# Quod Erat Demonstrandum

## 2009/03/19

### 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

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