Question

Let `f:[a,b]->RR, f` integrate and converges at `x_0` `F:[a,b]->RR,``F(x)=int_a^xf(t)dt` how do you show `F` differentiable at `x_0` ?

Answer 1

By definition of derivative, `F` is differentiable in `x_0` if the limit:

`lim_(x->x_0) (F(x)-F(x_0))/(x-x_0)`

exists.

Consider now any sequence `{x_n}` such that:

`lim_(n->oo) x_n = x_0`

note that:

`F(x_n)-F(x_0) = int_a^(x_n) f(t)dt - int_a^(x_0) f(t)dt`

and using the additivity of the integral:

`F(x_n)-F(x_0) = int_(x_0)^(x_n) f(t)dt`

Based on the mean value theorem, there must then be a point `xi_n in (x_0,x_n)` such that:

`int_(x_0)^x f(t)dt =(x_n-x_0)f(xi_n)`

so that:

`(F(x_n)-F(x_0) )/(x_n-x_0) = f(xi_n)`

Now, as `xi_n in (x_0,x_n)` and `lim_n x_n = x_0` it follows that necessarily also:

`lim_n xi_n = x_0`

and as `f(x)` converges (really we need here `ƒ(x)` to be continuous) in `x_0` then:

`lim_n f(xi_n) = f(x_0)`

and then:

`lim_n (F(x_n)-F(x_0) )/(x_n-x_0) = f(x_0)`

and because `{x_n}` is arbitrary:

`lim_(x->x_0) (F(x)-F(x_0))/(x-x_0) = f(x_0)`

which is basically the proof of the fundamental theorem of calculus.