Suppose `x` and `x+h` are values in the open interval `(a,b)`.

Since we defined `F(x)` as `int_a^xf(t)dt`, we can write:

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

We can re-express the first integral on the right as the sum of 2 integrals (note the upper and lower limits), and simplify the whole thing as follows:

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

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

Now divide both sides by `h>0`:

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

Now, for any curve in the interval `(x,x+h)` there will be some value `c` such that `f(c)` is the absolute minimum value of the function in that interval, and some value `d` such that `f(d)` is the absolute maximum value of the function in that interval. (This is a consequence of what is called the Extreme Value Theorem.)

So we can write:

`f(c) <= (F(x+h)-F(x))/h <= f(d)`

Now if `h` becomes very small, both `c` and `d` approach the value `x`. (They get "squeezed" closer to `x` as `h` gets smaller).

So we can write the following limits:

`lim_(h->0)f(c) = lim_(c->x)f(c) = f(x)`


`lim_(h->0)f(d) = lim_(d->x)f(d) = f(x)`

Since our expressions are being squeezed on both sides to the value `f(x)`, we can conclude:

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

But we recognize the limit on the left is the definition of the derivative of `F(x)`, so we have proved that `F(x)` is differentiable, and that `F'(x) = f(x)`. Also, since `F(x)` is differentiable at all points in the interval `(a,b)`, it is also continuous in that interval.

Note: When integrating, it doesn't really make any difference what variable we use, so it's OK to use `t` or `x` interchangeably, as long as we are consistent.