Evaluating $\int_{-a}^a x^{2n+1}\mathrm{d}x$ for all non-negative integers $n$ simultaneously

Question

My assumption would be

$$\int_{-a}^a x\ dx=0$$

Am I on the right track here? Also, for indefinite integrals

$$\int (f)x\ dx$$

would this be correct as well?

Background

My professor raised this question in his lecture and I provided the following

\begin{align}\int_{-a}^{a}\left(x^3\right)dx&= 0\end{align}

and

\begin{align}\int_{-a}^{a}\left(x^7\right)dx&= 0\end{align}

to support that odd degrees will always equal to zero. The professor stated my evaluations were correct, however, I couldn't use the fact that it works for two positive odd exponents to deduce conclusively that the result will hold for all positive odd exponents. Thus, my assumption is that

$$\int_{-a}^a x\ dx=0$$

covers all non-negative integers $n$ simultaneously. Any help in this would be appreciated!

Answer 1

Here it is a more general result which may help you.

Consider that the function $f:\textbf{R}\to\textbf{R}$ is odd. This means that $f(-x) = -f(x)$. Thus one has that \begin{align*} \int_{-a}^{a}f(x)\mathrm{d}x & = \int_{-a}^{0}f(x)\mathrm{d}x + \int_{0}^{a}f(x)\mathrm{d}x\\\\ & = \int_{0}^{a}f(-x)\mathrm{d}x + \int_{0}^{a}f(x)\mathrm{d}x\\\\ & = -\int_{0}^{a}f(x)\mathrm{d}x + \int_{0}^{a}f(x)\mathrm{d}x = 0 \end{align*}

where it has been used the change of variable $u = -x$.

In particular, at your case, $f(x) = x^{2n+1}$, which is odd.

Answer 2

We have

$$\int_{-a}^ax^{2n+1}dx=\frac{1}{2n+2}x^{2n+2}\bigg\vert_{-a}^a=\frac{1}{2n+2}\left(a^{2n+2}-(-a)^{2n+2}\right)=\frac{a^{2n+2}}{2n+2}\left(1-(-1)^{2n+2}\right)$$

But $2n+2$ is always even. This implies $(-1)^{2n+2}=1$ which gives us

$$\frac{a^{2n+2}}{2n+2}\left(1-(-1)^{2n+2}\right)=\frac{a^{2n+2}}{2n+2}\left(1-1\right)=0$$