Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

bettyboop8904

  • 3 years ago

Can someone help me with this integral?

  • This Question is Closed
  1. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\int\limits_{0}^{2} \frac{ dx }{ e ^{\pi x} }\]

  2. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that e is to the power of \[\pi x\]

  3. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    any idea? I think i have to separate the fraction so that 1 is in the numerator and then maybe use "u" substitution? to be able to take the anti-derivative?

  4. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    don't need to substitute; the integrand is just e^(-pi*x) so the antiderivative is \[\frac{ -1 }{ \pi }e^{-\pi x}\]

  5. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no that wouldn't work because when you find du it equals 0 = (

  6. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    just plug in your limits and your done :)

  7. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it says the answer is \[\frac{ 1 }{ \pi } (1-e ^{-2\pi})\]

  8. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you show me how to do it step by step? in laments terms lol

  9. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    right! take the expression i gave for the antideriv above and evaluate it at 2 and 0 , ie. plug in 2 then plug in 0 and take the difference \[\frac{ -1 }{ \pi }e ^{-2 \pi}-\frac{ -1 }{ \pi }e ^{-2(0)} = \frac{ -1 }{ \pi } \left( e ^{-2 \pi } - 1\right)\] \[=\frac{ 1 }{ \pi } \left( 1 -e ^{-2 \pi } \right) \]

  10. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ur just muliplying thru by -1 as last step to reverse the terms inside the parentheses

  11. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok = ) can you show me step by step how you did the antiderivative. I seem to be having trouble = (

  12. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    whenever u hv e to the power of some constant * x then the anti-derivative is the same expoential function divided by the constant \[\int\limits_{}^{}e ^{a x}= \frac{ 1 }{ a } e ^{a x } \]

  13. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is that a rule? like what you learn from the book or class? I might just have skimmed over it in my notes and forgot = )

  14. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    definitely one of the standard integration rules but not really worth memorizing as u can see it makes intuitive sense: Dx(e^x) = e^x and Dx(e^ax) = ae^ax so when taking anti-deriv you hv to put in constant in denominator to get back original function. it's all about thinking in reverse :)

  15. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry one moment trying to make sense of all this lol

  16. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so i guess where I'm confused is that i thought the antiderivative formula was \[\frac{ n ^{x+1} }{ x+1 }\] Why is it so different with e and ln from other functions?

  17. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Tolio

  18. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    it is confusing; in your defn, n is a variable and x is a constant, x^2 or x^4, etc. in exponential fncs the n would be the constant and x the variable which requires different treatment the following whole webpage is great to explain the derivation but just to understand the difference a little more concretely scroll down to the very bottom of the page :) http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx

  19. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    http://tutorial.math.lamar.edu/Classes/CalcI/ComputingIndefiniteIntegrals.aspx

  20. bettyboop8904
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank you so much this page is helping me out a lot = ) @Tolio

  21. Tolio
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you're quite welcome :)

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy