Answer
$\frac{1}{4}(b^4-a^4)$
Work Step by Step
Step 1. Let $f(x)=x^3$ 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,
$x_k=a+k\Delta x, f(x_k)=(a+k\Delta x)^3=a^3+3a^2(k\Delta x)+3a(k\Delta x)^2+(k\Delta x)^3$
and the area of the rectangle of this division is given by
$A_k=f(x_k)\Delta x=a^3\Delta x+3a^2(\Delta x)^2k+3a(\Delta x)^3k^2+(\Delta x)^4k^3=a^3(\frac{b-a}{n})+3a^2(\frac{b-a}{n})^2k+3a(\frac{b-a}{n})^3k^2+(\frac{b-a}{n})^4k^3$
Step 3. Evaluate the Riemann Sum as
$A_R=\Sigma_{k=1}^nA_k=\Sigma_{k=1}^n(a^3(\frac{b-a}{n})+3a^2(\frac{b-a}{n})^2k+3a(\frac{b-a}{n})^3k^2+(\frac{b-a}{n})^4k^3)=a^3(\frac{b-a}{n})n+3a^2(\frac{b-a}{n})^2(\frac{n(n+1)}{2})+3a(\frac{b-a}{n})^3(\frac{n(n+1)(2n+1)}{6})+(\frac{b-a}{n})^4(\frac{n^2(n+1)^2}{4})=a^3(b-a)+a^2(b-a)^2(\frac{3(n+1)}{2n})+a(b-a)^3(\frac{(n+1)(2n+1)}{2n^2})+(b-a)^4(\frac{(n+1)^2}{4n^2})$
Step 4. Evaluate the integral as
$\int_a^bx^2dx=\lim_{||P||\to0}A_R=\lim_{n\to\infty}A_R=a^3(b-a)+\frac{3}{2}a^2(b-a)^2+a(b-a)^3+\frac{1}{4}(b-a)^4=\frac{1}{4}(4a^3b-4a^4+6a^2b^2-12a^3b+6a^4+4ab^3-12a^2b^2+12a^3b-4a^3+b^4-4ab^3+6a^2b^2-4a^3b+a^4)=\frac{1}{4}(b^4-a^4)$
