Answer
See proof
Work Step by Step
$\int_a^b{xdx}$
= $\lim\limits_{n \to {\infty}}{\frac{b-a}{n}}{\Sigma_{i=1}^n}\left[a+\frac{b-a}{n}i\right]$
= $\lim\limits_{n \to {\infty}}\left[{\frac{a(b-a)}{n}}{\Sigma_{i=1}^n}1+\frac{(b-a)^2}{n^2}{\Sigma_{i=1}^n}i\right]$
= $\lim\limits_{n \to {\infty}}\left[\frac{a(b-a)}{n}\cdot n+\frac{(b-a)^2}{n^2}\cdot \frac{n(n+1)}{2}\right]$
= $a(b-a)+\lim\limits_{n \to {\infty}}\left[\frac{(b-a)^2}{2}\left(1+\frac{1}{n}\right)\right]$
= $a(b-a)+\frac{1}{2}(b-a)^2$
= $\frac{1}{2}(b^2-a^2)$
