设f''(x)在[0,1]连续,且f(0)=1,f(2)=3,f'(2)=5,求∫[0,1]xf''(2x)dx

∫[0→1]xf''(2x)dx
=(1/2)∫[0→1]xdf'(2x)
=(1/2)xf'(2x)|[0→1]-(1/2)∫[0→1]f'(2x)dx
=(1/2)f'(2)-(1/4)f(2x)|[0→1]
=5/2-(1/4)[f(2)-f(0)]
=5/2-1/2
=2