# Fundamental Theorem of Calculus part 2 problem

I'm following the 10th edition of Calculus Early Transcendentals by Anton, Bivens and Davis.

They introduce the second part of the FTC in more or less the following way:

They let $$A(x)=\int_{a}^{x}f(t)dt$$ equal the area between the x-axis and the curve of the function from $$t = a$$ to $$t = x$$. Then by part one it follows that $$A'(x)=f(x)$$ and they let $$F(x)$$ be any antiderivative of $$f(x)$$. Because $$A$$ and $$F$$ are both antiderivatives of $$f(x)$$ they differ by a constant so: $$F(x)=A(x)+C$$.

Now $$F(b)=A(b)+C$$ and $$F(a)=A(a)+C$$. Take the difference and you get: $$F(b)-F(a)=A(b)+C-A(a)-C$$But $$A(a) = 0$$ so what you end up with is: $$F(b)-F(a)=A(b)$$ or otherwise written: $$\int_{a}^{b}f(t)dt = F(b)-F(a)$$

What bothers me here is the following: subtracting $$F(a)$$ amounts to just subtracting the constant $$C$$ because $$F(a)=A(a)+C$$ but $$A(a) = 0$$. We should then get the same result if instead of subtracting $$F(a)$$, we subtract just $$C$$: $$F(b)-C=A(b)+C-C=A(b)$$ So, $$\int_{a}^{b}f(t)dt=F(b)-C$$

For ex.:

$$\int_{0}^{1}xdx = \frac{x^2}{2} + C|_{x=b}-C=\frac{b^2}{2}+C-C=\frac{b^2}{2}$$ which is obviously true only if $$F(a)=0$$ (like in this integral).

Why then do I get that $$F(b)-F(a)=F(b)-C = \int_{a}^{b}f(t)dt$$?

• I'm not following your example. What are $a$ and $b$? Are they $0$ and $1$? If so, then I don't see the problem. Also, note that $F(a) = F(0) = \frac{0^2}{2} + C = C$, as expected. – Theo Bendit Mar 7 at 0:52
• The problem with your approach is that you don't know the constant before you compute the definite integral. And these two $C$ are in fact not the same $C$ - it should be written $\int_0^1 xdx = (x^2/2) + C_1 - C_2$ – enedil Mar 7 at 0:57

you are assuming that the constant already in the antiderivative is equal to its difference from the area, this is only true if the terms of x in the antiderivative evaluate to $$0$$, as it does in your example:
$$F(b) = \int_{a}^{b}f(x)dx + c$$ $$F(b)=g(b)+c_1=\int_{a}^{b}f(x)dx + c=A(b)+c$$, where $$g(b)$$ has only terms of x $$F(a)=g(a)+c_1=A(a)+c \iff g(a)+c_1=c$$,
$$c_1=c \iff g(a)=0$$, as it does in your example! Otherwise, this assumed relationship between $$c_1=c$$ cannot possibly hold.