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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.