已知f(x)具有二阶连续导数,且f(0)=1,f(2)=4,f'(2)=2 求∫xf''(2x)dx

∫(0→1) xƒ''(2x) dx= (1/2)∫(0→1) xƒ''(2x) d(2x)= (1/2)∫(0→1) x d[ƒ'(2x)]= (1/2)[xƒ'(2x)] |(0→1) - (1/2)∫(0→1) ƒ'(2x) dx= (1/2)xƒ'(2x) - (1/4)ƒ(2x) |(0→1)= [...