Question on integration of exponentials (1 Viewer)

karnage

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Just a small question im stuck on (last question for HW then im done! :cool: )

a) Differentiate xe^x

Pretty sure i got this part covered.
d/dx = xe^x + e^x
= e^x (x + 1)

b) Hence find ∫ 2 to 0 xe^x dx

Not sure on this part as i keep on getting the wrong answer (i assumed hence was the key word that part a had something to do with this part). If you maths guru's could please enlighten me on what to do as i am not the best at mathematics. Much appreciated :uhhuh:
 
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pLuvia

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Haven't answered maths questions in ages :p

Since this is 2 unit, just part a) to do part b)

d/dx(xex)=xex+ex
xex=d/dx(xex)-ex
int. {0 to 2}xexdx=[xex]{0 to 2}-int. {0 to 2}exdx

Then you just go from there :)
 

Mark576

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d/dx (xe^x) = xe^x + e^x
∫(xe^x + e^x) dx = ∫(xe^x) dx + ∫e^x dx
=> xe^x = ∫xe^x dx + e^x
=> ∫xe^x dx = [xe^x - e^x]

Remembering this is all considered with the upper and lower bounds omitted for simplification.

Finally just evaluate the expression [xe^x - e^x] from x = 2 to x = 0

EDIT: Damn, not fast enough :D
 
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karnage

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Thanks alot for the quick reply.

Just the remaining working out for the sake of it.
int.{0 to 2} xe^x - e^x dx
= [ xe^x - e^x ] {0 to 2}
= ( 2e^2 - e^2 ) - ( 0 - 1)
= e^2 + 1

Hope that looks right, because its the right answer :)

On a side note, how did you do the superscript for the x.

Hmmm brains. Where can i get one of those? :lol:
 
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pLuvia

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It's just the codes

[sup ] TEXT [/sup ] without the spaces
 

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