---
tag: 4SP1394Y2R
kind: proof
name: Proof of the integral of a power
msc: [26A36]
proves: 0Q3HYFT9RQ
uses: [0JZGW75C5Z, 0Z9JB71BS0, 07ZMX4ANJS, 06SESV7GQY, 0XYAFCN0PT, 021WEM680W]
sources:
  - url: https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
    locator: Computing a particular integral
---

## Proof

We find an antiderivative and apply the fundamental theorem of calculus.

1. Let $n \ge 0$ be an integer, and let $F(x) = \frac{x^{n+1}}{n+1}$ and $f(x) = x^n$ for real $x$, with $x^0 = 1$.
2. Since $n + 1 \ge 1$, the [[power rule|0JZGW75C5Z]] says that $x \mapsto x^{n+1}$ is differentiable on $\R$ with derivative $(n+1) x^n$.
3. By the [[constant multiple rule|0Z9JB71BS0]] with the constant $\frac{1}{n+1}$, $F$ is differentiable on $\R$ and $F'(x) = x^n = f(x)$.
4. [[Differentiable functions are continuous|07ZMX4ANJS]], so $F$ is continuous on $\R$.
5. The function $f$ is continuous on $\R$. For $n = 0$ it is the constant $1$. For $n \ge 1$ it is differentiable by the power rule, so it is continuous by the same result as in step 4.
6. Assume $a < b$. By step 5, $f$ is [[Riemann integrable on $[a, b]$|06SESV7GQY]].
7. By steps 3, 4 and 6, $F$ is continuous on $[a, b]$, differentiable on $(a, b)$, and $F' = f$ there with $f$ integrable. The [[second fundamental theorem of calculus|0XYAFCN0PT]] gives $\int_a^b x^n \, dx = F(b) - F(a)$.
8. Assume $a = b$. By the convention for the [[definite integral|021WEM680W]], $\int_a^a x^n \, dx = 0 = F(a) - F(a)$.
9. Assume $a > b$. By the same convention and step 7 on $[b, a]$, $\int_a^b x^n \, dx = -\int_b^a x^n \, dx = -(F(a) - F(b)) = F(b) - F(a)$.
10. In all cases, $\int_a^b x^n \, dx = F(b) - F(a) = \frac{b^{n+1} - a^{n+1}}{n+1}$.

## Idea

The power rule shows that $x^{n+1}/(n+1)$ has derivative $x^n$. The fundamental theorem of calculus then turns the integral into a difference of values of this antiderivative. The sign convention for reversed limits makes the formula hold for every order of $a$ and $b$.