Question

∫3x(-e^(-x)) dx

Hello, trying to solve this multiplication of integrals and get a wrong answer. Hopefully I can express the ? Lucidly enough that I can understand my mistake.


f(x) 3x f'(x) = 3 g(x) and g'(x) = (-e^(-x))

When I multiply f(x) * g(x) it equals 3x(-e^-x) whereas the text and wolfram give the equation as 3x (e^(-x))

Which is then subtracted from g(x) * f'(x) to solve the integral.

Why do I get 3x (-e^-x) for f(x) * g(x) rather than the correct 3x (e^(-x)) ?

Answer 1

trying to use integration by parts?

Answer 2

well 3∫e^(-x)xdx
u=x dv=e^(-x)
du= 1 v= -e^(-x)

should be straight forward now.

Answer 3

So u * v = x * (e^(-x)) not -e^(-x) * x like I figure

Answer 4

well im about to leave so i will leave you the solution for reference :)
\[\int 3x(-e^{-x})dx => \int -3xe^{-x}dx => -3\int xe^{-x}dx\]
u=x => du=1
dv= e^{-x} => v= -e^(x)
\[=>3e^{-x}-3\int e^{-x}dx => 3e^{-x}x+3e^{-x} + C\]

Answer 5

Ok thanks I will check out your answer.

Answer 6

use the table integral to integrate :
∫3x(-e^(-x)) dx =