fundamental theorem of calculus part iif fx is

Question

Fundamental Theorem of Calculus, Part I

If f(x) is continuous on [a,b] so,

g(x) = _a∫^x f(t) dt

is continuous on [a,b] and this is differentiable on (a, b) and as,

g′(x) = f(x)

 Proof

Assume that x and x + h are in (a, b).  We then get,

g(x + h) - g(x) = _a∫^x+h f(t) dt - _a∫^x f(t) dt

Here, by using Property 5 of the Integral Properties we can rewrite the very first integral and then make a little simplification as given below.

g(x + h) - g(x) = (_a∫^x f(t) dt  + _a∫^x+h f(t) dt) - _a∫^x f(t) dt

= _a∫^x+h f(t) dt

At last suppose that   h ≠0 and we find,

(g(x + h) - g(x))/h = (1/h) _a∫^x+h f(t) dt                                      (1)

Let's here suppose that h > 0 and as we are even assuming that x + h are in (a, b) we know that f(x) is continuous on [x, x + h]and therefore by the  Extreme Value Theorem we get that there are numbers c and d in [x, x + h] thus f(c) = m is the absolute minimum of f(x) in [x, x + h] and that f(d) = M is the absolute maximum of f(x) in [x, x + h].

Therefore, by Property 10 of the Integral Properties we then get,

mh < _a∫^x+h f(t) dt < Mh

or, f(c)h < _a∫^x+h f(t) dt < f(d)h

Then divide both sides of this with h to have,

f(c) < (1/h)_a∫^x+h f(t) dt < f(d)

and now use (1) to have,

f(c) < (((g(x + h) - g(x))/h)dt < f(d)                             (2)

Subsequently, if h < 0 we can go through similar argument above except we will be working on [x + h, x] to arrive at exactly similar inequality above. Conversely, (2) is true provided h ≠0.

Then here, if we take h → 0 we also have c → x and d → x since both c and d are among x and x + h. it means that we have the subsequent two limits.

lim_h→0 f(c) = lim_c→xf(c)                                     lim_h→0 f(d) = lim_d→xf(x)                                   

The Squeeze Theorem here tells us,

lim_h→0 =(((g(x + h) - g(x))/h) = f(x)

although the left side of this is exactly the definition of the derivative of g(x) and therefore we have, g′(x) = f(x)

Therefore, we've demonstrated that g(x) is differentiable on (a, b).

Here, the theorem at the end of the Definition of the Derivative section give us that g(x) is also continuous on (a, b). At last, if we take x = a or x = b we can go through a same argument we used to find (3) using one-sided limits to have similar result and therefore the theorem at the end of the Definition of the Derivative section will also lead us that g(x) is continuous at x = a or x = b and therefore really g (x) is also continuous on [a, b].