Question

求定积分∫（上限2，下限-1）∣x^2-x∣dx

Answer 1

令x^2-x=x(x-1)>0
x>1或 x<0
令x^2-x=x(x-1)<0
0<x<1
∴
∫(-1->2) |x^2-x|dx

=∫(-1->0) (x^2-x)dx +∫(0->1) -(x^2-x)dx +∫(1->2)(x^2-x)dx
=(x^3/3-x^2/2)|(-1->0) -(x^3/3-x^2/2)|(0->1) +(x^3/3-x^2/2)|(1->2)
=[(0-0)-(-1/3-1/2)]-[(1/3-1/2)-(0-0)]+[(8/3-4/2)-(1/3-1/2)]
=5/6+1/6 +5/6
=11/6

Answer 2

令x^2-x=0,则x=0或1
当-1<x<0,x^2-x>0
当0<x<1,x^2-x<0
当1<x<2,x^2-x>0
∫(-1,2)|x^2-x|dx=∫(-1,0)(x^2-x)dx+∫(1,2)(x^2-x)dx-∫(0,1)(x^2-x)dx
=(1/3*x^3-1/2x^2)|(-1,0)+(1/3*x^3-1/2x^2)|(1,2)-(1/3*x^3-1/2x^2)|(0,1)
=5/6+5/6+1/6
=11/6