Trapezoidal rule

$\begin{aligned} I & = \int_0^\pi \sin^2 x \space dx = \int_0^\pi \frac{1-\cos (2x)}{2} dx = \frac{1}{2} \left[x - \frac{\sin (2x)}{2} \right]_0^\pi = \frac{\pi}{2} \end{aligned}$

$\begin{aligned} T_{100} & = \frac{\pi - 0}{100} \left[ \frac{\sin^2 0 - \sin^2 \pi}{2} + \sum_{k=1}^{99} \sin^2 \left(\frac{k \pi}{100} \right) \right] \\ & = \frac{\pi}{100} \sum_{k=1}^{99} \sin^2 \left(\frac{k \pi}{100} \right) \quad \quad \small \color{#3D99F6}{\text{As } \sin x = \sin (\pi -x)} \\ & = \frac{\pi}{100} \left( 2 \sum_{k=1}^{49} \sin^2 \left(\frac{k \pi}{100} \right) + \sin^2 \left(\frac{50 \pi}{100} \right) \right) \\ & = \frac{\pi}{100} \left( 2 \sum_{k=1}^{24}\left[ \sin^2 \left(\frac{k \pi}{100} \right) + \sin^2 \left(\frac{(50-k) \pi}{100} \right) \right] + 2 \sin^2 \left(\frac{25 \pi}{100} \right) + \sin^2 \left(\frac{\pi}{2} \right) \right) \\ & = \frac{\pi}{100} \left( 2 \sum_{k=1}^{24}\left( \sin^2 \left(\frac{k \pi}{100} \right) + \cos^2 \left(\frac{k \pi}{100} \right) \right) + 2 \sin^2 \left(\frac{\pi}{4} \right) + \sin^2 \left(\frac{\pi}{2} \right) \right) \\ & = \frac{\pi}{100} \left( 2 \sum_{k=1}^{24} 1 + 1 + 1 \right) = \frac{\pi}{100} \left( 48 + 1 + 1 \right) = \frac{\pi}{2} \end{aligned}$

$\Rightarrow E_{100} = I - T_{100} = \boxed{0.000}$

We have $\begin{array}{rcl} T_N & = & \displaystyle \frac{\pi}{N}\sum_{j=1}^{N-1} \sin^2\big(\tfrac{j \pi}{N}\big) \; =\; \frac{\pi}{2N}\sum_{j=0}^{N-1}\Big(1 - \cos\big(\tfrac{2j\pi}{N}\big)\Big) \\ & = & \displaystyle \tfrac12\pi - \frac{\pi}{2N}\mathfrak{Re}\left[\sum_{j=0}^{N-1} e^{\frac{2j\pi i}{N}}\right] \; = \; \tfrac12\pi - \frac{\pi}{2N}\mathfrak{Re}\left[\frac{e^{2\pi i} - 1}{e^{\frac{2\pi i}{N}} - 1}\right] \\ & = & \tfrac12\pi \end{array}$ and so, since $I = \tfrac12\pi$ , we have $E_N = 0$ for all integers $N$ .