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bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{0}^{2} \frac{ dx }{ e ^{\pi x} }\]
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
that e is to the power of \[\pi x\]
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.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 antiderivative?
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.2
don't need to substitute; the integrand is just e^(pi*x) so the antiderivative is \[\frac{ 1 }{ \pi }e^{\pi x}\]
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
no that wouldn't work because when you find du it equals 0 = (
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.2
just plug in your limits and your done :)
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
it says the answer is \[\frac{ 1 }{ \pi } (1e ^{2\pi})\]
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
can you show me how to do it step by step? in laments terms lol
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.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 \[\frac{ 1 }{ \pi }e ^{2 \pi}\frac{ 1 }{ \pi }e ^{2(0)} = \frac{ 1 }{ \pi } \left( e ^{2 \pi }  1\right)\] \[=\frac{ 1 }{ \pi } \left( 1 e ^{2 \pi } \right) \]
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.2
ur just muliplying thru by 1 as last step to reverse the terms inside the parentheses
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
oh ok = ) can you show me step by step how you did the antiderivative. I seem to be having trouble = (
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.2
whenever u hv e to the power of some constant * x then the antiderivative is the same expoential function divided by the constant \[\int\limits_{}^{}e ^{a x}= \frac{ 1 }{ a } e ^{a x } \]
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.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 = )
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.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 antideriv you hv to put in constant in denominator to get back original function. it's all about thinking in reverse :)
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
sorry one moment trying to make sense of all this lol
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.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?
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
@Tolio
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.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
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.2
http://tutorial.math.lamar.edu/Classes/CalcI/ComputingIndefiniteIntegrals.aspx
 one year ago

bettyboop8904 Group TitleBest ResponseYou've already chosen the best response.0
thank you so much this page is helping me out a lot = ) @Tolio
 one year ago

Tolio Group TitleBest ResponseYou've already chosen the best response.2
you're quite welcome :)
 one year ago
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