4x.f(x2)+3f(1−x)=1−x2
⇔2.I=∫012x.f(x2)dx+3∫01f(1-x)dx=∫011−x2dx⇔2A+3B=∫011−x2dx (*)A=∫012x.f(x2)dx
Đặt t=x2⇒dt=2xdx; x=0⇒t=0; x=1⇒t=1; x=1⇒t=0 B=∫01f(t)dt=∫01f(x)dx
(*)⇔2∫01f(x)dx+3∫01f(x)dx=∫011−x2dx⇔5.∫01f(x)dx=∫011−x2dx
Đặt x=sint⇒dx=costdt∈(−2π;2π); x=0⇒t=0, x=1⇒t=2π
⇒∫011−x2dx=∫02π1−sin2t.costdt=∫02π21+cos2xdt=21.(t+21sin2t)∣∣02π=4π
Vậy ∫01f(x)dx=20π