How would you integrate the function $$ f(x) = \left( \int_{0}^{x} e^{-u} \; du \right)^{2} $$ without first solving the inner integral? I've tried using the chain rule, but there seems to be no variation of this that works.

  • $\begingroup$ we have $\int e^{-u}du=-e^{-u}+C$ $\endgroup$ – Dr. Sonnhard Graubner Dec 31 '15 at 15:41
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    $\begingroup$ You could turn it into a double integral: $ \iint_{[0,x] \times [0,x]} e^{-u-v} du dv$. Not sure if that is any easier, but it connects to to multivariable calculus, which appears to be the subject you are studying $\endgroup$ – Hans Engler Dec 31 '15 at 15:42
  • $\begingroup$ I don't know if this would qualify, but one could integrate by parts twice. Of course, one would then need to calculate $f$ at the end points of the interval over which it was integrated. So, I don't see the point of that approach. But, perhaps this helps. $\endgroup$ – Mark Viola Dec 31 '15 at 17:22
  • $\begingroup$ @HansEngler: The OP does not want $f(x)$, he wants $\int f(x) {\rm d} x$ without computing the integral that gives $f(x)$. $\endgroup$ – Alex M. Dec 31 '15 at 17:37
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    $\begingroup$ Your question is extremely unnatural: why would anybody try anything else than evaluating the inner integral? What is the motivation of this question? What do we learn from it? Is it supposed to just keep us busy on New Year's eve? If we may not evaluate the inner integral, it would be useful to know what we are allowed to use. $\endgroup$ – Alex M. Dec 31 '15 at 18:13

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