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