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Section 7.4 Integration by Parts and Change of Variables

The Fundamental Theorem of Calculus turns differentiation rules into integration formulas. Two of the most useful are integration by parts, which comes from the product rule, and change of variables, which comes from the chain rule.

Subsection 7.4.1 Integration by Parts

Proof.
Since differentiability implies continuity, both \(F\) and \(G\) are continuous on \([a,b]\text{.}\) Hence Continuous Functions Are Integrable shows that \(F,G \in R[a,b]\text{.}\) Because \(F'\) and \(G'\) are integrable by assumption, Proposition 7.2.4 implies that \(F'G\) and \(FG'\) are integrable.
By the product rule (Proposition 5.1.6),
\begin{equation*} (FG)'=F'G+FG'. \end{equation*}
Therefore \((FG)'\) is integrable, by part (1) of Basic Properties of the Integral. Applying Fundamental Theorem of Calculus (First Form) to \(FG\text{,}\) we get
\begin{equation*} F(b)G(b)-F(a)G(a) = \int_a^b (FG)'(x)\,dx = \int_a^b F'(x)G(x)\,dx + \int_a^b F(x)G'(x)\,dx. \end{equation*}
Rearranging yields the formula.

Subsection 7.4.2 Change of Variables

Proof.
Since \(f\) is continuous on \(\varphi([a,b])\text{,}\) the second form of the Fundamental Theorem of Calculus (Theorem 7.3.1) gives a function
\begin{equation*} U(y)=\int_{\varphi(a)}^{y} f(u)\,du \qquad (y \in \varphi([a,b])) \end{equation*}
such that \(U'(y)=f(y)\) for every \(y \in \varphi([a,b])\text{.}\)
Define \(H(x)=U(\varphi(x))\) on \([a,b]\text{.}\) By the chain rule,
\begin{equation*} H'(x)=U'(\varphi(x))\varphi'(x)=f(\varphi(x))\varphi'(x). \end{equation*}
Because \(\varphi\) and \(\varphi'\) are continuous on \([a,b]\text{,}\) and \(f\) is continuous on \(\varphi([a,b])\text{,}\) the function \(f \circ \varphi\) is continuous on \([a,b]\text{.}\) Hence \(H'\) is continuous, in particular integrable, on \([a,b]\text{.}\)
Applying Fundamental Theorem of Calculus (First Form) to \(H\text{,}\) we obtain
\begin{equation*} \int_a^b f(\varphi(x))\varphi'(x)\,dx = \int_a^b H'(x)\,dx = H(b)-H(a) = U(\varphi(b))-U(\varphi(a)). \end{equation*}
Since \(U(\varphi(a))=0\text{,}\) the right-hand side is
\begin{equation*} \int_{\varphi(a)}^{\varphi(b)} f(u)\,du. \end{equation*}
This proves the formula.

Example 7.4.3.

Consider
\begin{equation*} \int_{-1}^{2} \frac{x}{1+x^2}\,dx. \end{equation*}
Let \(\varphi(x)=1+x^2\) and \(f(u)=1/(2u)\text{.}\) Then
\begin{equation*} \varphi'(x)=2x \qquad\text{and}\qquad f(\varphi(x))\varphi'(x) = \frac{1}{2(1+x^2)}\cdot 2x = \frac{x}{1+x^2}. \end{equation*}
Therefore the change-of-variables theorem gives
\begin{equation*} \int_{-1}^{2} \frac{x}{1+x^2}\,dx = \int_{2}^{5} \frac{1}{2u}\,du = \frac12 \ln\frac{5}{2}. \end{equation*}
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