• anonymous
How do you take the integral of e^(2x-x^2)
  • Stacey Warren - Expert
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  • katieb
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  • anonymous
Hi, I believe you need to provide more information for this problem. If you are looking for an antiderivative \[\int e^{2x-x^2} dx\] then I believe you are out of luck. Here is the reason: using a u-substitution of u=1-x, this becomes \[-\int e^{1-u^2} du \ = \ -\int e\cdot e^{-u^2} du=-e\int e^{-u^2} du.\] You will recognize this last form as a Gaussian integral (look it up on Wolfram or something). We cannot find a nice antiderivative of this integral because one does not exist ("nice" meaning made of polynomials, root functions, trig functions etc.). If your problem involves bounds of integration, then you may be in luck (depending on the bounds of integration). For example, Laplace proved that the following integral converges: \[\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}\] This derivation is also available at Wolfram. You can also read about it here: I hope this helps.

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