Fundamental Theorem of Calculus: how the arbitrary C cancels out?

Question

From my textbook:

$$F(b)-F(a) = [A(b)+C]-[A(a)+C] = A(b)-A(a)$$

F is an antiderivative of f.

Probably a very basic question to ask. My teacher didn't explain.

So my textbook is saying that since A(x) is an antiderivative of f(x), every other antiderivative of f(x) on [a,b] can obtained by adding a constant.

A being the area under the graph of function f, continuous on [a,b] on interval [a, x]. "a" being a fixed number and "x" being a general variable and a<=x<=b.

I understand most of the operations done in the picture. But I think because C is arbitrary, it's possible for it to have different values. So can't we always cancel it?

http://math.stackexchange.com/questions/877907/is-this-proof-of-the-fundamental-theorem-of-calculus-correct — , Jan 12 '16 at 02:39
@Doeser when you say "F is the anti-derivative of f" you are a little off. F is an anti-derivative of f, a very specific one with a very specific but unknown C value. Since F is a specific antiderivative, the two C's are exactly the same. If we had two different antiderivatives then we would have the problem you are suggesting. — Sean English, Jan 12 '16 at 02:48

score 6 · Accepted Answer · answered Jan 12 '16 at 02:44

If I understand your question correctly, it's that if the $C$ is the above proof is arbitrary, how can we always cancel it? Well, the idea is that the $C$ could be any value, but the two $C$'s mentioned in the proof are the exact same and so can always be cancelled.

score 1 · Answer 2 · answered Jan 12 '16 at 02:51

1

In the expression $F(a)-F(b)$, each $F$ refers to the same antiderivative. So, each $C$ must refer to the same number in the expression after the first equal sign.

answered Jan 12 '16 at 02:51

Tim Raczkowski

20,046

Fundamental Theorem of Calculus: how the arbitrary C cancels out?

2 Answers2