Integration by Parts/Primitive

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Theorem

Let $f$ and $g$ be real functions which are continuous on the closed interval $\closedint a b$.

Let $f$ and $g$ have primitives $F$ and $G$ respectively on $\closedint a b$.


Then:

$\ds \int \map f t \map G t \rd t = \map F t \map G t - \int \map F t \map g t \rd t$

on $\closedint a b$.


Corollary

Let $u$ and $v$ be real functions which are integrable on their domain.

Then:

$\ds \int u v \rd x = \paren {\int u \rd x} v - \int \paren {\int u \rd x} \dfrac {\d v} {\d x} \rd x$


Proof $1$

\(\ds \map {\dfrac \d {\d t} } {\map F t \map G t}\) \(=\) \(\ds \map f t \map G t + \map F t \map g t\) Product Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds \int \paren {\map f t \map G t + \map F t \map g t} \rd t\) \(=\) \(\ds \map F t \map G t\) Fundamental Theorem of Calculus: integrating both sides with respect to $t$
\(\ds \leadsto \ \ \) \(\ds \int \map f t \map G t \rd t + \int \map F t \map g t \rd t\) \(=\) \(\ds \map F t \map G t\) Linear Combination of Primitives
\(\ds \leadsto \ \ \) \(\ds \int \map f t \map G t \rd t\) \(=\) \(\ds \map F t \map G t - \int \map F t \map g t \rd t\) rearranging

$\blacksquare$


Proof $2$

Let $\map u x$ and $\map v x$ be integrable functions defined on $\closedint a b$.

Then:

\(\ds \map {\dfrac \d {\d x} } {u v}\) \(=\) \(\ds u \dfrac {\d v} {\d x} + v \dfrac {\d u} {\d x}\) Product Rule for Derivatives
\(\ds \leadsto \ \ \) \(\ds v \dfrac {\d u} {\d x}\) \(=\) \(\ds \map {\dfrac \d {\d x} } {u v} - u \dfrac {\d v} {\d x}\) rearranging
\(\ds \leadsto \ \ \) \(\ds \int v \dfrac {\d u} {\d x} \rd x\) \(=\) \(\ds \int \paren {\map {\dfrac \d {\d x} } {u v} - u \dfrac {\d v} {\d x} } \rd x\) integrating both sides with respect to $x$
\(\ds \) \(=\) \(\ds \int \map {\dfrac \d {\d x} } {u v} \rd x - \int u \dfrac {\d v} {\d x} \rd x\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds u v - \int u \dfrac {\d v} {\d x} \rd x\) Fundamental Theorem of Calculus

$\blacksquare$


Also presented as

Integration by Parts is often seen presented in this sort of form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

or:

$\ds \int u \rd v = u v - \int v \rd u$

where it is understood that $u$ and $v$ are functions of the independent variable.


Proof Technique

The technique of solving an integral in the form $\ds \int \map f t \map G t \rd t$ in this manner is called integration by parts.

Its validity as a solution technique stems from the fact that it may be possible to choose $f$ and $G$ such that $G$ is easier to differentiate than to integrate.

Thus the plan is to reduce the integral to one such that $\ds \int \map F t \map g t \rd t$ is easier to solve than $\ds \int \map f t \map G t \rd t$.

It may be, of course, that one or more further applications of this technique are needed before the solution can be extracted.


Sources

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