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

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The discussion revolves around finding the integral of the function (x-100)(0.002e^(-0.002x)) from 100 to infinity. The initial calculations provided by the user led to an incorrect answer of 14.84, while the expected result is 409.37. Participants pointed out that the integration was not performed correctly and emphasized the need to use integration by parts. The user attempted to integrate the function but struggled with the steps, leading to confusion and frustration over the calculations. Clarifications were made regarding the proper setup for integration by parts, indicating that further corrections are necessary to arrive at the correct answer.
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Homework Statement

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



Homework Equations





The Attempt at a Solution

.(xe^-(.002x))(e^(-.002x))-(-100e^(-.002x)) |from 100 to infiniti = -(100e^-.2)(e^-.2)+100e^(-.2)=14.84
the answer should be 409.37. What am I missing, please?
 
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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).
 


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...
 


Reformatted a bit to make it easier to read.
stevecallaway said:
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).
 
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