So, we are now ready to prove the FTC part 1. We’re going to follow the proof in Stewart and add in some discussion as we go along to motivate what we are doing. What we are going to prove is that:


\frac{d}{dx} \int_a^x f(t) dt=f(x)


for x\in [a,b] when f is continuous on [a,b].




we define g(x)=\int_a^x f(t)dt and we want to find the derivative of g. We will do this by using the fundamental definition of the derivative, so let’s look at calculating this function at x and x+h – ie. how much does it change when we change x by a little bit?


g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt


But remember that the definite integral is just the area, so this difference is the area between a and x+h minus the area between a and x. Which is just the area between x and x+h. Using the properties of integrals, we can write this formally as:


g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt=\left(\int_a^{x}f(t)+\int_x^{x+h}f(t)\right)-\int_a^{x}f(t)=\int_x^{x+h}f(t)dt


and we can write, for h\ne 0:




Restated, we can think of this as the area between x and x+h divided by h. Let’s put up a plot to understand this:


Remember we want to know how quickly the area is growing, and we have found that it’s related to the area between x and x+h divided by h. We are going to show that as h goes to zero, the quotient is just the value of the function itself at x.


For now we will assume that h>0.


By the extreme value theorem (and indeed it makes intuitive sense I hope), since f is continuous on [x,x+h], the function will have some position at which it is a minimum and some position at which it is a maximum – we call these positions u and v, and the values of the functions at these positions are f(u)=m and f(v)=M.


We also have a property of integrals which tells us about the bounds of an integral. The integral from x to x+h can take on a minimum value of h times the minimum value that f takes in this region, and a maximum value of h times the maximum value that f takes in this region. That is:


mh\le\int_x^{x+h}f(t)dt\le Mh


or in other words:


f(u)h\le\int_x^{x+h}f(t)dt\le f(v)h


As h is greater than zero, dividing this whole thing by h doesn’t change the directions of inequalities, so we have:


f(u)\le\frac{1}{h}\int_x^{x+h}f(t)dt\le f(v)


and the thing in the middle is just the difference quotient we had above:


f(u)\le\frac{g(x+h)-g(x)}{h}\le f(v)


You can do the same thing for negative h but you just have to swap around the inequality directions.


Now if we let h\rightarrow 0, then the interval [x,x+h] shrinks to a point: x, and both u and v which are in this interval must both simply be x. So we have:


\lim_{h\rightarrow 0} f(u)=f(x)    and    \lim_{h\rightarrow 0} f(v)=f(x)


So by the squeeze theorem:


g'(x)=\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}=f(x).


Here we can approach the limit from the left or the right (ie the limits from 0^+ or 0^-) and get the same answer, but if x=a or x=b, then we can only take one of these limits. This shows that g is continuous on [a,b].


And we are done. We have shown that the rate of change of the area under the function is equal to the value of the function itself at the value of x where we are changing the bound.



How clear is this post?