Integration by Substitution/Definite Integral

Theorem
Let $\phi$ be a real function which has a derivative on the closed interval $\closedint a b$.

Let $I$ be an open interval which contains the image of $\closedint a b$ under $\phi$.

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

If $\map \phi a \le \map \phi b$, then the definite integral of $f$ from $a$ to $b$ can be evaluated by:


 * $\ds \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

where $t = \map \phi u$.

If $\map \phi a > \map \phi b$, then the definite integral of $f$ from $a$ to $b$ can be evaluated by:


 * $\ds - \int_{\map \phi b}^{\map \phi a} \map f t \rd t = \int_a^b \map f {\map \phi u} \dfrac \d {\d u} \map \phi u \rd u$

The technique of solving an integral in this manner is called integration by substitution.

Proof
Let $F$ be an antiderivative of $f$.

We have:

Hence $\map F {\map \phi u}$ is an antiderivative of $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$.

Thus:

If $\map \phi a \le \map \phi b$, we also have:

which was to be proved.

If instead $\map \phi a > \map \phi b$, we have:

which was to be proved.