分段函数,f(x)=x+1,当x≦1时;f(x)= x²,当x>1时;求【0,2】∫f(x)dx【0,2】∫f(x)dx=【0,1】∫(x+1)dx+【1,2】∫x²dx=[(x+1)²/2]【0,1】+(x³/3)【1,2】=[2-(1/2)]+(8/3-1/3)=(3/2)+(7/3)=23/6.【x=1是间断点】
f(x) = x + 1/x, f'(x) = 1 - 1/x^2f(x) = (1+x)/x = 1/x + 1, f'(x) = -1/x^2