Here is a statement of the second part of the Fundamental Theorem of Calculus (FTC2), from a well-known calculus text (James Stewart, Calculus, 4th ed):
If $f$ is continuous on $[a,b]$, then $\int_a^b f(x) \, dx = F(b)-F(a)$, where $F$ is any [emphasis mine] antiderivative of $f$, that is, a function such that $F'=f$.
The following, however, seems to give a counterexample.*
Can someone resolve this for me?:
Let $f(x) = \frac{1}{4 \sin (x)+5}$.
$f$ is continuous on $[0, 2 \pi]$:
Consider two antiderivatives of $f$, $F_1$ and $F_2$:
$$F_1(x) = \frac{x}{3}+\frac{2}{3} \tan^{-1}\left(\frac{\cos (x)}{\sin (x)+2}\right)$$
$$F_2(x)=\frac{1}{3} \left(\tan ^{-1}\left(2-\frac{3}{\tan \left(\frac{x}{2}\right)+2}\right)-\tan^{-1}\left(2-\frac{3}{\cot \left(\frac{x}{2}\right)+2}\right)\right).$$
Using Mathematica, I've confirmed that both $F_1'= f$ and $F_2'= f$. According to my reading of the above statement of FTC(2), $\int_0^{2\pi} f (x) \, dx = F_1(2\pi)-F_1(0)= F_2(2\pi)-F_2(0)$
However,
$F_1(2\pi)-F_1(0)=2\pi/3$
$F_2(2\pi)-F_2(0)=0$
Note from the plots below that $F_1$ is continuous on $[a,b]$, while $F_2$ is not. Given all of this, it seems the sufficient condition for $\int_a^b f (x) \, dx = F(b)-F(a)$ is that the antiderivative be continuous on $[a,b]$, not the integrand.
$F_1 =$
$F_2=$
*I've taken this example function from a wolfram.com blog.
Best Answer
A differentiable function is continuous. If your "antiderivative" is not continuous, it is not an antiderivative of a continuous function.