The problem involves finding the integral of the function (x-100)(0.002e^(-0.002x)) from 100 to infinity. The context is within the subject area of calculus, specifically focusing on integration techniques.
The discussion is ongoing, with participants providing feedback on each other's calculations and suggesting corrections. Some participants express frustration over the complexity of the problem and the results obtained, indicating that multiple interpretations of the integration steps are being explored.
There is a noted discrepancy between the calculated result and the expected answer, prompting participants to question the setup and execution of the integration process. The discussion reflects a collaborative effort to identify and rectify misunderstandings in the calculations.
That's wrong. Fix this and the value for the first integral should come out okay.stevecallaway said:0.002 x e^(-0.002x) - 0.2 e^(-0.002x)
u = 0.002x
du = dx
This is only the value of the first integral. You still need to do the integral of 0.2 e^(-0.002x).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...