# How do you evaluate the integral int_0^1x^2dx ?

Sep 25, 2014

${\int}_{0}^{1} {x}^{2} \mathrm{dx} = \frac{1}{3}$

Remember one of the most important theorems in Calculus:

Fundamental Theorem of Calculus (Part 2)

${\int}_{a}^{b} f \left(x\right) \mathrm{dx} = {\left[F \left(x\right)\right]}_{a}^{b} = F \left(b\right) - F \left(a\right)$,

where $F \left(x\right)$ is an antiderivative of $f \left(x\right)$.

Let us the theorem above to evaluate the definite integral.

${\int}_{0}^{1} {x}^{2} \mathrm{dx}$

by finding an antiderivative of ${x}^{2}$ using Power Rule,

$= {\left[{x}^{3} / 3\right]}_{0}^{1}$

by plugging in the upper and the lower limits,

$= {\left(1\right)}^{3} / 3 - {\left(0\right)}^{3} / 3$

by simplifying,

$= \frac{1}{3}$