If $f(x^2)$ is $\mathscr{R} [0,a]$, then prove $f(x^2)$ and $xf(x^2)$ are $\mathscr{R}[-a,a]$.

Question

I want to show that if $f:[0,a^2] \rightarrow \mathbb{R}$ is bounded function and if $f(x^2)$ is Riemann integrable on $[0,a]$, then $f(x^2)$ and $xf(x^2)$ are Riemann integrable on $[-a, a]$ and

$$\int_{-a}^{a} f(x^2) dx = 2 \int_{0}^{a} f(x^2)dx, \int_{-a}^{a}xf(x^2)dx=0$$

I don't even know where to begin on this proof. I know that if $f(x^2)$ is RI, then $$\exists\ \text{a partition P s.t }\ U(f,p) - \epsilon < I < L(f,p) + \epsilon$$ where I is the Riemann Integral.

Help or hints would be much appreciated

Answer 1

If we know that $\int_{0}^{a}f(x^2) \> dx < \infty$, what can we say about $\int_{-a}^{0} f(x^2) \> dx$? In particular, if we write $g(x)=f(x^2),$ is there a relationship between $g(c)$ and $g(-c)$? If that hint doesn't make sense maybe try graphing some functions of $x^2$. For the second one perhaps you can make a substitution.