AP Calculus AB and BC

Integral of x^2: Answer, Proof, and Steps

Q: Does the integral of x^2 need u-substitution?

No. x^2 is a single power of x, so the reverse power rule handles it directly. Substitution is only needed when the integrand is a composite, such as (2x+1)^2.

The integral of x^2 is x^3/3 + C. By the reverse power rule you add 1 to the exponent to get x^3, then divide by the new exponent 3. Differentiating x^3/3 gives 3x^2/3 = x^2, which confirms the answer.

\int x^2\,dx = \frac{x^3}{3} + C

Applying the reverse power rule

The power rule for antiderivatives runs the derivative power rule backward. To integrate a power of $x$ , add $1$ to the exponent and divide by that new exponent.

\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1

Here $n = 2$ , so the exponent rises to $3$ and the divisor is $3$ .

\int x^2\,dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C

Divide by the new exponent

After raising $x^2$ to $x^3$ , you divide by $3$ , the exponent you just created, never by the original $2$ .

Checking by differentiating

The fastest way to trust an antiderivative is to differentiate it and confirm you land back on the integrand.

\frac{d}{dx}\left(\frac{x^3}{3}\right) = \frac{1}{3}\cdot 3x^2 = x^2

The power rule brings the $3$ down in front, where it cancels the $\frac{1}{3}$ , leaving exactly $x^2$ . The constant $C$ differentiates to $0$ , which is why every antiderivative carries one.

Common mistakes

Two errors account for almost every lost point on this integral.

Dividing by the old exponent. The answer is $\frac{x^3}{3}$ , not $\frac{x^3}{2}$ . You always divide by the new exponent.
Dropping the constant of integration. An indefinite integral has infinitely many antiderivatives, so $+ C$ is part of the answer, not an afterthought.
Confusing the integral with the derivative. The derivative of $x^2$ is $2x$ , while the integral is $\frac{x^3}{3} + C$ . They move the exponent in opposite directions.

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you.

Frequently asked questions

What is the integral of x^2 from 0 to 3?

$\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9$ . Evaluate the antiderivative $\frac{x^3}{3}$ at the top and bottom limits and subtract.

Why do you divide by 3 and not by 2?

Because the reverse power rule divides by the new exponent. Raising $x^2$ to $x^3$ makes the new exponent $3$ , so you divide by $3$ . Differentiating $\frac{x^3}{3}$ brings that $3$ back down and returns $x^2$ .

Does the integral of x^2 need u-substitution?

No. $x^2$ is a single power of $x$ , so the reverse power rule handles it directly. Substitution is only needed when the integrand is a composite, such as $(2x+1)^2$ .