(Theorem) Integration by parts

Theorem: Integration by parts for a single variable

Let $F,G:[a,b]\to \mathbb{R}$ be differentiable. If $f=F’$ and $g=G’$ are both Darboux integrable on $[a,b]$, then

$$\begin{array}{rl}{\int }_a^{b}F(x)g(x)dx& =F(b)G(b)-F(a)G(a)-{\int }_a^{b}f(x)G(x)dx.\end{array}$$

Proof from Modern Analysis I.

Define $H:[a,b]\to \mathbb{R}$ by $H(x)=F(x)G(x)$. By the product rule for derivatives, we have

$$H^{\prime }(x)=F^{\prime }(x)G(x)+F(x)G^{\prime }(x)=(fG)(x)+(Fg)(x).$$

Applying FTC II, we have

$$\begin{array}{rl}{\int }_a^b{H}^{\prime }(x)dx& =H(b)-H(a)=F(b)G(b)-F(a)G(a)\\ & ={\int }_a^b[f(x)G(x)+F(x)g(x)]dx,\end{array}$$

where the final equality follows by substituting the definition of $H$. Then the statement follows by rearranging the expression. $\square$