# What is the antiderivative of x^2?

May 23, 2018

It is ${x}^{3} / 3 + C$

#### Explanation:

Note that $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$ if $n \setminus \ne - 1$

May 24, 2018

$\frac{1}{3} {x}^{3} + C$

#### Explanation:

The antiderivative of a function is the integral of that function. So here we have:

$\int {x}^{2} \setminus \mathrm{dx}$

Using the rule that $\int {x}^{n} \setminus \mathrm{dx} = \frac{{x}^{n + 1}}{n + 1} + C , n \ne - 1$, we find that it equals:

$= \frac{{x}^{2 + 1}}{2 + 1} + C$

$= \frac{1}{3} {x}^{3} + C$