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If we have:

Let $I := [a,b]$ and let $f: I\to\mathbb ℝ$ be continuous on $I$. Also let $J := [c,d]$ and let $u: J\to\mathbb ℝ$ be differentiable on $J$ and satisfy $u(J)\subseteq I$. Show that if $G: J\to\mathbb ℝ$ is defined by $ G(x) :=\int_a^{u(x)} f(t)\,dt $ for $x$ in $J$, then $G'(x) = (f \circ u)(x)u'(x)$ for all $x\in J$.

Can we just say use the Fundamental Thm of Calculus or do we need to break it up over two integrals?

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    You$ $might want to consider changing the title o$f$ your question. It doesn't seem very help$f$ul i$f$ someone else is searching for an answer to your problem.2012-11-29

1 Answers 1

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I'd say you can use the FTC: since $\,f\,$ is continuous we know that

$\int_a^{u(x)}f(t)\,dt=F(u(x))-F(a)$

where $\,F\,$ is a primitive of $\,f\,$ on $\,[a,b]\;:\;\;F'(t)=f(t)\,\,,\,\forall\,\,t\in [a,b]\,$ , and thus applying the chain rule

$G(x)=F(u(x))-F(a)\Longrightarrow G'(x)=F'(u(x))\cdot u'(x)=f(u(x))\cdot u'(x)$