x = t − 2 sin t, y = 1 − 2 cos t, 0 ≤ t ≤ 6π

Set up an integral that represents the length of the curve.

s = ∫[0,6π] √((dx/dt)^2 + (dy/dt)^2) dt

= ∫[0,6π] √((1-2cost)^2 + (2sint)^2) dt
= ∫[0,6π] √(1-4cost+4cos^2t + 4sin^2t) dt
= ∫[0,6π] √(5-4cost) dt

To find the length of the curve, we need to evaluate the integral of the magnitude of the derivative of the parametric equations with respect to t. The formula for the length of a curve given by parametric equations is:

L = ∫[a,b] √[(dx/dt)² + (dy/dt)²] dt

In this case, x = t − 2 sin(t) and y = 1 − 2 cos(t). So, we need to find dx/dt and dy/dt to calculate the integral.

To find dx/dt, we can differentiate x with respect to t:

dx/dt = 1 - 2cos(t)

To find dy/dt, we can differentiate y with respect to t:

dy/dt = 2sin(t)

Now, we can substitute these values into the formula:

L = ∫[a,b] √[(dx/dt)² + (dy/dt)²] dt
L = ∫[0,6π] √[(1 - 2cos(t))² + (2sin(t))²] dt

Therefore, the integral that represents the length of the curve is:

L = ∫[0,6π] √[(1 - 2cos(t))² + (2sin(t))²] dt

Note: The limits of integration are 0 and 6π since the given range for t is 0 ≤ t ≤ 6π.

To find the length of the curve defined by the parametric equations, we can use the arc length formula for parametric curves. The formula for calculating the arc length of a parametric curve is given by:

L = ∫(sqrt(dx/dt)^2 + (dy/dt)^2) dt

In this case, we are given the parametric equations x = t - 2sin(t) and y = 1 - 2cos(t). We need to find dx/dt and dy/dt to calculate the integral.

1. Differentiate x = t - 2sin(t) with respect to t:
dx/dt = 1 - 2cos(t)

2. Differentiate y = 1 - 2cos(t) with respect to t:
dy/dt = 2sin(t)

Now, substitute dx/dt and dy/dt into the arc length formula:

L = ∫(sqrt((1 - 2cos(t))^2 + (2sin(t))^2)) dt

Simplify the expression inside the square root:

L = ∫(sqrt(1 - 4cos(t) + 4cos^2(t) + 4sin^2(t))) dt
L = ∫(sqrt(5 - 4cos(t))) dt

Now, integrate the expression with respect to t over the given interval 0 ≤ t ≤ 6π:

L = ∫(sqrt(5 - 4cos(t))) dt | from t=0 to t=6π

To evaluate this integral, you can use numerical methods or approximation techniques such as numerical integration or calculators that can perform definite integrals.

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