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Could you, please, help with this problem? I need to use the definition of the Riemann integral to show that

$$ \int_{-a}^af(x^2)dx=2\int_0^a f(x^2)dx $$

You may assume that both integrals exist. I do not know how to solve this problem, because I do not understand what $$ f(x^2)$$ means shouldn't it be $$x^2$$? If the problem is correct, could you give help me with this problem? Thank you.

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    $\begingroup$ You have a mistake in the lower limit of the first integral... $\endgroup$ – DonAntonio Apr 5 '18 at 21:10
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    $\begingroup$ shouldn't it be $\int_{-a}^a f(x^2) dx$? $\endgroup$ – Ricky Apr 5 '18 at 21:10
  • $\begingroup$ yes, you're right, I corrected it $\endgroup$ – Learningmath Apr 5 '18 at 21:16
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    $\begingroup$ Since $f(x^2)$ is even, this is a special case of math.stackexchange.com/q/2718697/148510 $\endgroup$ – RRL Apr 5 '18 at 21:18
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Note $f(x^2)=f((-x)^2)$ hence $$\int^a_{-a}f(x^2)\,dx=\int^0_{-a}f(x^2)\,dx+\int^a_0f(x^2)\,dx=\int^0_af((-x)^2)\,d(-x)+\int^a_0f(x^2)\,dx\\=-\int^0_af(x^2)\,dx+\int^a_0f(x^2)\,dx=\int^a_0f(x^2)\,dx+\int^a_0f(x^2)\,dx\\ =2\int^a_0f(x^2)\,dx$$

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