Differenciability of $F(x) = \int_{0}^{v(x)} f(t)dt$

Question

Let $f, \ u, \ v: \mathbb{R} \to \mathbb{R}$ with $f$ continuous, $u$ and $v$ differentiable.

(1) Let $F(x) = \int_{0}^{v(x)} f(t)dt$, prove that $F$ is differentiable and calculate $F'(x)$.

(2) Now let $F(x) = \int_{u(x)}^{v(x)} f(t)dt$ and prove again that $F$ is differentiable and calculate $F'(x)$.

I tried to prove it using the fundamental theorem of calculus. In this way, i got:

As f is continuous, so F is derivable and its derivative is $F'(x) = f(v(x))$. This way i didn't use that u is differentiable (a bit weird i guess). Am i right? I either tried to use this method in second item, but again it falls into the same problem.

That derivative is incorrect because you forgot to take chain rule into account. — Ninad Munshi, Dec 07 '19 at 21:01
Aaaah! make sense. I was stuck because of having a function in the integration limit, but in fact it is quite simple — Fredgrub, Dec 07 '19 at 21:09

score 1 · Accepted Answer · answered Dec 07 '19 at 21:02

1

Let $G(x)=\int_0^xf(t)dt$, $F(x)=G(v(x))$, $F'(x)=G'(v(x))v'(x)=f(v(x))v'(x)$.

answered Dec 07 '19 at 21:02

Tsemo Aristide

87,475

Differenciability of $F(x) = \int_{0}^{v(x)} f(t)dt$

1 Answers1