Find the average value of f:[1,2] - (-infinity, +infinity) where f(x) = xe^(-210x) for all x is an element of [1,2]

So far I have:

$$1/(2-1)\int\limits_{1}^{2}xe^{-210x}dx = x(e^{-210x}/-210) - \int\limits_{1}^{2} e^{-210x}dx$$

Am I using the correct method to solve this?

Question

Find the average value of f:[1,2] -> (-infinity, +infinity) where f(x) = xe^(-210x) for all x is an element of [1,2]

So far I have:

$$1/(2-1)\int\limits_{1}^{2}xe^{-210x}dx = x(e^{-210x}/-210) - \int\limits_{1}^{2} e^{-210x}dx$$

Am I using the correct method to solve this?

australopithecus · Answer

I have simplified this integral down to

$$(-2(210)e^{-420} + e^{-420} + 210e^{-210}  - e^{-210} )/210^{2}$$

anonymous · Answer

you will have the answer to your question here http://tutorial.math.lamar.edu/Classes/CalcI/AvgFcnValue.aspx

australopithecus · Answer

so what I'm doing is correct?

anonymous · Answer

looks good to me...

anonymous · Answer

and ill try to do the calculations 2 just to make sure , yes ur method is right

australopithecus · Answer

so much simplification blah lol

anonymous · Answer

if that simplification is correct that is.... :)

australopithecus · Answer

but yeah thanks now I need to get cracking on simplifying this ugly mess of an answer so i can plug it into my calculator

anonymous · Answer

ask mr wolfram... is your simplification is good or not

anonymous · Answer

**if your simpli....

australopithecus · Answer

for sure thanks for help dpanic :)

australopithecus · Answer

wish I could givem ore than one medal

anonymous · Answer

yw...

anonymous · Answer

5 mins earlier i was certain that 39-37 dosent equal 2 , so i dont think i am qualified enough to do this 4 u..maybe 2moro

australopithecus · Answer

I checked it my answer was incorrect I forgot a minus, the answer is extremely tiny https://www.wolframalpha.com/input/?i=integrate+from+1%2C2+f%28x%29+%3D+xe^%28-210x%29dx I simplified it down but yeah no ordinary scientific calc can handle this

australopithecus · Answer

My prof is evil though so yeah