设f(x)={x^2(x属于[0,1]) 2-x(x属于[1,2]),则f(x)的0到2的定积分等于?
问题描述:
设f(x)={x^2(x属于[0,1]) 2-x(x属于[1,2]),则f(x)的0到2的定积分等于?
答
[0,2] ∫ f(x) dx=[0,1] ∫ x²dx + [1,2] ∫ (2-x)dx=1/3 x³ | [0,1] + (2x - ½ x²) | [1,2]=1/3 (1 - 0) + (2*2 - ½ 2² - 2*1 + ½)=5/6