## University Calculus: Early Transcendentals (3rd Edition)

a) $\int_{0}^{2 \pi} 2 \sin^2 x dx =\pi$ b) $\int_{0}^{2 \pi} 2 \sin^2 x dx =\pi$
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, $\int_{0}^{2 \pi} 2 \sin^2 x dx \approx T =\dfrac{\frac{\pi}{6}}{3}(2 \sin^2 (0)+4[2 \sin^2 (\pi/6)]+2[2 \sin^2 (\pi/3)]+......)=\dfrac{\pi}{18}(0+2+3+8+.....)=\pi$