5.24 Proof of integration by substitution theorem

(Not examinable.) Suppose that $F$ has $F^{\prime}=f$; then by the chain rule

$${{d}\over{dx}}F(u(x))={{dF}\over{du}}{{du}\over{dx}}=f(u(x)){{du}\over{dx}}$$

and we integrate to get

$$\int_{a}^{b}{{d}\over{dx}}F(u(x))dx=\int_{a}^{b}f(u(x)){{du}\over{dx}}dx$$

so

$$\bigl[F(u(x))\bigr]_{a}^{b}=\int_{a}^{b}f(u(x)){{du}\over{dx}}dx$$

We also have by the fundamental theorem of calculus

$$\int_{u(a)}^{u(b)}f(u)\,du=\bigl[F(u)\bigr]_{u(a)}^{u(b)},$$

so, comparing the terms in square brackets,

$$\bigl[F(u)\bigr]_{u(a)}^{u(b)}=\bigl[F(u(x))\bigr]_{a}^{b}$$

we get

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