Definition of the Definite Integral — Question 2

Question 2 - Solution

Divide $[1,3]$ into $n$ equal subintervals of width $\Delta x=2/n$.

The midpoints are $x_i^*=1+(2i-1)/n$.

$$M_n=\frac2n\sum_{i=1}^n\left(1+\frac{2i-1}{n}\right)^2.$$

Use $\sum(2i-1)=n^2$ and $\sum(2i-1)^2=n(4n^2-1)/3$:

$$M_n=2+4+\frac{2(4n^2-1)}{3n^2}=\frac{26}{3}-\frac{2}{3n^2}.$$

$$\boxed{\int_1^3x^2\,dx=\lim_{n\to\infty}M_n=\frac{26}{3}.}$$