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