구간 [0,\pi]에서 \sin x \ge \sin^2 x이다. (예: x = \dfrac{π}{2}에서 1 \ge 1^2)

넓이 = \displaystyle \int_{0}^{\pi} (\sin x - \sin^2 x) dx = \int_{0}^{\pi} \sin x dx - \int_{0}^{\pi} \sin^2 x dx

1. \displaystyle \int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos\pi) - (-\cos 0) = -(-1) - (-1) = 2

2. \displaystyle \int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} \dfrac{1-\cos(2x)}{2} dx = \left[\dfrac{1}{2}x - \dfrac{1}{4}\sin(2x)\right]_{0}^{\pi} = \left(\dfrac{\pi}{2} - 0\right) - (0) = \dfrac{\pi}{2}

넓이 = \displaystyle 2 - \dfrac{\pi}{2}