Final Answer:
The exact length of the curve x = e^y + 1/4 e^(-y) for 0 ≤ y ≤ 2 is 3.25(e^2 - 1).
Step-by-step explanation:
To find the exact length of the curve, we can use the arc length formula:
L = ∫√(dx/dy)^2 + (dy/dx)^2 dy
where x = e^y + 1/4 e^(-y) and y is the variable of integration.
First, we need to find dx/dy and dy/dx:
dx/dy = e^y - 1/4 e^(-y)
dy/dx = (e^y - 1/4 e^(-y))^-1
Now, we can substitute dx/dy and dy/dx into the arc length formula and evaluate the integral:
L = ∫√(e^y - 1/4 e^(-y))^2 + ((e^y - 1/4 e^(-y))^-1)^2 dy
Simplifying the expression under the radical, we get:
L = ∫√(e^(2y) - 1/2 + 1/16e^(-2y)) dy
Using the trigonometric substitution u = e^y, du = e^y dy, we can evaluate the integral:
L = ∫√(u^2 - 1/2 + 1/16u^(-2)) du
Completing the square in the radical, we get:
L = ∫√((u - 1/4)^2 + 3/16) du
Now, we can use the inverse sine function to evaluate the integral:
L = arcsin((u - 1/4)/√(3/16)) | from y=0 to y=2
Substituting u = e^y and simplifying, we get:
L = arcsin((e^y - 1/4)/√(3/16)) | from y=0 to y=2
Evaluating the integral, we get:
L = 3.25(e^2 - 1)
Therefore, the exact length of the curve is 3.25(e^2 - 1).