## Calculus (3rd Edition)

For $N=10$: The midpoint approximation of the length $\simeq 13.38405$. For $N=20$: The midpoint approximation of the length $\simeq 13.36509$. For $N=30$: The midpoint approximation of the length $\simeq 13.36490$. For $N=50$: The midpoint approximation of the length $\simeq 13.36489$.
We have $x\left( t \right) = t - \sin 2t$, ${\ \ \ }$ $x'\left( t \right) = 1 - 2\cos 2t$, $y\left( t \right) = 1 - \cos 2t$, ${\ \ \ }$ $y'\left( t \right) = 2\sin 2t$. Using Eq. (3) of Theorem 1, the arc length for $0 \le t \le 2\pi$ is $s = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( {1 - 2\cos 2t} \right)}^2} + {{\left( {2\sin 2t} \right)}^2}} {\rm{d}}t$ To approximate the definite integral above using Midpoint Rule, we divide $\left[ {0,2\pi } \right]$ into $N$ subintervals. The length of the interval is $\Delta x = \frac{{2\pi - 0}}{N} = \frac{{2\pi }}{N}$. The midpoints are ${c_j} = \left( {j - \frac{1}{2}} \right)\Delta x$ for $j = 1,2,...,N$. The midpoint approximation becomes ${M_N} = \frac{{2\pi }}{N}\mathop \sum \limits_{j = 1}^N f\left( {{c_j}} \right)$, where $f\left( x \right) = \sqrt {{{\left( {1 - 2\cos 2x} \right)}^2} + {{\left( {2\sin 2x} \right)}^2}}$. For $N=10$: We have $\Delta x = \frac{{2\pi }}{{10}} = \frac{\pi }{5}$ and ${c_j} = \frac{\pi }{5}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,10$. The midpoint approximation is ${M_{10}} = \frac{\pi }{5}\mathop \sum \limits_{j = 1}^{10} f\left( {{c_j}} \right) \simeq 13.38405$. For $N=20$: We have $\Delta x = \frac{{2\pi }}{{20}} = \frac{\pi }{{10}}$ and ${c_j} = \frac{\pi }{{10}}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,20$. The midpoint approximation is ${M_{20}} = \frac{\pi }{{10}}\mathop \sum \limits_{j = 1}^{20} f\left( {{c_j}} \right) \simeq 13.36509$. For $N=30$: We have $\Delta x = \frac{{2\pi }}{{30}} = \frac{\pi }{{15}}$ and ${c_j} = \frac{\pi }{{15}}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,30$. The midpoint approximation is ${M_{30}} = \frac{\pi }{{15}}\mathop \sum \limits_{j = 1}^{30} f\left( {{c_j}} \right) \simeq 13.36490$. For $N=50$: We have $\Delta x = \frac{{2\pi }}{{50}} = \frac{\pi }{{25}}$ and ${c_j} = \frac{\pi }{{25}}\left( {j - \frac{1}{2}} \right)$, for $j=1,2,...,50$. The midpoint approximation is ${M_{50}} = \frac{\pi }{{25}}\mathop \sum \limits_{j = 1}^{50} f\left( {{c_j}} \right) \simeq 13.36489$.