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 $F(b)-F(a)=F(b)-C = \int_{a}^{b}f(t)dt$?

  • $\begingroup$ 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. $\endgroup$ – Theo Bendit Mar 7 at 0:52
  • 3
    $\begingroup$ 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 $\int_0^1 xdx = (x^2/2) + C_1 - C_2$ $\endgroup$ – 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.


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