What is the arc length of the function 1/2(e^x+e^-x) on the interval [0,2]?

[fd25t6]

Hello everyone, I am having a bit of an issue evaluating the arc length of the function 1/2(e^x+e^-x) interval [0,2]. We were instructed to solve using the arc length formula square root of 1+(dy/dx)^2. The solution should be 3.627 according to my CAS. However my calculations are yielding 3.511.


arc length of the function 1/2(e^x+e^-x) interval [0,2]

Homework Equations

square root of 1+(dy/dx)^2

The Attempt at a Solution

F(x)= 1/2(e^x+e^-x)
F'(X)= 1/2(e^x-e^-x)
F'(x)^2= (e^2x - 2 - e^-2x)/4 ==> ((e^2x)/4) - (1/2) - (1/4e^2x)

So I try to evaluate the integral from the interval [0,2] of the square root of 1+[(e^2x)/4 - (1/2) - (1/4e^2x)] and end up with 3.511.

any help is greatly appreciated. Thank you in advance

 

[haruspex]

F'(X)= 1/2(e^x-e^-x)
F'(x)^2= (e^2x - 2 - e^-2x)/4 ==> ((e^2x)/4) - (1/2) - (1/4e^2x)

Check the signs.

 

[fd25t6]

such a silly mistake.. it has been a long day. Thank you much