find the exact arc length of the curve: y=0.5(e^x + e^-x) for 0 is less than or equal to x and x is less than or equals to ln2.
To find the exact arc length of a curve, we need to use the arc length formula. The arc length formula for a curve given by the equation y = f(x) over the interval [a, b] is:
L = ∫[a, b] √(1 + (f'(x))^2) dx
Where f'(x) represents the derivative of f(x) with respect to x.
In this case, we have the equation y = 0.5(e^x + e^(-x)) over the interval 0 ≤ x ≤ ln(2).
Step 1: Find the derivative of f(x)
To calculate f'(x), we need to find the derivative of y = 0.5(e^x + e^(-x)) with respect to x.
f'(x) = 0.5(e^x - e^(-x))
Step 2: Evaluate the integral
Using the arc length formula, we can calculate the arc length as:
L = ∫[0, ln(2)] √(1 + (0.5(e^x - e^(-x)))^2) dx
Simplifying the expression under the square root:
L = ∫[0, ln(2)] √(1 + 0.25(e^(2x) - 2 + e^(-2x))) dx
L = ∫[0, ln(2)] √(1 + 0.25(e^(2x) + e^(-2x) - 2)) dx
L = ∫[0, ln(2)] √(0.25e^(2x) + 0.25e^(-2x) + 0.5) dx
Step 3: Integrate
Integrate the expression using the appropriate integration techniques. In this case, you can use substitution. Let's substitute u = e^x:
du = e^x dx
dx = du/u
The limits of integration also change. When x = 0, u = e^0 = 1, and when x = ln(2), u = e^(ln(2)) = 2.
Now we can rewrite the integral using u as the variable:
L = ∫[1, 2] √(0.25u^2 + 0.25/u^2 + 0.5) (du/u)
L = ∫[1, 2] (1/u)√(0.25u^4 + 0.25 + 0.5u^2) du
The integral can be solved using appropriate techniques such as partial fractions or integration by parts. However, this process might involve complex calculations.
Unfortunately, it's not feasible to provide the exact value of the arc length of the curve without performing the integral. You can use numerical methods or integral approximation techniques, such as numerical integration or software like Mathematica, to calculate the approximate value.
To find the exact arc length of the curve y = 0.5(e^x + e^(-x)) for 0 ≤ x ≤ ln2, we need to use the formula for the arc length of a curve:
L = ∫[a to b] √(1 + (dy/dx)^2) dx
Here, a = 0 and b = ln2.
First, let's find dy/dx:
dy/dx = d/dx [0.5(e^x + e^(-x))]
dy/dx = 0.5(e^x - e^(-x))
Now, let's substitute this into the arc length formula:
L = ∫[0 to ln2] √[1 + (0.5(e^x - e^(-x)))^2] dx
Simplifying the integrand:
L = ∫[0 to ln2] √[1 + 0.25(e^2x - 2 + e^(-2x))] dx
L = ∫[0 to ln2] √[1 + 0.25(e^2x + e^(-2x) - 2)] dx
Using the property that cosh(2x) = (e^2x + e^(-2x))/2, we can rewrite the integrand as:
L = ∫[0 to ln2] √[1 + 0.25(2cosh(2x) - 2)] dx
L = ∫[0 to ln2] √[1 + 0.5cosh(2x) - 0.5] dx
Now, let's integrate:
L = ∫[0 to ln2] √[0.5cosh(2x) + 0.5] dx
Using the identity cosh^2(x) - sinh^2(x) = 1, we can rewrite cosh(2x) as:
cosh(2x) = 1 + sinh^2(2x)
Substituting back into the integral:
L = ∫[0 to ln2] √[0.5(1 + sinh^2(2x)) + 0.5] dx
L = ∫[0 to ln2] √[0.5 + 0.5sinh^2(2x)] dx
Next, we can use the identity sinh(2x) = 2sinh(x)cosh(x):
L = ∫[0 to ln2] √[0.5 + 0.5(2sinh(x)cosh(x))^2] dx
L = ∫[0 to ln2] √[0.5 + 0.5(2sinh^2(x)cosh^2(x))] dx
Using the identity cosh^2(x) - sinh^2(x) = 1 again:
L = ∫[0 to ln2] √[0.5 + 0.5(2sinh^2(x)(1 + sinh^2(x)))] dx
L = ∫[0 to ln2] √[0.5 + sinh^2(x)(1 + sinh^2(x))] dx
Now, we can make a substitution, letting sinh(x) = t:
L = ∫[0 to ln2] √[0.5 + t^2(1 + t^2)] dx
L = ∫[0 to ln2] √[0.5 + t^2 + t^4] dx
This integral does not have a simple closed-form solution using elementary functions. Therefore, the exact arc length of the curve y = 0.5(e^x + e^(-x)) for 0 ≤ x ≤ ln2 cannot be expressed in terms of elementary functions.