# Evaluating definite integrals whose limits are absolutely equal

The question that I'm stuck on reads as follows:

The substitution rule can sometimes be used to simplify the evaluation of an integral of an even or odd function. What can you say about the integral $\int\limits_{-a}^a{f(x)dx}$ if

a) $f$ is an even function

b) $f$ is an odd function

I ignored the part about the substitution rule, and tried to prove for a) that $\int\limits_{-a}^a{f(x)dx}=2F(a)$ (which I assumed using common sense). However, my solution for a) consistently came out as 0. For example,

$\int\limits_{-a}^a{f(x)dx}=\int\limits_{-a}^0{f(x)dx}+\int\limits_{0}^a{f(x)dx}\\ =F(0)-F(-a)+F(a)-F(0)=-F(a)+F(a)=0$

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So, I then tried a Riemann sum: $\displaystyle\lim_{n \to \infty}\left[\frac{0-(-a)}{n}\sum_{i=0}^{n}{f\left(-a+\frac{0-(-a)i}{n}\right)}\right]+\displaystyle\lim_{n \to \infty}\left[\frac{a}{n}\sum_{i=0}^{n}{f\left(\frac{ai}{n}\right)}\right]\\ =\displaystyle\lim_{n \to \infty}\left[\frac{a}{n}\left(f(-a)+f\left(-a+\frac{a}{n}\right)+f\left(-a+\frac{2a}{n}\right)+...+f(-a+a)\right)\right]+\displaystyle\lim_{n \to \infty}\left[\frac{a}{n}\left(f(0)+f\left(\frac{a}{n}\right)+...+f\left(\frac{a(n-1)}{n}\right)+f(a)\right)\right]\\ =\displaystyle\lim_{n \to \infty}\left[\frac{a}{n}\left(f(0)-f(0)+f\left(\frac{a}{n}\right)-f\left(\frac{a}{n}\right)+...+f\left(a-\frac{a}{n}\right)-f\left(a-\frac{a}{n}\right)+f(a)-f(a)\right)\right]\\ =0$

$\$

What am I doing wrong?

• Well for one thing, your LaTeX is all messed up! – M. Vinay Jun 9 '14 at 8:02
• Ugh!? Your post hurts my eyes. – Tunk-Fey Jun 9 '14 at 8:03
• Sorry. It's my first time on this site. Does  not work? I don't know how to do previews either. – ahorn Jun 9 '14 at 8:04