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Let $a, b \in \mathbb{R}$ with $a<b$.

Let $g$ be a function differentiable on $[a, b]$ with $g'$ continuous on $[a, b]$.
Let $f$ be a function continuous on the range of $g$.

How can I show that $A(x) = \int_a^xf(g(t))g'(t)\,dt$ and $B(x) = \int_{g(a)}^{g(x)} f(t)\,dt$ has the same derivative $\forall x \in [a,b]$?

I understand that $A'(x) = f(g(x))g'(x)$ as a result of the FTC 1, where all the assumptions for FTC 1 are met. But how do I show that $B(x)$ has the same derivative as well, that is, how to show that $B(x)$ meets the requirement for using FTC?

Additionally, is it possible to add on to that and prove $A(x) = B(x)$ for all $x\in [a,b]$

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  • $\begingroup$ Compute $B'(x)$ using the chain rule. And, for your second question, how are $A(a)$ and $B(a)$ related? $\endgroup$ Commented Jan 27, 2020 at 2:24
  • $\begingroup$ You may find this proof to be helpful. $\endgroup$ Commented Jan 27, 2020 at 2:25

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