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I'm having trouble trying to solve the following system.

$f(x) = 1 + \int_0^x{g(t)\mathrm{d}t}$

$g(x) = x(x - 1) + \int_{-1}^{1}{f(t)}\mathrm{d}t$

I have in mind to substitute this integrations for constants, but I'm kind lost. Can someone help ?

Thanks in advance

aajjbb
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1 Answers1

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Hint: if $\int_{-1}^1 f(t)\ dt = c$, you have $g(x) = x(x-1)+c$. What then is $f(x)$?

Robert Israel
  • 448,999
  • I think I got it, making $f(x) = 1 + a$ and $g(x) = x(x - 1) + b$, resolving, $a = \dfrac{x^3}{3} - \dfrac{x^2}{2} + bx$ and $b = \dfrac{5}{3}$. Thank you. – aajjbb Sep 22 '13 at 19:13