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anonymous
 3 years ago
Can someone help me with this integral?
anonymous
 3 years ago
Can someone help me with this integral?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{2} \frac{ dx }{ e ^{\pi x} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that e is to the power of \[\pi x\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0any 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 antiderivative?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0don't need to substitute; the integrand is just e^(pi*x) so the antiderivative is \[\frac{ 1 }{ \pi }e^{\pi x}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no that wouldn't work because when you find du it equals 0 = (

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just plug in your limits and your done :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it says the answer is \[\frac{ 1 }{ \pi } (1e ^{2\pi})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you show me how to do it step by step? in laments terms lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0right! 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 \[\frac{ 1 }{ \pi }e ^{2 \pi}\frac{ 1 }{ \pi }e ^{2(0)} = \frac{ 1 }{ \pi } \left( e ^{2 \pi }  1\right)\] \[=\frac{ 1 }{ \pi } \left( 1 e ^{2 \pi } \right) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ur just muliplying thru by 1 as last step to reverse the terms inside the parentheses

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok = ) can you show me step by step how you did the antiderivative. I seem to be having trouble = (

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whenever u hv e to the power of some constant * x then the antiderivative is the same expoential function divided by the constant \[\int\limits_{}^{}e ^{a x}= \frac{ 1 }{ a } e ^{a x } \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is that a rule? like what you learn from the book or class? I might just have skimmed over it in my notes and forgot = )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0definitely 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 antideriv you hv to put in constant in denominator to get back original function. it's all about thinking in reverse :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry one moment trying to make sense of all this lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0http://tutorial.math.lamar.edu/Classes/CalcI/ComputingIndefiniteIntegrals.aspx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you so much this page is helping me out a lot = ) @Tolio

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you're quite welcome :)
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