Integration by Substitution/Definite Integral

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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:

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

where $t = \map \phi u$.


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


Corollary

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

Let $f$ be integrable.

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


Then:

$\ds \int_{a - c}^{b - c} \map f t \rd t = \int_a^b \map f {t - c} \rd t$


Proof

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

We have:

\(\ds \map {\frac \d {\d u} } {\map F t}\) \(=\) \(\ds \map {\frac \d {\d u} } {\map F {\map \phi u} }\) Definition of $\map \phi u$
\(\ds \) \(=\) \(\ds \dfrac \d {\d t} \map F {\map \phi u} \dfrac \d {\d u} \map \phi u\) Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \map f {\map \phi u} \dfrac \d {\d u} \map \phi u\) as $\map F t = \ds \int \map f t \rd t$

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


Thus:

\(\ds \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u\) \(=\) \(\ds \bigintlimits {\map F {\map \phi u} } a b\) Fundamental Theorem of Calculus: Second Part
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \map F {\map \phi b} - \map F {\map \phi a}\)


However, also:

\(\ds \int_{\map \phi a}^{\map \phi b} \map f t \rd t\) \(=\) \(\ds \bigintlimits {\map F t} {\map \phi a} {\map \phi b}\)
\(\ds \) \(=\) \(\ds \map F {\map \phi b} - \map F {\map \phi a}\)
\(\ds \) \(=\) \(\ds \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u\) from $(1)$

which was to be proved.

$\blacksquare$


Proof Technique

The usefulness of the technique of Integration by Substitution stems from the fact that it may be possible to choose $\phi$ such that $\map f {\map \phi u} \dfrac \d {\d u} \map \phi u$ (despite its seeming complexity in this context) may be easier to integrate.

If $\phi$ is a trigonometric function, the use of trigonometric identities to simplify the integrand is called integration by trigonometric substitution (or simply trig substitution).

Care must be taken to address the domain and image of $\phi$.

This consideration frequently arises when inverse trigonometric functions are involved.


Sources