Sho Kano said:
By taking the derivative of that definite integral, and by showing that in the end, ##F'=f##, means that I took a derivative of something and that something is the anti-derivative of ##f##, the original function (what I started with). What does this guarantee?
Yes, we are taking the derivative of a
function ##F(y) = \int_{0}^{y} f(x) \ dx ## that is defined as a definite integral of another function ##f(x)##. (That's somewhat clearer than saying we are taking "the derivative of definite integral", which makes it sound like we are taking the derivative of a single numerical value.)
The proof shows ##F'(y) = f(y)##. (This is not a simple consequence of the definition of ##F##. The definition of ##F## does not require that ##F'(y) = f(y)##.)
Now suppose we need to compute ##\int_a^b f(x) \ dx##. From the properties of integration, we have:
##\int_a^b f(x)\ dx = \int_0^b f(x)\ dx - \int_0^a f(x)\ dx##
In terms of areas, this expresses the area ##\int_a^b f(x)\ dx## as the difference between two overlapping areas.
Writing things in terms of ##F(y)## instead of ##f(x)## we have:
##\int_a^b f(x) \ dx = F(b) - F(a)## where ##F## has the property that ##F'(b) = f(b)## and ##F'(a) = f(a)##.
For example, suppose if we are faced with the problem of finding ##\int_a^b sin(x) dx## and nobody had given us any information about what function ##F(y) = \int_0^y sin(x)\ dx## is. We know that ##F(y)## has the property that ##F'(y) = f(y)##. So if we know rules for taking the derivatives of trigonometric functions, we may be able to find a specific function ##F(y)## whose derivative is ##sin(x)## without actually approximating the area ##\int_0^y sin(x) \ dx## as the area of rectangles.
An intuitive way of seeing why ##F'(y) = f(y)## is to begin by thinking of the integrals involved as being approximated by Riemann sums where the base of the rectangles each have the same length ##h##. (This is not rigorous math because the modern definition of Riemann sum does not require the base of rectangles each have same length and it certainly does not require that when approximating two different integrals, we use the same length base in each approximation.)
We have, by the definition of a derivative, ##F'(y) = \lim_{h \rightarrow 0} \frac{ \int_0^{x+h} f(x)\ dx - \int_0^x f(x)\ dx}{h}##
Suppose we approximate each of the integrals as Riemann sums using rectangles with base ##h## and ##x = nh## for some integer ##n##. Then the sum for ##\int_0^{x+h} f(x)\ dx \approx \sum_{k=1}^{n+1} h f(kh) = S(n+1)## contains one more rectangle than the sum for ##\int_0^x f(x)\ dx \approx \sum_{k=1}^n h f(kh) = S(n)##.
So ##\frac{ lim_{h \rightarrow 0} \int_0^{x+h} f(x)\ dx - \int_0^x f(x)\ dx}{h} \approx \frac{ S(n+1) - S(n)}{h}##
## = \frac { (\ S(n) + h f((n+1)h)\ ) - S(n)} {h} = \frac{ h f((n+1)h)}{h} = f((n+1)h)##.
If ##f## is a continuous function , then ##f((n+1)h)## is nearly equal to ##f(nh) = f(x)## when ##h## is small.
