AP Calculus AB and BC

Integral of the Cube Root of x: Answer and Steps

Q: How do you write the cube root of x for integration?

As the power x^{1/3}. Once in that form, the reverse power rule adds 1 to the exponent to get \frac{4}{3} and divides, giving \frac{3}{4}x^{4/3} + C.

Q: Why is the coefficient 3/4 and not 4/3?

Because you divide by the new exponent \frac{4}{3}, and dividing by \frac{4}{3} equals multiplying by its reciprocal \frac{3}{4}.

The integral of the cube root of x is (3/4)x^(4/3) + C. Writing the cube root as x^(1/3), the reverse power rule raises the exponent to 4/3 and divides by 4/3, which is the same as multiplying by 3/4. Differentiating (3/4)x^(4/3) returns x^(1/3).

\int \sqrt[3]{x}\,dx = \frac{3}{4}x^{4/3} + C

Rewrite the root as a power

A radical is a fractional power. Writing the cube root as $x^{1/3}$ turns the problem into a plain reverse power rule.

\sqrt[3]{x} = x^{1/3}

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

Here $n = \frac{1}{3}$ , so the new exponent is $\frac{1}{3} + 1 = \frac{4}{3}$ , and dividing by $\frac{4}{3}$ is the same as multiplying by $\frac{3}{4}$ .

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

Divide by a fraction, multiply by its reciprocal

Dividing by $\frac{4}{3}$ becomes multiplying by $\frac{3}{4}$ . That reciprocal is where the $\frac{3}{4}$ comes from.

Checking by differentiating

Differentiate the result with the power rule and watch the fractional exponents cancel.

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

The $\frac{4}{3}$ from the power rule cancels the $\frac{3}{4}$ out front, and the exponent drops from $\frac{4}{3}$ back to $\frac{1}{3}$ , which is $\sqrt[3]{x}$ .

Common mistakes

Leaving the answer as $\frac{x^{4/3}}{4/3}$ without simplifying. Clearing the complex fraction gives $\frac{3}{4}x^{4/3}$ .
Miscomputing the new exponent. Adding $1$ to $\frac{1}{3}$ gives $\frac{4}{3}$ , not $\frac{2}{3}$ .
Treating $\sqrt[3]{x}$ as if the reverse power rule did not apply. It does, once you rewrite the root as $x^{1/3}$ .

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 the cube root of x from 0 to 8?

$\int_0^8 x^{1/3}\,dx = \left[\frac{3}{4}x^{4/3}\right]_0^8 = \frac{3}{4}\cdot 16 = 12$ , since $8^{4/3} = (8^{1/3})^4 = 2^4 = 16$ .

How do you write the cube root of x for integration?

As the power $x^{1/3}$ . Once in that form, the reverse power rule adds $1$ to the exponent to get $\frac{4}{3}$ and divides, giving $\frac{3}{4}x^{4/3} + C$ .

Why is the coefficient 3/4 and not 4/3?

Because you divide by the new exponent $\frac{4}{3}$ , and dividing by $\frac{4}{3}$ equals multiplying by its reciprocal $\frac{3}{4}$ .