Quod Erat Demonstrandum


Reply to a F.7 student

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 11:42 上午

To estimate \int_0^1e^{-x^2}dx, please refer to the normal table, you may see

A(z) = \int_0^{z}\frac{1}{\sqrt{2\pi}}e^{\frac{-u^2}{2}}du

Now, to evaluate \int_0^1e^{-x^2}dx, what substitution need to make?

Just look at the power of e, ask yourself, what substitution we need so as to convert x^2 into \frac{u^2}{2}?

Intentionally, we set

x^2 = \frac{u^2}{2}, or take x = \frac{u}{\sqrt{2}}, then

= \int_0^{\sqrt{2}}e^{\frac{-u^2}{2}}(\frac{1}{\sqrt{2}})du
= (\frac{\sqrt{2\pi}}{\sqrt{2}})\int_0^{\sqrt{2}}\frac{1}{\sqrt{2\pi}}e^{\frac{-u^2}{2}}du
= \sqrt{\pi}A(\sqrt{2})
= \sqrt{\pi} \times 0.4213
= 0.7467

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