about improper integrals?

Question

integrate xe^(-x)dx ,(0,positve infinity)?.......i can integrate this shall i use the integration by when i set the u is equal to -x the differential is -dx,theres a dx but how can i make the du equal to xdx ?

Answer 1

Here we use integration by parts
∫uv' = uv - ∫v'u
Set u=x, v=-e^(-x)
∫xe^(-x) dx = -xe^(-x) + ∫e^(-x) dx
∫xe^(-x) dx = -xe^(-x) - e^(-x)
Now evaluating from 0 to ∞:
[x→∞]lim (-x/(e^x) - e^(-x)) - (0*e^0 - e^0)
Evaluate the right hand side, and use L'hopital's rule to simplify the indeterminate form on the left:
[x→∞]lim (-1/(e^x)) +1
1

Thus [0, ∞]∫xe^(-x) dx = 1

Answer 2

it is not integration by substitution,
it will not work that way
you need to do integration by parts
so u=x
du=dx
and then dv=e^{-x}
so v=-e^{-x}
then the integral
is :
uv- integral of (vdu)
-xe^{-x} - integral -e^{-x} dx
=-xe^{-x} -e^{-x}
now you can evaluate and then take the limit

say from 0 to d ( and then take the limit when d--> infinity)
so you get:
-de^{-d}-e^{-d} + 0e^{-0} +e^{-0}
when d--> infinity the first two terms --->0,
so the only remaing term is
e^{-0}=1

Answer 3

Use integration by parts.
Let u = x and dv = e^(-x)dx, then
du = dx and v = -e^(-x)dx
You can take it from there. Remember the formula is
integral of u dv = uv - integral of v du

Answer 4

To integrate this, you cannot do a u substitution. You must use integration by parts. If you have learned that, try it. If not, it is too hard to explain so don't worry about it.-unless you really want me to.
By the way the integral is -e^(-x)(x+1)

Answer 5

set e**(-x)=y
take ln of borh sides

e**(-x)dx=-du

integral becomes

int( (-lnu du)=- ( ulnu-u)+c

lnu=-x

at x=0;u=1

at x=infinity;u=0

0*ln0+0+1ln1-1=(ln1)-1


= -1