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

  • Thread starter stevecallaway
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Also, your limits of integration should be from 100 to infinity, not 0 to infinity. Keep going and you should get the correct answer.In summary, the integral from 100 to infinity for (x-100)(.002e^(-.002*x)) is equal to -(100e^(-.2))(e^(-.2))+100e^(-.2), which simplifies to 14.84. The correct answer should be 409.37. The mistake lies in not integrating the first part correctly. By using the integration by parts method, the correct integral should be 0.2e^(-.002x) from 100 to infinity.
  • #1
stevecallaway
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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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  • #2


You didn't integrate it correctly. Post the details of your calculations so we can see where you went wrong.
 
  • #3


(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).
 
  • #4


The first integral you did is wrong. You need to do it by parts.
 
  • #5


.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...
 
  • #6


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).
 

1. What is the purpose of finding the integral of this function?

The integral of a function represents the total area under the curve of that function. In this case, finding the integral of (x-100)(.002e^(-.002*x)) can help us calculate the cumulative effect of the function over a certain range of values.

2. How do you solve for the integral of (x-100)(.002e^(-.002*x))?

To solve for the integral, you can use integration techniques such as substitution or integration by parts. Alternatively, you can use a calculator or computer software to calculate the integral numerically.

3. What is the domain and range of the function (x-100)(.002e^(-.002*x))?

The domain of the function is all real numbers, while the range depends on the values of x and the coefficient of the exponential term. Generally, the range will be a subset of the real numbers.

4. Can the integral of (x-100)(.002e^(-.002*x)) be used to find the average value of the function?

Yes, the average value of the function can be calculated by dividing the integral by the total range of the function.

5. What is the physical interpretation of the integral of (x-100)(.002e^(-.002*x))?

The integral represents the total accumulated effect of the function over a certain range. In a physical context, this could represent the total amount of a substance or the total energy over a period of time.

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