# find the average value of the function - with an e

• Oct 4th 2009, 12:05 PM
calcbeg
find the average value of the function - with an e
Hi

The goal is to find the average value of the function h (t) = 1/(e^2t) over the interval (0,1)

So havg (1-0) = S a=0; b=1 e^-2t dt

So I am assuming I should do some substitution

let u = e^-2t then du =-2e^-2t dt so du/-2 = e^-2t dt

Then havg = S a=0;b=1 du/-2 = -1/2 S du = -1/2 (u) a=0;b=1

So then -1/2 (e^-2t) a=0;b=1

= -1/2 (e^-2 - e^0)

= -1/(2e) + 1

Did I do this correctly???

Thanks

Calculus beginner - who no longer likes the letter e
• Oct 4th 2009, 03:27 PM
redsoxfan325
Quote:

Originally Posted by calcbeg
Hi

The goal is to find the average value of the function h (t) = 1/(e^2t) over the interval (0,1)

So havg (1-0) = S a=0; b=1 e^-2t dt

So I am assuming I should do some substitution

let u = e^-2t then du =-2e^-2t dt so du/-2 = e^-2t dt

Then havg = S a=0;b=1 du/-2 = -1/2 S du = -1/2 (u) a=0;b=1

So then -1/2 (e^-2t) a=0;b=1

= -1/2 (e^-2 - e^0) So far, so good

= -1/(2e) + 1 you simplified incorrectly

Did I do this correctly???

Thanks

Calculus beginner - who no longer likes the letter e

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

And $\displaystyle e$ is great: $\displaystyle \frac{d}{dx}[e^x]=\int e^x\,dx=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x$

Learn to love it, because it never goes away.