Xét ∫−121f'(2x)dx
Đặt 2x=t⇒dx=2dt
Khi x=-1⇒t=-2; x=21⇒t=1
Ta có I1=∫−121f'(2x)dx=∫−21f'(t)2dt=21∫−21f'(t)dt=21∫−21f'(x)dx
Xét I2=∫032f′(x+1)dx
Đặt x+1=u⇒dx=du
Khi x=0⇒u=1; x=3⇒u=4
Ta có I2=∫032f′(x+1)dx=∫142f(u)du=21∫14f'(u)du=21∫14f'(x)dx
Có I=I1+I2
=21∫−21f'(x)dx+21∫14f'(x)dx
=21∫−24f'(x)dx=21f(x)∣∣−24=2f(4)−f(2)=28−2=3