Answer
$\frac{1}{3}(b^3-a^3)$
Work Step by Step
Step 1. Let $f(x)=x^2$ and divide the interval $[a,b]$ into $n$ parts with equal width of $\Delta x=\frac{b-a}{n}$; we denote the partition as $||P||$ and as $n\to\infty, ||P||\to0$.
Step 2. For the $k$th division, $x_k=a+k\Delta x, f(x_k)=(a+k\Delta x)^2$ and the area of the rectangle of this division is given by $A_k=f(x_k)\Delta x=(a+k\Delta x)^2\Delta x=a^2\Delta x+2ak(\Delta x)^2+k^2(\Delta x)^3=a^2(\frac{b-a}{n})+2ak(\frac{b-a}{n})^2+k^2(\frac{b-a}{n})^3$
Step 3. Evaluate the Riemann Sum as $A_R=\Sigma_{k=1}^nA_k=\Sigma_{k=1}^n(a^2(\frac{b-a}{n})+2ak(\frac{b-a}{n})^2+k^2(\frac{b-a}{n})^3)=a^2(\frac{b-a}{n})n+2a(\frac{b-a}{n})^2(\frac{n(n+1)}{2})+(\frac{b-a}{n})^3(\frac{n(n+1)(2n+1)}{6})=a^2(b-a)+a(b-a)^2(\frac{n+1}{n})+(b-a)^3\frac{(n+1)(2n+1)}{6n^2}$
Step 4. Evaluate the integral as $\int_a^bx^2dx=\lim_{||P||\to0}A_R=\lim_{n\to\infty}A_R=a^2(b-a)+a(b-a)^2+(b-a)^3\frac{2}{6}=\frac{1}{3}(3a^2b-3a^3+3ab^2-6a^2b+3a^3+b^3-3ab^2+3a^2b-a^3)=\frac{1}{3}(b^3-a^3)$
