View Full Version : Integration challenge
galactus
Mar 1, 2008, 05:35 AM
Here's another problem if anyone is interested.
Can you perform the integration? I am curious to see if anyone has a cool method.
Anyone who knows integration knows that e^{\frac{x^{2}}{2}} is not easily integrated.
\int{(x^{2}+1)e^{\frac{x^{2}}{2}}}dx
galactus
Mar 10, 2008, 04:41 PM
It appears no one was interested in trying the problem. Not that anyone cares, but I will post my solution.
Parts or substitution do not work too well. Here is a rather obscure method that works well
when you have something of the form p(x)e^{x}
This is known as integration by recognition.
\frac{d}{dx}[p(x)e^{\frac{x^{2}}{2}}]=(x^{2}+1)e^{\frac{x^{2}}{2}}
Let p(x)=ax^{2}+bx+c
Then \frac{d}{dx}(ax^{2}+bx+x)e^{\frac{x^{2}}{2}}
By the product rule:
(2ax+b)(e^{\frac{x^{2}}{2}})-(ax^{2}+bx+c)xe^{\frac{x^{2}}{2}}=
(x^{2}+1)e^{\frac{x^{2}}{2}}
Factor, group like terms and equate coefficients, we get:
a=0, \;\ b=1, \;\ 2a+c=0
\frac{d}{dx}(xe^{\frac{x^{2}}{2}})=(x^{2}+1)\cdot{ e^{\frac{x^{2}}{2}}}
Now integrate both sides and we get:
\fbox{xe^{\frac{x^{2}}{2}}=\int{(x^{2}+1)e^{\frac{ x^{2}}{2}}dx}
Capuchin
Mar 10, 2008, 05:24 PM
Hi gal, just wanted to say that I enjoy these, don't get disheartened that nobody attempted it.
Also, clean out your inbox, I might need to message you in the future :)
galactus
Mar 10, 2008, 05:32 PM
It's cleaned. It wasn't full. About half.
Capuchin
Mar 10, 2008, 05:43 PM
Last time I wanted to send you a message it was full! :o