum.

Use the definition $\int_0^a f(x)dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1}f(rh)$ with $h=a/n$, and standard summation formulas for $\sum r$ and $\sum r^2$.

Let $f(x)=3x^2+2x$, $a=2$, and divide $[0,2]$ into $n$ equal parts of width $h=2/n$. By definition:

$\int_0^2 f(x)\,dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1} f(rh)$

$f(rh) = 3(rh)^2+2(rh) = 3r^2h^2+2rh$

$h\sum_{r=0}^{n-1}f(rh) = 3h^3\sum_{r=0}^{n-1}r^2 + 2h^2\sum_{r=0}^{n-1}r$

Using $\displaystyle\sum_{r=0}^{n-1}r = \frac{n(n-1)}{2}$ and $\displaystyle\sum_{r=0}^{n-1}r^2 = \frac{(n-1)n(2n-1)}{6}$, and $h=2/n$ (so $h^2=4/n^2$, $h^3=8/n^3$):

$3h^3\sum r^2 = 3\cdot\frac{8}{n^3}\cdot\frac{(n-1)n(2n-1)}{6} = \frac{4(n-1)(2n-1)}{n^2}$

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