# Integration by Substitution

## Theorem

Let $\phi$ be a real function which has a derivative on the closed interval $\left[{a \,.\,.\, b}\right]$.

Let $I$ be an open interval which contains the image of $\left[{a \,.\,.\, b}\right]$ under $\phi$.

Let $f$ be a real function which is continuous on $I$.

Then:

$\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \ \mathrm d t = \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u$

and:

$\displaystyle \int f \left({x}\right) \ \mathrm d x = \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u$

where $x = \phi \left({u}\right)$.

Because the most usual substitution variable used is $u$, this method is often referred to as u-substitution in the source works for a number of introductory-level calculus courses.

### Corollary

Let $f : \R \to \R$ be a real function.

Let $f$ be integrable.

Let $a$, $b$, and $c$ be real numbers.

Then:

$\displaystyle \int_{a - c}^{b - c} f \left({t}\right) \ \mathrm d t = \int_a^b f \left({t - c}\right) \ \mathrm d t$

## Proof for Definite Integrals

Let $\displaystyle F$ be an antiderivative of $f$.

$\dfrac {\mathrm d} {\mathrm d u} F \left({\phi \left({u}\right)}\right) = F' \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) = f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)$

Hence $F \left({\phi \left({u}\right)}\right)$ is an antiderivative of $f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)$.

Thus:

 $\displaystyle \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u$ $=$ $\displaystyle \left[{F \left({\phi \left({u}\right)}\right)}\right]_a^b$ Fundamental Theorem of Calculus: Second Part $(1):\quad$ $\displaystyle$ $=$ $\displaystyle F \left({\phi \left({b}\right)}\right) - F \left({\phi \left({a}\right)}\right)$

However, also:

 $\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) \ \mathrm d t$ $=$ $\displaystyle \left[{F \left({t}\right)}\right]_{\phi \left({a}\right)}^{\phi \left({b}\right)}$ $\displaystyle$ $=$ $\displaystyle F \left({\phi \left({b}\right)}\right) - F \left({\phi \left({a}\right)}\right)$ $\displaystyle$ $=$ $\displaystyle \int_a^b f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u$ from $(1)$

which was to be proved.

$\blacksquare$

## Proof for Indefinite Integrals

Let $\displaystyle F \left({u}\right) = \int f \left({u}\right) \ \mathrm d u$.

By definition $F \left({u}\right)$ is an antiderivative of $f \left({u}\right)$.

Thus by the Chain Rule:

 $\displaystyle \frac {\mathrm d} {\mathrm d u} F \left({\phi \left({u}\right)}\right)$ $=$ $\displaystyle F' \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right)$ $(1):\quad$ $\displaystyle$ $=$ $\displaystyle f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)$

So $F \left({\phi \left({u}\right)}\right)$ is an antiderivative of $f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right)$.

Therefore:

$\displaystyle \int f \left({\phi \left({u}\right)}\right) \phi' \left({u}\right) \ \mathrm d u = F \left({\phi \left({u}\right)}\right) = \int f \left({x}\right) \ \mathrm d x$

where $x = \phi \left({u}\right)$.

$\blacksquare$