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:
 * $\displaystyle \int_{\phi \left({a}\right)}^{\phi \left({b}\right)} f \left({t}\right) dt = \int_a^b f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) du$

Proof
Let $\displaystyle F \left({x}\right) = \int_{\phi \left({a}\right)}^x f \left({t}\right) dt$.

From Derivative of a Composite Function and the first part of the Fundamental Theorem of Calculus:


 * $\displaystyle \frac d {du} F \left({\phi \left({u}\right)}\right) = F^{\prime} \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) = f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right)$.

Thus from the second part of the Fundamental Theorem of Calculus:


 * $\displaystyle \int_a^b f \left({\phi \left({u}\right)}\right) \phi^{\prime} \left({u}\right) du = \left[{F \left({\phi \left({u}\right)}\right)}\right]_a^b$

which is what we wanted to prove.