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$$\left | \int_ {-a}^{0} f - \int_{0}^{a} f \right | \geq \left | \left | \int_ {-a}^{0} f \right |- \left | \int_{0}^{a} f \right| \right | \geq \left | \int_ {-a}^{0} |f| - \int_{0}^{a}| f |\right | $$

Can I apply the reverse and triangle inequalities like this? Assume $f$ is continuous everywhere

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The first inequality is true, the second often isn't. Take $f$ to be $0$ on $[0,a]$, and defined on $[-a,0]$ in such a way that $\int_{-a}^0f=0$ but $\int_{-a}^0|f|>0$.

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Huh? what do you mean? –  sidht Mar 16 '13 at 8:55
    
I mean the second inequality is not generally true. I mean that if $f:[-a,a]\to \mathbb R$ is any function of your choosing satisfying the properties that $f(x)=0$ for all $x\in[0,a]$, $\int_{-a}^0f(x)dx = 0$, and $\int_{-a}^0|f(x)|dx>0$, then the second inequality will be false for that $f$. And implicitly I mean that you can come up with such an example without too much trouble. $f$ will be sometimes positive, sometimes negative. Actually any example where $f$ is $0$ on $[0,a]$ and changes signs on $[-a,0]$ will give a counterexample, but I wanted to point out a simple extreme case. –  Jonas Meyer Mar 16 '13 at 8:57
    
Oh okay, but there are cases when it applies right? –  sidht Mar 16 '13 at 9:01
    
@sizz: Yes, for example when $f(x)=0$ for all $x$. But it is not typical that it will hold. Consider that it implies that $\int_{-a}^0|f(x)|dx\leq\left|\int_{-a}^0f(x)dx\right|$ when $f$ is zero on $[0,a]$, but this inequality is backwards. –  Jonas Meyer Mar 16 '13 at 9:03

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