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 $I$.

The definite integral of $f$ from $a$ to $b$ can be evaluated by:


 * $\displaystyle \int_{\map \phi a}^{\map \phi b} \map f t \rd t = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u$

where $x = \map \phi u$.

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

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

From Derivative of Composite Function:


 * $\dfrac \d {\d u} \map F {\map \phi u} = \map {F'} {\map \phi u} \map {\phi'} u = \map f {\map \phi u} \map {\phi'} u$

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

Thus:

However, also:

which was to be proved.