>>340
0≦x≦1だから|x|=x　よってx-aの符号だけ見れば良い
(i)a≦0のときx-a≧0
よって∫[0,1]|x(x-a)|dx=∫[0,1]x(x-a)dx
=∫[0,1](x^2-ax)dx
=[(1/3)x^3-(1/2)a*x^2][0,1]
=(1/3)-(1/2)a
(ii)0≦a≦1のとき
∫[0,1]|x(x-a)|dx=∫[0,a]x(a-x)dx+∫[a,1]x(x-a)dx
=∫[0,a](ax-x^2)dx+∫[a,1](x^2-ax)dx
=[(1/2)a*x^2-(1/3)x^3][0,a]+[(1/3)x^3-(1/2)a*x^2][a,1]
=(1/2)a^3-(1/3)a^3+(1/3)-(1/2)a-(1/3)a^3+(1/2)a^3
=(1/3)a^3-(1/2)a+(1/3)
(iii)a≧1のときx-a≦0
よって∫[0,1]|x(x-a)|dx=∫[0,1]x(a-x)dx
=∫[0,1](ax-x^2)dx
=[(1/2)a*x^2-(1/3)x^3][0,1]
=(1/2)a-(1/3)