The Substitution Rule for Definite Integrals

Proposition 2.3.5 (Substitution Rule for Definite Integrals)   We have

$$\int_{a}^{b} f(g(x)) g'(x) dx = \int_{g(a)}^{g(b)} f(u) du,$$

assuming that $u=g(x)$ is a function that is differentiable and whose range is an interval on which $f$ is continuous.

Proof. If $F'= f$, then by the chain rule, $F(g(x))$ is an antiderivative of $f(g(x)) g'(x)$. Thus

$$\int_{a}^{b} f(g(x)) g'(x) dx = \Bigl[ F(g(x))\Bigr]_a^b = F(g(b)) - F(g(a)) = \int_{g(a)}^{g(b)} f(u) du.$$

$\qedsymbol$

Example 2.3.6  

$$\int_{0}^{\sqrt{\pi}} x \cos(x^2) dx$$

We let $u=x^2$, so $du=2xdx$ and $xdx = \frac{1}{2} du$ and the integral becomes

$$\frac{1}{2} \cdot \int_{(0)^2}^{(\sqrt{\pi})^2} \cos(u) du = \frac{1}{2}\cdot \left[ \sin(u) \right]_{0}^{\pi} = \frac{1}{2} \cdot (0 - 0) = 0.$$