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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

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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.

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  • $\begingroup$ I accidentally left out part of the question stating that $\int_{-a}^{a} f(x^2) dx = 2 \int_{0}^{a} f(x^2)dx, \int_{-a}^{a}xf(x^2)dx=0$. Was this what you were hinting at? Also, it seems that a hint was given to show that $U(xf, p) - L(xf,p) \leq a U(f,p) - L(f,p)$? But I don't see how this is helpful. $\endgroup$ – Nikitau Dec 12 '16 at 3:33
  • $\begingroup$ The point if basically that $f(x^2)$ is even and $xf(x)^2$ is odd. So then $f((-x)^2)=f(x^2),$ and $(-x)f((-x)^2)=-xf(x^2).$ $\endgroup$ – tylm5678 Dec 12 '16 at 3:39
  • $\begingroup$ Oh I see. And because $xf(x)^2$ is odd we have that the integral of it is 0. Thanks! I think I know how to prove $\int_{-a}^{a} xf(x^2)=0$. I'm still a little stuck on proving Riemann integrability, but I'll mull it over. $\endgroup$ – Nikitau Dec 12 '16 at 3:57
  • $\begingroup$ It's a fact that a function is Riemann integrable if and only if it is bounded and its set of discontinuities has measure zero. This is the so-called Riemann-Lebesgue theorem, and it certainly suffices to prove your result. I'm guessing that the exercise you have probably wants you to just play around with partitions. $\endgroup$ – tylm5678 Dec 12 '16 at 5:28

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