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Integration of exponential function

paulmdrdo

Active member
May 13, 2013
386
just want to confirm if i did set up my integral correctly and got a correct answer.


$\displaystyle\int_0^a (e^{\frac{x}{a}}-e^{-\frac{x}{a}})$

using substitution for the first term in my integrand

$\displaystyle u=\frac{x}{a}$ $\displaystyle du=\frac{1}{a}dx$; $\displaystyle dx=adu
$

for the second term of my integrand,

$\displaystyle v=-\frac{x}{a}$; $\displaystyle dv=-\frac{1}{a}dx$; $\displaystyle dx=adv$

my integrand will now be,

$\displaystyle a\int_0^a e^udu+a\int_0^a e^{-v}dv$

$\displaystyle ae^{\frac{x}{a}}+ae^{-\frac{x}{a}}|_0^a$

plugging in limits of integration,

$\displaystyle ae^{\frac{a}{a}}+ae^{-\frac{a}{a}}-(ae^{\frac{0}{a}}+ae^{-\frac{0}{a}})$

simplifying we have,

$\displaystyle ae^1+ae^{-1}-ae^0-ae^{0}=ae+ae^{-1}-2a=a(e+\frac{1}{e}-2)$

please kindly check if i have any errors. thanks!

 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: integration of exponential function

Correct :) .