If f is continuously differentiable function then∫01.5 x2f′(x)dx is

∫01.5 x2f′(x)dx=∫01 x2f′(x)dx+∫12 x2f′(x)dx+∫21.5 x2f′(x)dx=0+∫12 f′(x)dx+2∫21.5 f′(x)dx=f(2)−f(1)+2[f(1.5)−f(2)]=2f(1.5)−f(2)−f(1)