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seen Jun 24 '12 at 23:48

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answered Calculating triple integral $z^2$ over an ellipsoid
Jun
2
comment Inverse Laplace transform computation
That $s$ at the front suggests something by itself.
Feb
15
revised $f(x^2) = xf(x)$ implies that $ f(x) = mx$?
modified title
Feb
15
comment $f(x^2) = xf(x)$ implies that $ f(x) = mx$?
Of course. Thank you.
Feb
15
accepted $f(x^2) = xf(x)$ implies that $ f(x) = mx$?
Feb
15
asked $f(x^2) = xf(x)$ implies that $ f(x) = mx$?
Feb
2
comment Integrating: $\int \limits_{0}^{1} (1+xe^x)e^{e^x}dx$ (wolfram alpha doesn't provide the steps for this one)
That's why you practice, to help build up your tool kit.
Feb
2
comment Integrating: $\int \limits_{0}^{1} (1+xe^x)e^{e^x}dx$ (wolfram alpha doesn't provide the steps for this one)
I didn't know before I started the procedure it would end so neatly. But now that it had, we can stop. If it hadn't, we would have had to have kept on looking for a method that works.
Feb
2
comment Integrating: $\int \limits_{0}^{1} (1+xe^x)e^{e^x}dx$ (wolfram alpha doesn't provide the steps for this one)
Obviously I know that. What I'm showing you is how the solution suggested itself to me. As I looked at the integral originally, I didn't recognize the answer. But what I did recognize is that $e^xe^{e^x}$ was the derivative of $e^{e^x}$. Hence I then thought: why not try and integrate it? Now in the original integral, it's attached to $x$. So what happens if I integrate the entire term $$x.e^xe^{e^x}$$ Having done that, I see what I'm left with can be rearranged into the original integral, and then we have the answer.
Feb
2
comment Integrating: $\int \limits_{0}^{1} (1+xe^x)e^{e^x}dx$ (wolfram alpha doesn't provide the steps for this one)
In the integral I evaluated by parts, we have $u = x$ and $dv = e^x.e^{e^x} dx$
Feb
2
comment Integrating: $\int \limits_{0}^{1} (1+xe^x)e^{e^x}dx$ (wolfram alpha doesn't provide the steps for this one)
If nothing else, now you've seen this method you'll be able to cope with integrals such as $$\int (\sec x . \tan x + 1)e^{\sin x} dx $$
Feb
2
comment Integrating: $\int \limits_{0}^{1} (1+xe^x)e^{e^x}dx$ (wolfram alpha doesn't provide the steps for this one)
I'm sorry you don't like it. It's a very natural thing to notice because the derivative of $e^{f(x)}$ is $f'(x)e^{f(x)}$.