Question

Question #16cc0

Answer 1

Consider a real function of real variable `f(x)` defined and continuous over the interval `[a,b]`.

For every `x in [a,b]` we can define the function:

`F(x) = int_a^x f(t)dt`

and then for any `h` such that `x+h in [a,b]`:

`F(x+h) -F(x) = int_a^(x+h) f(t)dt - int_a^x f(t)dt`

Based on the additivity of the integral:

`F(x+h) -F(x) = int_a^x f(t)dt + int_x^(x+h)f(t)dt - int_a^x f(t)dt = int_x^(x+h)f(t)dt`

Now, the mean value theorem ensures that there is a point `xi_h in (x,x+h)` such that:

`1/h int_x^(x+h)f(t)dt = f(xi_h)`

and then:

`(F(x+h) -F(x))/h = f(xi_h)`

If we now let `h->0`, necessarily `xi_h ->x`, so that:

`lim_(h->0) (F(x+h) -F(x))/h = f(x)`

which by definition of derivative means that:

`d/dx F(x) = f(x)`

We can conclude that:

`d/dx (int_a^x f(t)dt) = f(x)`

that is the differentiation is the reciprocal of the integration, and symmetrically, if the function `f(x)` has a continuous derivative:

`int_a^x f'(t)dt = f(x)-f(a)`

Answer 2

The Fundamental Theorem of Calculus tells us so.

The FTM I states that:

If `f` is a continuous function defined on an interval `[a,b]`, and we define`F` in `[a,b]` by:

`F(x)=int _{a}^{x} \ f(t) \ dt`

Then, `F` is differentiable on `(a,b)` and:

`F'(x) = f(x)`

(ie the derivative of an antiderivative of an integrand is the same as the initial integrand)