# Integral of (x-100)(.002e^(-.002*x))

## Homework Statement

find the integral from 100 to infiniti for (x-100)(.002e^(-.002*x))

## The Attempt at a Solution

.(xe^-(.002x))(e^(-.002x))-(-100e^(-.002x)) |from 100 to infiniti = -(100e^-.2)(e^-.2)+100e^(-.2)=14.84

vela
Staff Emeritus
Homework Helper

You didn't integrate it correctly. Post the details of your calculations so we can see where you went wrong.

(x-100)(.002e^(-.002*x))
I first multiplied out these two and got .002xe^(-.002x) - .2e^(-.002x). Then integrating the .002xe^(-.002x) I integrated the e part first and then the x part to get -xe^(-.002x) * e^(-.002x) and then I integrated the .2e^(-.002x) and got -100e^(-.002x).

vela
Staff Emeritus
Homework Helper

The first integral you did is wrong. You need to do it by parts.

.002xe^(-.002x) - .2e^(-.002x). Ok. u=.002x du=dx
v= -(1/.002)e^(-.002x) dv=e^(-.002x)
so then, (.oo2x)(-(1/.002)e^(-.002x)- Integral of -(1/.002)e^(-.002x) dx from 100 to infiniti. (.oo2x)(-(1/.002)e^(-.002x) + (1/(.002*.002))e^(-.002x) from 100 to infiniti
0 - .2*(-409.3654)+(204682.69)==way wrong answer...Oh my goodness this is the ultimate in frustration...

vela
Staff Emeritus
Homework Helper

Reformatted a bit to make it easier to read.
0.002 x e^(-0.002x) - 0.2 e^(-0.002x)

u = 0.002x
du = dx
That's wrong. Fix this and the value for the first integral should come out okay.
dv = e^(-0.002x) dx
v = -(1/0.002) e^(-0.002x)

so then,

(0.002x)(-(1/0.002) e^(-0.002x) - Integral of -(1/0.002)e^(-0.002x) dx from 100 to infinity.
(0.002x)(-(1/0.002) e^(-0.002x) + (1/(0.002*0.002)) e^(-0.002x) from 100 to infinity
[0 - 0.2*(-409.3654)] + (204682.69) == way wrong answer

Oh my goodness this is the ultimate in frustration...
This is only the value of the first integral. You still need to do the integral of 0.2 e^(-0.002x).