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Let $f:[a,b]\to \mathbb{R}$ be a continuous function and $\alpha,\beta:I\to [a,b]$ differentiable. Define $\phi:I\to \mathbb{R}$ by $\phi(x) = \int_{\alpha(x)}^{\beta(x)} f(t)\ dt$. Prove that $\phi$ is differentiable and show $\phi'$.

I'm thinking about $\phi(x)$ as $\phi(x) = F(\beta(x))-F(\alpha(x))$ where $F(x)$ is the antiderivative of $f$. Therefore by the chain rule I could find $\phi'(x) = f(\beta(x))\beta'(x)-f(\alpha(x))\alpha'(x)$. Am I right? If I am, why it exists?

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  • $\begingroup$ yes $F(x) = \int_a^x f(t) dt$ that exists since $f(x)$ is continuous, and $F'(x) = \lim_{h \to 0} \frac{F(x+h)-F(x)}{h} = \lim_{h \to 0} \frac{1}{h} \int_x^{x+h} f(t)dt = \lim_{h \to 0} f(x) + \frac{1}{h} \int_x^{x+h} (f(t)-f(x)) dt = f(x)$ since $\frac{1}{h}|\int_x^{x+h} (f(t)-f(x)) dt| < \sup_{t \in [x,x+h]} |f(t)|$ $\endgroup$ – reuns Oct 2 '16 at 4:55
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consider, $\psi (x)=\int_a^x f(t) dt$ on $[a,b]$ now as f is continuous so by fundamental theorem $\psi$ is differentiable. And $\psi'(x)=f(x)$. Now $\phi(x)=\int_{\alpha(x)}^{\beta(x)}f(t)dt=\int_a^{\beta(x)}f(t)dt-\int_a^{\alpha(x)}f(t)dt= \psi(\beta(x))-\psi(\alpha(x))$ \ now as both $\alpha$ and $\beta $ are also differentiable so we have $\phi'(x)=\psi'(\beta(x))\beta'(x)-\psi'(\alpha(x))\alpha'(x)$ by chain rule. which is further equal to what u have written.

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