题目

设f(x)=sinx,(0≤x≤π/2)；f(x)=1/2,(π/2≤x≤π) 求定积分∫f(t)dt 积分上限x ;积分下限0

提问时间：2020-11-03

答案

0 <= x <= π/2,
∫_{0}^{x}f(t)dt = ∫_{0}^{x}sin(t)dt = 1 - cos(x)
π/2 ≤ x ≤ π,
∫_{0}^{x}f(t)dt = ∫_{0}^{π/2}f(t)dt + ∫_{π/2}^{x}f(t)dt
= ∫_{0}^{π/2}sin(t)dt + ∫_{π/2}^{x}dt/2
= 1 + (x - π/2)/2