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How to integrate:

$$\int_0^1(1+e^{-x^2}) dx$$

I have tried it so many times, but couldn't find any clue Please help

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  • $\begingroup$ Do you know the error function ? $\endgroup$ Jan 1, 2015 at 9:12

2 Answers 2

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$$\int_0^1 1+e^{-x^2}=\int_0^1 1+\int_0^1 e^{-x^2}=1+\int_0^1 e^{-x^2}$$ Now $$\int e^{-x^2}=\frac{\sqrt\pi\text{erf(x)}}{2}\implies I=1+\frac{\sqrt\pi\text{erf}(1)}2\approx 1.746$$ Where $$\text{erf(x)}=\frac{2}{\sqrt\pi}\int_0^x e^{-t^2} \,dt$$

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Hint:

Use the Gaussian error function

$$erf(x) =\int \limits_{0}^{x}e^{-t^2}dt. $$

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