Q: Proving $\phi(x) = \int_{\alpha(x)}^{\beta(x)} f(t)\ dt$ when $f$ continuous, $\alpha(x),\beta(x)$ differentiable, is differentiable 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?

A: 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.
