Question

Write an integral that represents the arc length of the curve on the given interval. Do not evaluate the integral. $$ x=t+\sin t, \quad y=t-\cos t \quad 0 \leq t \leq \pi $$

Short answer

The integral that represents the arc length of the curve over the interval [0, \(\pi\)] is \(L=\int_{0}^{\pi}{\sqrt{2 + 2\cos t + 2\sin t}\ dt}\).

Step-by-step solution

Find the Derivatives

Firstly find the derivatives of x and y with respect to t, \(dx/dt\) and \(dy/dt\). Based on the given parametric equations, \(x=t+\sin t\) and \(y=t-\cos t\), derive \(dx/dt = 1 + \cos t\) and \(dy/dt = 1 + \sin t\).

Substitute into the Formula

Substitute \(dx/dt\) and \(dy/dt\) into the arc length formula. This gives: \(L=\int_{0}^{\pi}{\sqrt{(1 + \cos t)^{2}+(1 + \sin t)^{2}}\ dt}\).

Simplify the Integral

Now, simplify the integral: \(L=\int_{0}^{\pi}{\sqrt{2 + 2\cos t + 2\sin t}\ dt}\). Remember, that the task is not to evaluate the integral, therefore the final answer is left as it is.