Easy Integration Question (1 Viewer)

[.ben]

Question:
Evaluate
2∏∫2 to 0 (xe^-x2)dx

thanks

 

[pLuvia]

2π∫2 to 0 (xe^-x2)dx
Let u=-x²
du=-2xdx

2π(-1/2)∫{0->2}-2xe^udx
=-π∫{0->2}e^udu
=-π[e^-x2]{0->2}
=-π[1-e^-4]

 

[.ben]

Oh bugger, i screwed the bounds. thanks pluvia

 

[Raginsheep]

I think the integration bit is do-able by "inspection".

 

[.ben]

not for me, i'm dumb lol

 

[Riviet]

not for me, i'm dumb lol

With some practice and patience, you should be able to identify which method is best for a given integral.