Question

已知f(0)=1,f(2)=4,f '(2)=2,求∫ x f ' '(2x)dx,上限1,下限0.

Answer 1

∫xf"(2x)dx

=(1/2)∫xf"(x)d2x

=(1/2)∫xdf'(2x)

=(1/2)xf'(2x)-(1/2)∫f'(2x)dx

=(1/2)xf'(2x)-(1/4)∫f'(2x)d2x

=(1/2)xf'(2x)-(1/4)f(2x)+C

x=1,(1/2)xf'(2x)-(1/4)f(2x)=(1/2)*1*f'(2)-(1/4)*f(2)=1-1=0

x=0,(1/2)xf'(2x)-(1/4)f(2x)=(1/2)*0*f'(0)-(1/4)*f(0)=0-1/4=-1/4

所以定积分=0-(-1/4)=1/4