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I have a question about the proof for integration by substitution. It starts the following way:

Let $ \phi: [a, b] \rightarrow [c, d]$ be continuously differentiable and $f: [c, d] \rightarrow \Bbb R$ continuous. Furthermore, let $F: [c, d] \rightarrow \Bbb R$ be an antiderivative. Applying the chain rule gives

$(F \circ \phi)' = (f \circ \phi) \phi'.$

Since $F \circ \phi$ is an antiderivative of $(f \circ \phi) \phi'$, applying the fundamental theorem of calculus gives

$\int_a^b f(\phi(t))\phi'(t) dt = F(\phi(b)) - F(\phi(a)).$

Now, how does he conclude the last step? I don't see how he is applying the theorem actually.

Julian
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  • maybe you are looking for the change of limits: https://proofwiki.org/wiki/Integration_by_Substitution – gbox Aug 26 '16 at 10:06

2 Answers2

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look: $$ \int_a^b f(\phi(t))\phi'(t)dt=\int_a^b F(\phi(t))'dt=[F(\phi(t))]_a^b=F(\phi(b))-F(\phi(a)). $$

Best regards.

Serdar
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Maybe writing it as a simpler looking function will help. Let $F(x)$ be an antideriviative of $f(x)$. Let $H(x) = F(\phi(x))$. Then

$$H'(x) = F'(\phi(x))\cdot \phi'(x) = f(\phi(x))\cdot \phi'(x)$$

So, $H(x) = F(\phi(x))$ is an antiderivative of $f(\phi(x))\cdot \phi'(x)$. By the Fundamental Theorem of Calculus:

$$\begin{align*}\int_a^b f(\phi(t)) \phi'(t) dt &= H(b) - H(a) \\ &= F(\phi(b)) - F(\phi(a))\end{align*}$$

David P
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