Okay, I am in the middle of learning single variable calculus and I can understand most of the topics I've learned so far. There are some things that are not obvious or easy understand to me, e.g. the Fundamental Theorem of Calculus: int(a,b, f(x) dx) = F(b) - F(a) , where F is an antiderivative of f.

 

I can understand that in order to find the area between two intervals one would subtract the smaller area from the larger. So I guess this would imply that taking the antiderivative of a function and evaluating it at some point (we'll call it b) would give you the area from 0 to b. (Someone verify this)

 

I see computing the area under a curve from the development of the Riemann Sum, but how is that the same as evaluating the difference between two antiderivatives ( F(b), and F(a) ) ?

 

If anyone can give a nice explanation or reference on FTC I would appreciate it. Sorry for any typos. Thanks.

The fundamental theorem is just saying the area under a curve, from a to b, is equal to the integrals value with b minus a

 

Take f(x) = x

F(x) = (1/2)x^2

 

If we want to find the area under y=x from x=0 to x=1, we take (1/2)1^2 = 1/2 and subtract (1/2)0^2=0

So, the area is 1/2

 

Look at f(x) as the slope at that point, so you find the 'original' equation.

 

A Riemann sum is an approximation. Some functions cannot be integrated, such as x^x. These must be approximated using end-points and the mid-point to estimate the area. This is usually not an exact value, unlike normal integration.

Right, I know that one can compute the area under a curve using FTC.

 

However I don't see the connection by just subtracting the two antiderivatives. It seems like you should just make a different notation for antidifferentation and a different notation for computing definite integrals ...

 

Sorry about that. I should have said the limit of a riemann sum as the number of intervals approach infinity. (Which would yield an exact answer assuming you can find its limit [whether or not the sums are left, right, or mid-point sums])