Urgent Integration Q Help! (1 Viewer)

香港!

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Swalbs said:
Guys and gals, how do you integrate xe^x?

Looks easy i know, but im stumped..
I=int. xe^x dx
let u=x and dv\dx=e^x
du=dx and v=e^x
.: I=xe^x-e^x+C using 'Integration by Parts'
 
I

icycloud

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Integration by parts is part of the 4U syllabus. Integrals of this kind would not be examined in the 3U exam. However, you might be given a lead-in,

e.g. Differentiate xln(x). Hence or otherwise, integrate xe^x. :)
In which case,

d/dx [xln(x)] = ln(x) + 1
∫ln(x) + 1 dx = xln(x) + C
∫ln(x) dx = xln(x) - x + C

Now, I = ∫xe^x dx
Let u = e^x, du = e^x dx

I = ∫x du
But x = ln(u)
Thus, I = ∫ln(u) du
= uln(u) - u + C (from previous part)
= e^x*ln(e^x) - e^x + C
= xe^x - e^x + C #
 
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Swalbs

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Thx guys, glad this is 4unit only, although i found the Q in a 2unit textbook!

But it was a simpsons rule question, not pure integration, so that wud explain it.



Cheers
 

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