Evaluating definite integrals whose limits are absolutely equal

Question

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

Give reasons for your answers.

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,

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

$\ $

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?

Answer 1

The statement $$\int\limits_{-a}^a{f(x)dx}=\int\limits_{-a}^0{f(x)dx}=\int\limits_{0}^a{f(x)dx}\\ =F(0)-F(-a)+F(a)-F(0)=-F(a)+F(a)=0$$ is false because it is the integrand $f(x)$ that is even, not its antiderivative $F(x)$. That is to say, $f(x) = f(-x)$ for all $x \in \mathbb R$ does not imply that $F(x) = F(-x)$ where $F$ is a function that satisfies $F'(x) = f(x)$. Rather, you should write with the substitution $u = -x$, $du = -dx$, $$\int_{x=-a}^0 f(x) \, dx = \int_{u=a}^0 f(-u) \, (-du) = \int_{u=0}^a f(-u) \, du = \int_{u=0}^a f(u) \, du,$$ and the only place where we use the evenness of $f$ is in the rightmost equality above.

Answer 2

There was a mistake across the second-last equality of my Riemann sum:

$=\displaystyle\lim_{n \to \infty}\left[\frac{a}{n}\left(f(-a)+f\left(-a+\frac{a}{n}\right)+f\left(-a+\frac{2a}{n}\right)+...+f(-a+a)\right)\right]+\displaystyle\lim_{n \to \infty}\left[\frac{a}{n}\left(f(0)+f\left(\frac{a}{n}\right)+...+f\left(\frac{a(n-1)}{n}\right)+f(a)\right)\right]\\ =\displaystyle\lim_{n \to \infty}\left[\frac{a}{n}\left(f(0)+f(0)+f\left(\frac{a}{n}\right)+f\left(\frac{a}{n}\right)+...+f\left(a-\frac{a}{n}\right)+f\left(a-\frac{a}{n}\right)+f(a)+f(a)\right)\right]\\ =2\displaystyle\int_0^a f(x) dx$