# Integral of x²: Formula, Derivation and Examples

> The integral of x² is x³/3 + C. See the reverse power rule, a Riemann sum proof, definite integrals, the (ax + b)² form and the mistakes students make.

Source: https://practicecalculus.com/integrals/x-squared
Updated: 2026-11-06

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

## At a glance

- **Antiderivative:** $\dfrac{x^3}{3} + C$
- **Rule used:** Reverse power rule with $n = 2$
- **Area from 0 to b:** $\dfrac{b^3}{3}$
- **Area from 0 to 1:** $\dfrac{1}{3}$
- **Check:** $\dfrac{d}{dx}\dfrac{x^3}{3} = x^2$

## Key takeaways

- The integral of $x^2$ is $\frac{x^3}{3} + C$.
- The area under $y = x^2$ from $0$ to $b$ is $\frac{b^3}{3}$, one third of the $b$ by $b^2$ rectangle around it.
- The integral of $3x^2$ is $x^3 + C$, which is the easiest way to remember the pattern.
- With a linear inside, $\int (ax + b)^2\,dx = \frac{(ax + b)^3}{3a} + C$.

The **integral of $x^2$ is $\dfrac{x^3}{3} + C$**. Add one to the exponent and divide by the new exponent.

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

## Graph: the area under y = x²

```graph
title: 'Area under y = x² from 0 to 2, with the antiderivative y = x³/3'
caption: 'The shaded area is $\frac{2^3}{3} = \frac{8}{3} \approx 2.67$. The dashed antiderivative $y = \frac{x^3}{3}$ passes through $\left(2, \frac{8}{3}\right)$.'
x: [-2, 2.6]
y: [-2.5, 5]
functions:
  - expr: x^2
    label: y = x²
  - expr: x^3/3
    label: y = x³/3 (antiderivative)
    dashed: true
shade: {expr: x^2, from: 0, to: 2}
points:
  - [2, 2.667, '(2, 8/3)']
```

Notice that the antiderivative is flat at $x = 0$ (where $x^2 = 0$) and never decreases, because $x^2$ is never negative.

## Derivation with the reverse power rule

1. Start from the [reverse power rule](/rules/reverse-power-rule): $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ for $n \ne -1$.
2. Set $n = 2$: the new exponent is $3$.
3. Divide by $3$: $\frac{x^3}{3}$.
4. Add $C$.

## A proof from Riemann sums

You can also find the area under $x^2$ from $0$ to $b$ without knowing any antiderivative. Split $[0, b]$ into $n$ strips of width $\frac{b}{n}$ and use right endpoints $x_i = \frac{bi}{n}$:

$$\sum_{i=1}^{n}\left(\frac{bi}{n}\right)^2\frac{b}{n} = \frac{b^3}{n^3}\sum_{i=1}^{n} i^2 = \frac{b^3}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}$$

As $n \to \infty$, the fraction $\frac{n(n+1)(2n+1)}{6n^3}$ tends to $\frac{2}{6} = \frac{1}{3}$, so the area is $\frac{b^3}{3}$. That matches the antiderivative. This is the same computation Archimedes did for the parabola, and you can practise it on the [Riemann sums](/blog/riemann-sums) guide.

## Check by differentiating

$$\frac{d}{dx}\left(\frac{x^3}{3} + C\right) = \frac{3x^2}{3} = x^2 \checkmark$$

See also the [derivative of x²](/derivatives/x-squared).

## Substitution variant: integral of (ax + b)²

With $u = ax + b$ and $du = a\,dx$:

$$\int (ax + b)^2\,dx = \frac{1}{a}\int u^2\,du = \frac{(ax + b)^3}{3a} + C$$

For example, $\int (2x + 1)^2\,dx = \frac{(2x + 1)^3}{6} + C$. Check: the derivative is $\frac{3(2x+1)^2\cdot 2}{6} = (2x + 1)^2$.

The same idea handles an extra factor of $x$: $\int x(x^2 + 1)^2\,dx = \frac{(x^2 + 1)^3}{6} + C$, using $u = x^2 + 1$. More on this on the [u-substitution rule](/rules/u-substitution-rule) page.

## Worked examples

**Example 1.** $\int_1^3 x^2\,dx = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$.

**Example 2.** $\int (x^2 + 3x - 2)\,dx = \frac{x^3}{3} + \frac{3x^2}{2} - 2x + C$.

**Example 3.** $\int_{-2}^{2} x^2\,dx = \frac{8}{3} - \left(-\frac{8}{3}\right) = \frac{16}{3}$. Since $x^2$ is even, this is twice $\int_0^2 x^2\,dx$.

**Example 4: volume.** Spinning $y = x$ around the $x$-axis on $[0, h]$ gives a cone with volume $\pi\int_0^h x^2\,dx = \frac{\pi h^3}{3}$, the familiar $\frac{1}{3}\pi r^2 h$ with $r = h$.

## Common mistakes

1. **Writing $2x$.** That is the derivative, not the integral.
2. **Forgetting to divide by $3$.** $x^3$ has derivative $3x^2$, not $x^2$.
3. **Integrating $(2x + 1)^2$ as $\frac{(2x+1)^3}{3}$.** You must also divide by the inside derivative, $2$.
4. **Squaring the integral of $x$.** Integrals do not distribute over products.

Practice with the [antiderivative problems](/practice-problems/antiderivatives) and [definite integral problems](/practice-problems/definite-integrals).

## Frequently asked questions

### What is the integral of x²?

The integral of $x^2$ is $\frac{x^3}{3} + C$. Raise the exponent from $2$ to $3$, then divide by $3$.

### What is the integral of x² from 0 to 1?

It is $\frac{1^3}{3} - 0 = \frac{1}{3}$.

### What is the integral of 3x²?

It is $x^3 + C$, because $3\cdot\frac{x^3}{3} = x^3$.

### Is the integral of x² the same as (integral of x)²?

No. $\left(\frac{x^2}{2}\right)^2 = \frac{x^4}{4}$, which is not $\frac{x^3}{3}$. You cannot square the integral of $x$ to get the integral of $x^2$.