Editing Integration by substitution (section)

=== Proof ===

Integration by substitution can be derived from the [[fundamental theorem of calculus]] as follows.  Let <math>f</math> and <math>g</math> be two functions satisfying the above hypothesis that <math>f</math> is continuous on <math>I</math> and <math>g'</math> is integrable on the closed interval <math>[a,b]</math>.  Then the function <math>f(g(x))\cdot g'(x)</math> is also integrable on <math>[a,b]</math>.  Hence the integrals
<math display="block">\int_a^b f(g(x))\cdot g'(x)\ dx</math>
and
<math display="block">\int_{g(a)}^{g(b)} f(u)\ du</math>
in fact exist, and it remains to show that they are equal.

Since <math>f</math> is continuous, it has an [[antiderivative]] <math>F</math>. The [[function composition|composite function]] <math>F \circ g</math> is then defined. Since <math>g</math> is differentiable, combining the [[chain rule]] and the definition of an antiderivative gives:
<math display="block">(F \circ g)'(x) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x).</math>

Applying the [[fundamental theorem of calculus]] twice gives:
<math display="block">\begin{align}
\int_a^b f(g(x)) \cdot g'(x)\ dx
&= \int_a^b (F \circ g)'(x)\ dx \\
&= (F \circ g)(b) - (F \circ g)(a) \\
&= F(g(b)) - F(g(a)) \\
&= \int_{g(a)}^{g(b)} f(u)\, du,
\end{align}</math>
which is the substitution rule.