is $\int^b_a |f| = |F(b)|-|F(a)|$?

Question

sorry if this seems trivial or obvious. But is $\int^b_a |f| = |F(b)|-|F(a)|$? I'm currently trying to prove that $|F(b)-F(a)|\leq \int^b_a|f|$. Using the fundamental theorem of calculus, we have

$$\int^b_af=F(b)-F(a)\leq \int^b_a|f|$$ if $\int^b_a |f| = |F(b)|-|F(a)|$, then via the triangle inequality $$\int^b_af=F(b)-F(a)\leq \int^b_a|f|=|F(b)|-|F(a)|\leq |F(b)-F(a)|$$ which is the opposite of what I want to prove, so I suppose this would suggest that $\int^b_a |f| \neq |F(b)|-|F(a)|$,

Answer 1

Take $a=-1,b=1, F(x)=x^{2}$ which gives $f(x)=2x$. Clearly, $0=F(b)-F(a)\neq \int_a^{b} |f(x)| dx$.

Answer 2

If by $F$, you mean the antiderivative of $f$, then no.

Take $f(x)=\sin(x), a=\pi, b=2\pi$. Then $F(x) = -\cos x$.

Then $$\int_a^b |f(x)|dx = \int_\pi^{2\pi}|\sin x| dx = -\int_\pi^{2\pi}\sin xdx = -\left(-\cos x|_\pi^{2\pi}\right) = -\left(-\cos 2\pi - (-\cos \pi)\right) = -(-1 - 1) = 2$$

while $|F(b)| - |F(a)| = 0$

Answer 3

Let's say we restrict to continuous $f$ for simplicity. A giveaway is the fact that the quantity $\lvert F(b)\rvert-\lvert F(a)\rvert$ depends on the choice of an antiderivative $F$ of $f$. For instance, if we had chosen an antiderivative $F$ such that $F(a)$ and $F(b)$ are both positive, then this quantity would acutally be $\int_a^bf(x)\,dx$, but if we had it such that one is negative and the other is positive, then $\lvert F(b)\rvert-\lvert F(a)\rvert=\begin{cases}F(b)+F(a)&\text{if }F(b)>F(a)\\ -F(b)-F(a)&\text{if }F(b)<F(a)\end{cases}$.

Therefore, your statement would be that for all continuous $f$, for all $a,b$ and for all antiderivatives $F$ of $f$ we have $\int_a^b\lvert f(x)\rvert\,dx=\lvert F(b)\rvert-\lvert F(a)\rvert$. Which doesn't make sense, because there are two antiderivatives $F_1$ and $F_2$ that result in different RHS.

Answer 4

To solve your actual problem:

I'm currently trying to prove that $|F(b)-F(a)|\leq \int^b_a|f|$

That is true because $$ |F(b)-F(a)| = \left|\int_a^b f(x) \, dx \right| \leq \int^b_a|f(x)| \, dx $$ if the fundamental theorem of calculus applies, e.g. if $f$ is continuous and $F' = f$.