Arc Length of e^5x from 0 to ln(4)

[Oblakastouf]

Homework Statement

http://i47.tinypic.com/1z6naa.jpg

Note... I used wolfram alpha to get the answer, I did not get it myself... So I still need help. The answer shown is correct, so you'll know if you got it.

Homework Equations

Integral [0, ln(4)] sqrt(1+(dy/dx)^2)

The Attempt at a Solution

2pi Integral [0, ln(4)] y*sqrt(1+(dy/dx)^2)

2pi Integral [0, ln(4)] (e^5x)*sqrt(1+5e^5x^2)dx

u = 5e^5x
du = 25e^5x dx
dx = du/25e^5x

2pi Integral [0, ln(4)] (e^5x)*sqrt(1+u^2)du/25e^5x

2pi Integral [0, ln(4)] sqrt(1+u^2)du

u = tan(t)

2pi/25 Integral [0, ln(4)] sqrt(1+tan^2(t))du

2pi/25 Integral [0, ln(4)] sqrt(sec^2(t))du

2pi/25 Integral [0, ln(4)] sec(t)du

du = sec^2(t)dt
dt = du*cos^2(t)

2pi/25 Integral [0, ln(4)] cos^2(t)/cos(t)dt

2pi/25 Integral [0, ln(4)] cos(t)dt

Edit bounds...

[arctan(5), arctan(5e^(5*ln(4)))]

Then get ****ed over with an answer of .0048...

What did I do wrong.

 

[Roni1985]

the question asks the surface area and not the arc length. Perhaps this is your mistake

 

[mmmboh]

I have never heard of wolfram alpha before...very interesting. What did you type into get the answer to your question? I can't figure it out...