OpenStudy (anonymous):

Can someone help me with this integral?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

that e is to the power of $\pi x$

6 years ago
OpenStudy (anonymous):

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?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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 $\frac{ -1 }{ \pi }e ^{-2 \pi}-\frac{ -1 }{ \pi }e ^{-2(0)} = \frac{ -1 }{ \pi } \left( e ^{-2 \pi } - 1\right)$ $=\frac{ 1 }{ \pi } \left( 1 -e ^{-2 \pi } \right)$

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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 = )

6 years ago
OpenStudy (anonymous):

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 :)

6 years ago
OpenStudy (anonymous):

sorry one moment trying to make sense of all this lol

6 years ago
OpenStudy (anonymous):

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?

6 years ago
OpenStudy (anonymous):

@Tolio

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous): 6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

you're quite welcome :)

6 years ago