If $\int_a^b f(x)dx=\left[F(x)\right]_a^b$, Is $\int_a^b \lvert f(x)\rvert dx= \left[\lvert F(x)\rvert \right]_a^b$ true?

Question

I was brushing up on some basic inequalities and I attempted to derive an alternate form of the regular triangle inequality by using $(\left|b\right|-\left|a\right|)^2$. $$ \begin{align*}(\left|b\right|-\left|a\right|)^2 &= \left|b\right|^2 + \left|a\right|^2 - 2|a||b|\\ &=b^2+a^2-2|a||b|\\ &\leq(b-a)^2\\&=|b-a|^2\end{align*} $$ I arrived at the conclusion that $|b|-|a|\le|b-a| $. This however, does not seem to agree with the following: $$\biggl|\int_{a}^{b} f(x) \, dx\biggr| \leq \int_{a}^{b} |f(x)|\,dx$$ $$\biggr|F(b)-F(a)\biggl| \leq \biggr|F(b)\biggl|-\biggr|F(a)\biggl| $$

Edit: I believe there may be a mistake going from line 3 to line 4. It should be $|b|-|a| \le (b-a)$ which is only true for $\bigg||b|-|a|\bigg|\le\bigg|b-a\bigg|$.

I was wondering if maybe there was a mistake in my initial derivation or if I am using the absolute value signs incorrectly when applying them to the integrals through the fundamental theorem of calculus. Possibly a condition or assumption I am unaware of. It just seems like an odd result to get and I can't figure out why this is.

Answer 1

We know that $|b-a|\geq |b|-|a| $. This however, does not seem to agree with the following: $$\biggl|\int_{a}^{b} f(x) \, dx\biggr| \leq \int_{a}^{b} |f(x)|\,dx\tag{1}$$ $$\biggr|F(b)-F(a)\biggl| \leq \biggr|F(b)\biggl|-\biggr|F(a)\biggl| \tag{2}$$ Where am I going wrong?

We know that equation $(1)$ is correct. It is the equation $(2)$ that has the problems. If $\int_a^b f(x)dx=\left[F(x)\right]_a^b$, $\int_a^b |f(x)|dx\color{red}{\neq} \left[|F(x)|\right]_a^b$. One has to break the integral to get the latter. For example, \begin{align*} \left|\int_{-1}^{+1}x^3\,dx\right|&=\left|\frac14[x^4]_{-1}^{+1}\right|=0\\ \int_{-1}^{+1}|x^3|\,dx=-\int_{-1}^{0}x^3\,dx+\int_{0}^{+1}x^3\,dx=-\frac14[x^4]_{-1}^{0}+\frac14[x^4]_{0}^{+1}=\frac12&\color{red}{\neq}\frac14[|x^4|]_{-1}^{+1}=0\\ \end{align*}