Integration by Parts/Primitive
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
\(\ds \map {\dfrac \d {\d t} } {\map F t \map G t}\) | \(=\) | \(\ds \map f t \map G t + \map F t \map g t\) | 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 |
$\blacksquare$
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
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Integration: Integration by Parts
- 1967: Michael Spivak: Calculus ... (previous) ... (next): Part $\text {III}$: Derivatives and Integrals: Chapter $18$: Integration in Elementary Terms: Theorem $1$ (Integration by Parts)
- 1971: Wilfred Kaplan and Donald J. Lewis: Calculus and Linear Algebra ... (previous) ... (next): Appendix $\text I$: Table of Indefinite Integrals $3$.
- For a video presentation of the contents of this page, visit the Khan Academy.