AP Calculus AB and BC

Integral of x^3: Answer, Proof, and Steps

Q: Is the antiderivative of x^3 even or odd?

The antiderivative \frac{x^4}{4} is even, since x^4 is even. As a consequence \int_{-a}^{a} x^3\,dx = 0 for any a, because x^3 itself is odd and the signed areas cancel.

Q: How is integrating x^3 different from integrating x^2?

The rule is identical, only the exponent changes. For x^3 you get \frac{x^4}{4}, and for x^2 you get \frac{x^3}{3}. Each raises its exponent by one and divides by the result.

The integral of x^3 is x^4/4 + C. The reverse power rule raises the exponent from 3 to 4 and divides by the new exponent 4. Checking backward, the derivative of x^4/4 is 4x^3/4 = x^3.

\int x^3\,dx = \frac{x^4}{4} + C

Applying the reverse power rule

Integrating a power of $x$ reverses the derivative power rule: raise the exponent by $1$ and divide by the new exponent.

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

With $n = 3$ , the exponent climbs to $4$ and the divisor is $4$ .

\int x^3\,dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C

Add first, then divide

The divisor is $4$ , the new exponent, not the original $3$ .

Checking by differentiating

Differentiating the result should reproduce the integrand exactly.

\frac{d}{dx}\left(\frac{x^4}{4}\right) = \frac{1}{4}\cdot 4x^3 = x^3

The power rule pulls the $4$ down in front, where it cancels the $\frac{1}{4}$ , leaving $x^3$ . That cancellation is the whole reason the antiderivative divides by the new exponent.

Common mistakes

Dividing by $3$ instead of $4$ . The new exponent sets the divisor, so the answer is $\frac{x^4}{4}$ .
Writing $3x^2$ , which is the derivative of $x^3$ , not its integral. Integration raises the exponent, differentiation lowers it.
Forgetting $+ C$ . Every indefinite integral needs the constant of integration.

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 definite integral of x^3 from 0 to 2?

$\int_0^2 x^3\,dx = \left[\frac{x^4}{4}\right]_0^2 = \frac{16}{4} - 0 = 4$ .

Is the antiderivative of x^3 even or odd?

The antiderivative $\frac{x^4}{4}$ is even, since $x^4$ is even. As a consequence $\int_{-a}^{a} x^3\,dx = 0$ for any $a$ , because $x^3$ itself is odd and the signed areas cancel.

How is integrating x^3 different from integrating x^2?

The rule is identical, only the exponent changes. For $x^3$ you get $\frac{x^4}{4}$ , and for $x^2$ you get $\frac{x^3}{3}$ . Each raises its exponent by one and divides by the result.