The answer to that question is basically the
fundamental theorem of calculus.
To show enough to make this plausible, imagine you break the interval [a,b] into pieces. Call the pieces [a,x
1], [x
1,x
2], [x
2,x
3] ... [x
N-1,x
N], [x
N,b].
Assume they are equally large intervals and that x
i+1-x
i=h. Now you have a function F(x) and its derivative is F'(x)=f(x).
To show that the summation definition of the integral is in a sense the opposite of the derivative, we need to show that
\int_{a}^{b}f(x)dx=\int_{a}^{b}F'(x)dx = F(b)-F(a)
Use the definition of the derivative as an approximation (becomes exact as h goes to 0):
F'(x_i)\approx\frac{F(x_{i+1})-F(x_i)}{h}
Now look at the summation definition of the integral (again, only becomes exactly the integral as h goes to 0). Also, for convenience, let a=x
0 and b=x
N+1.
\int_{a}^{b}F'(x)dx =\sum_{i=0}^{N}F'(x_i)*(x_{i+1}-x_i)
Remember that x
i+1-x
i=h, so
\int_{a}^{b}F'(x)dx =\sum_{i=0}^{N}F'(x_i)*h
\approx\sum_{i=0}^{N}\frac{F(x_{i+1})-F(x_i)}{h}*h
=\sum_{i=0}^{N}F(x_{i+1})-F(x_i)
This type of sum is called a 'telescoping sum'. Watch what happens when you write the first few terms:
F(x
1) -F(x
0)+F(x
2) -F(x
1)+F(x
3) -F(x
2)+F(x
4) -F(x
3)
Notice that every term cancels another one except -F(x
0) and +F(x
4). No matter how many terms you add, all of them will cancel except the first and last. So that integral is
\int_{a}^{b}F'(x)dx = F(x_{N+1})-F(x_0)=F(b)-F(a)
So hopefully that shows how they are opposites.
