设函数f在[1]上存在二阶连续导数,且满足f(0)=f(1)=0,证明∫(1,0)f(x)dx=1/2∫(1,0)x(x-1)f"(x)dx

(1/2)∫[0→1] x(x - 1)ƒ''(x) dx= (1/2)∫[0→1] (x² - x) d[ƒ'(x)]= (1/2)(x² - x)ƒ'(x) |[0→1] - (1/2)∫[0→1] ƒ'(x) d(x² - x)= (- 1/2)∫[0→1] ƒ'(x)(2x - 1) dx= ...