Integration by Parts/Definite Integral

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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:

$\ds \int_a^b \map f t \map G t \rd t = \bigintlimits {\map F t \map G t} a b - \int_a^b \map F t \map g t \rd t$


Proof

By Product Rule for Derivatives:

$\map D {F G} = f G + F g$

Thus $F G$ is a primitive of $f G + F g$ on $\closedint a b$.

Hence, by the Fundamental Theorem of Calculus:

$\ds \int_a^b \paren {\map f t \map G t + \map F t \map g t} \rd t = \bigintlimits {\map F t \map G t} a b$

The result follows.

$\blacksquare$


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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