change of variable in definite integralChangeOfVariableInDefiniteIntegral2008-12-23 12:29:332009-12-31 12:36:51Theoremdefinite integral% this is the default PlanetMath preamble. as your knowledge
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\textbf{Theorem.}\, Let the real function \,$x \mapsto f(x)$\, be continuous on the interval \,$[a,\,b]$.\, We introduce via the the equation
$$x \;=\; \varphi(t)$$
a new variable $t$ satisfying
\begin{itemize}
\item $\varphi(\alpha) \,=\, a, \quad \varphi(\beta) \,=\, b$,
\item $\varphi$ and $\varphi'$ are continuous on the interval with endpoints $\alpha$ and $\beta$.
\end{itemize}
Then
$$\int_a^b\!f(x)\,dx \;=\; \int_\alpha^\beta\!f(\varphi(t))\,\varphi'(t)\,dt.$$\\
{\em Proof.}\, As a continuous function, $f$ has an antiderivative $F$.\, Then the compound function $F\circ\varphi$ is an antiderivative of $(f\circ\varphi)\cdot\varphi'$, since by the chain rule we have
$$\frac{d}{dt}F(\varphi(t)) \;=\; F'(\varphi(t))\,\varphi'(t) \;=\; f(\varphi(t))\,\varphi'(t).$$
Using the \PMlinkname{Newton--Leibniz formula}{NewtonLeibniz} we obtain
$$\int_a^b\!f(x)\,dx \;=\; F(b)-F(a) \;=\; F(\varphi(\beta))-F(\varphi(\alpha))
\;=\; \int_\alpha^\beta\!f(\varphi(t))\,\varphi'(t)\,dt,$$
Q.E.D.