Answer
See proof
Work Step by Step
$\int _{a}^{b}x^{2}~dx$
We know that the antiderivative of $f(x)=x^{n}$ is $F(x)=\frac{x^{n+1}}{n+1}+C$ where $n \geq 0$ and $C \in \mathbb R$.
So the antiderivative of $f(x)=x^{2}$ is
$F(x)=\frac{x^{2+1}}{2+1}+C=\frac{x^{3}}{3}+C$
So by the fundamental theorem of calculus it follows:
$\int _{a}^{b}x^{2}~dx=\frac{b^{3}}{3}+C-\left(\frac{a^{3}}{3}+C\right)$
$\int _{a}^{b}x^{2}~dx=\frac{b^{3}}{3}+C-\frac{a^{3}}{3}-C$
$\int _{a}^{b}x^{2}~dx=\frac{b^{3}}{3}-\frac{a^{3}}{3}$
$\int _{a}^{b}x^{2}~dx=\frac{b^{3}-a^{3}}{3}$
