[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
[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