Suppose we have the integral $\int_a ^b f(x)dx$ under what condition can we substitute $x=\phi(t)$ and write $\int_a^bf(x)dx=\int_\alpha^\beta f(\phi(t))\phi'(t)dt$,where $\phi(\alpha)=a,\phi(\beta)=b$.I should say what is a sufficient condition that the above substitution is valid,also give me a counterexample where this does not hold. What is the condition of the above to occur if $f$ is continuous and $\phi$ is also continuous?I want to understand why actually we can substitute because these are not taught in school level calculus,but as I am now in undergraduate level,I need to understand rigorously,so when we substitute $x=\sin t$ in a differential equation in an integral expression,we should know it actually is correct.
Also when can we substitute $x=\phi(t)$ in an indefinite integral?
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$\begingroup$We can substitute the variable if we can find a differentiable function $\phi(t)$ for which $f(\phi(t)) \phi'(t)$ is integrable over that interval. So a counterexample should be some $\phi$ continuous that does not meet the above conditions.
There's a good proof in Wikipedia, using only the Fundamental Theorem of Calculus (I'm not reproducing it here since I'm on my phone) and the chain rule.
$\endgroup$ 3 $\begingroup$The result is a consequence of the fundamental theorem of calculus. If $F$ is an antiderivative of $f$ in the interval $[\phi(\alpha), \phi(\beta)]$, then the chain rule says $F(\phi(t))$ is an antiderivative of $f(\phi(t)) \phi'(t)$, hence the fundamental theorem implies\begin{equation} \int_\alpha^\beta f(\phi(t)) \phi'(t) dt = F(\phi(\beta)) - F(\phi(\alpha)) =\int_{\phi(\alpha)}^{\phi(\beta)} f(x) dx \end{equation}You see that a simple condition is that $\phi$ is differentiable and that $f(\phi(t))\phi'(t)$ and $f(x)$ are integrable with $F$ being an antiderivative of $f$.
In particular, the formula is true if $f$ is continuous in $[\phi(\alpha), \phi(\beta)]$ and $\phi'$ is continuous in $[\alpha, \beta]$.
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