Answer
(a) $\int_{0}^{2 \pi} 2 \sin^2 x \space dx =\pi$
(b) $\int_{0}^{2 \pi} 2 \sin^2 x \space dx =\pi$
Work Step by Step
a) Apply the Trapezoidal Rule: $\int_a^b f(x) dx \approx T =\dfrac{\triangle x}{2}(y_0+y_1+y_2+......+y_n)$
Consider $\int_{0}^{2 \pi} 2 \sin^2 x dx \approx T =\dfrac{\frac{\pi}{6}}{2}(2 \sin^2 (0)+2[2 \sin^2 (\dfrac{\pi}{6})]+2[2 \sin^2 (\pi/3)]+..]=\dfrac{\pi}{12} \times (0+1+3+4+.....)=\pi$
b) Apply Simpson's Rule: $\int_a^b f(x) dx \approx T =\dfrac{\triangle x}{3}(y_0+4y_1+2y_2+4y_3+......+y_n)$
Consider $\int_{0}^{2 \pi} 2 \sin^2 x dx \approx T =\dfrac{\pi/6}{3}(2 \sin^2 (0)+4[2 \sin^2 (\dfrac{\pi}{6})]+2[2 \sin^2 (\dfrac{\pi}{3})]+...]=\dfrac{\pi}{18} \times (0+2+3+8+.....)=\pi$
