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

發表迴響 »

仍無迴響。

RSS feed for comments on this post. TrackBack URI

發表迴響

在下方填入你的資料或按右方圖示以社群網站登入:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / 變更 )

Twitter picture

You are commenting using your Twitter account. Log Out / 變更 )

Facebook照片

You are commenting using your Facebook account. Log Out / 變更 )

Google+ photo

You are commenting using your Google+ account. Log Out / 變更 )

連結到 %s

在 WordPress.com 建立免費網站或網誌.

%d 位部落客按了讚: