Area Under Curves

The area enclosed under a curve that lies above the axis equals the definite integral of the function over the interval, but only when the function stays non-negative throughout, otherwise signed cancellation distorts the geometric area. On [0, \pi] the sine function satisfies \sin x \ge 0, so the geometric area genuinely equals \int_0^\pi \sin x,dx = [-\cos x]_0^\pi = -\cos\pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2. Option 1 results from evaluating only over the half-arch [0, \pi/2]. Option \pi confuses the interval length with the enclosed area. Option 0 would apply over a full period [0, 2\pi], where the positive and negative arches cancel exactly. The non-negativity check is what licenses equating the signed integral with the true geometric area here. The two evaluations of cosine at the endpoints both yield magnitude one, and they add rather than cancel because of the negative sign in the antiderivative. As a final plausibility check, the arch fits inside a bounding rectangle of width \pi \approx 3.14 and height 1, giving area about 3.14, so an enclosed area of 2 sits sensibly below that rectangle, confirming geometric consistency.