Is the Integration of xex Correctly Solved?

[ƒ(x)]

Please forgive any formatting errors.

\int x^exdx = ?

Is this correct?

\int x^exdx

= \int e^ln(x)exdx
= \int e^exln(x)dx

u = e^x
x = ln(u)
du = e^xdx
du/u = dx

= ∫ ln(u)^u*(1/u)*du

a = ln(u)^u
EDIT: da = ln(u)^u-1du
dv = 1/u du
v = ln(u)

= ln(u)^u+1 - ∫ [ln(u)]^udu

u = e^x

= ln(e^x)^ex+1 - ∫ [ln(e^x)]^exdx

∫x^exdx = x^ex+1-∫x^exdx

2∫x^exdx = x^ex+1
∫x^exdx = x^ex+1*(1/2)+C

 

[gabbagabbahey]

a = ln(u)^u
da = ln(u)^u-1

eeks, that's not really how you differentiate that function is it?

∫x^exdx = x^ex+1*(1/2)+C

Have you tried differentiating this result?

More to the point, does this integral actually exist/converge for all values of x?...Surely you are given integration limits?

 

[ƒ(x)]

eeks, that's not really how you differentiate that function is it?

Well...I skipped some steps.

a = ln(u)^u
a = 1/u (chain rule) * u (bring down the exponent) * ln(u)^u-1 (subtract from the exponent)
a = ln(u)^(u-1) (simplify)

 

[Unknot]

You can't do that when the exponent is the variable.

 

[ƒ(x)]

You can't do that when the exponent is the variable.

Oh...Hm...what would you do then?

 

[Unknot]

make the function e^(something). Then you can use chain rule.

 

[gabbagabbahey]

Again, there is an even bigger issue here than you incorrectly differentiating 'a':

More to the point, does this integral actually exist/converge for all values of x?...Surely you are given integration limits?

To illustrate the problem, I've plotted the integrand for x=[0,3]:

http://img16.imageshack.us/img16/2144/hmmmp.th.jpg

Does it look as though the area under the curve is bounded?

 

[ƒ(x)]

You can't do that when the exponent is the variable.

Wait...what's wrong with the way I did it? I'm differentiating with respect to u.

 

[Unknot]

(d/dx)3^x is not x3^(x-1).

 

[ƒ(x)]

what is the integral of x^(e^x) then?

 

[Unknot]

I really doubt you can integrate it indefinitely.

 

[gabbagabbahey]

what is the integral of x^(e^x) then?

It is unreasonable to expect an antiderivative to exist, when the area under the curve of the integrand becomes unbounded at around x=5!...That means for almost half of the real number line, the integral does not exist.

There is no antiderivative for this function, but if you are given specific integration limits, and the integral is bounded on that interval, you may still be able to evaluate analytically by using a trick or two, and if that fails numerical integration is always available.

So, I ask for the 3rd time; Are you given limits of integration?

 

[gabbagabbahey]

Wait...what's wrong with the way I did it? I'm differentiating with respect to u.

\frac{d}{dx} f(x)^{g(x)}= \frac{d}{dx} e^{\ln(f(x)^{g(x)})}=\frac{d}{dx} e^{g(x)\ln(f(x))}=e^{g(x)\ln(f(x))}\left(g'(x)\ln(f(x))+\frac{g(x)}{f(x)}f'(x)\right)

=f(x)^{g(x)}\left(g'(x)\ln(f(x))+\frac{g(x)}{f(x)}f'(x)\right)\neq g'(x)f(x)^{g(x)-1}