My title can be a bit confusing, so here's a bit of background.

The corollary to the Fundamental Theorem of Calculus says that $\int_a^bf(x)dx = F(b)-F(a)$, assuming that *F*'(*x*) = *f* (*x*), or that the area under the curve *f* (*x*) from *x* = *a* to *x* = *b* is equal to the difference of values of the antiderivative of *f* (*x*) at *a* and *b*.

The following is my attempt to prove it.

1: The area under the curve of *f* (*x*) from *x* = *a* to *x* = *b* is equal to the area of the rectangles under the curve as you take more and more rectangles. See this image:

Mathematically speaking, it's $\int_a^bf(x)dx = \lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot f(a+hn)$

2: Let us replace our measly *f* (*x*) with its definition, in terms of the derivative of *F* (*x*), namely $f(x) = \lim_{j\to 0}\frac{F(x+j)-F(x)}{j}$. Thus, our first equation becomes

$\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot \lim_{j\to 0}\frac{F(a+hn+j)-F(a+hn)}{j}$

Now, my question is, since both *h* and *j* are going to zero via a limit, can I assume that they are effectively the same? Can I simply replace all instances of *j* with an *h* and rid myself of an unnecessary second limit? If I could, my proof would continue as follows:

3: Replacing all *j*'s with *h*'s yields:

$\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}h\cdot \frac{F(a+h(n+1))-F(a+hn)}{h}$, and the *h*s can cancel out: $\lim_{h\to 0} \sum_{n=1}^{(b-a)/h}F(a+h(n+1))-F(a+hn)$.

4: Thankfully, this becomes a telescoping series, as seen here:

$F(a+h(1))-F(a+0h) + F(a+h(2))-F(a+1h) + F(a+h(3))-F(a+2h) + ... = -F(a-h) + F(b-h)$

$ + F(a+h(\frac{b-a}{h}))-F(a+h(\frac{b-a}{h}-1) = F(b) - F(b-h)$

which, together, yields -*F* (*a* - *h*) + *F* (*b*) as the sum.

Putting this back in, we get $ \lim_{h \to 0} -F(a - h) + F(b) = F(b) - F(a) = \int_a^bf(x)dx = F(b)-F(a) $

However, steps 3 and 4 require the ability for me to assume that *h* and *j* are the same thing. My teacher (who admittedly doesn't deal with this too often), whom I asked first, said that perhaps *h* and *j* are going to 0 at different rates. However, I do not think that the concept of a limit to 0 *at a rate* actually means anything.

So the question I bring to you is: Is the operation that I performed to go from step 2 to step 3 a valid operation? If so, why? If not, why not?

Thanks for your help!

-Gabriel Benamy