Expert: Scotto - 2/4/2009

Question
QUESTION: Proove that the derivative of xe^x= (x+1)e^x by integrating (x+1)e^x. (e^x is an exponential function)

ANSWER: There are two ways you could go and both work.  
Both involve a u-v problem.

[1] Let u=x+1 and dv = e^x dx.  From here, we can see that
du = dx and v = e^x.

So, from this, we know that
⌠
⌡(x+1)e^x dx = uv -
⌠
⌡v du, which I will put on the next line so the
⌠
⌡ (which is two characters) lines up.

= (x+1)e^x - e^x since we know
⌠
⌡e^x dx = e^x + C.

Now if we take (x+1)e^x - e^x and multiply it out, we get
xe^x + e^x - e^x = xe^x, which is the function we were suppose
to get.

[2] Multiply out (x+1)e^x into xe^x + e^x.
Do xe^x by u-v, like I just did, and you get xe^x - e^x.
Note that we have an e^x added on at the end after this
integral was done, so we get xe^x.


---------- FOLLOW-UP ----------

QUESTION: Can you explain to me the u-v method? Ive learned of integrals by
subsitution (the u part) but I don't know what v will equal and why we
subtract it? The question was the integration of (x+1)e^x in case you can't
see it or something.

Answer
⌠
⌡ dx will be repaced by I()dx

I(v du) = uv - I(u dv).

Adding the I(u dv) to both sides gives I(u dv)dx + I(v du)dx = uv.

Taking the derivatve of both sides gives u v' + v u' = (uv)',
which we know is the product rule.

You might want a few examples:

(1) I(xe^x)dx, so u=x, dv=e^xdx.  
This makes du=dx and v=e^x, so we have
I(xe^x)
= uv - I(v du)dx
= xe^x - I(e^x dx)
= xe^x - e^x + C.

(2) I(ln(x))dx, so u=ln(x), dv=dx.  
This makes du=dx/x and v=x,
so I(xln(x))dx
= uv - I(v du)
= xln(x) - I(x/x)dx
= xln(x) - I(1)dx
= xln(x) - x.