Question

Let calculate the values of integral. For α>0.

Answer 1

Let calculate the values of integral. For \(\alpha > 0\)

\[\int\limits_{0}^{\infty} -x e^{- \alpha x} dx\]

Answer 2

use product rule : (uv)'=u'v+uv'
let u=-x and dv=e^{-ax}dx
make this substitution and use the above product rule

Answer 3

is this for Laplace transform? :)

Answer 4

how about this
\[-\frac{ 1 }{ \alpha^2 } \left( - \alpha x - 1 \right) e^{- \alpha x} |_{0}^{\infty}\]

Answer 5

no isn't :)

Answer 6

@myko

Answer 7

hbu @RadEn ??

Answer 8

use the table integral, the result of that integration is