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
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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
= (- 1/2)∫[0→1] (2x - 1) d[ƒ(x)]
= (- 1/2)(2x - 1)ƒ(x) |[0→1] + (1/2)∫[0→1] ƒ(x) d(2x - 1)
= (- 1/2){[2(1) - 1]ƒ(1) - [2(0) - 1]ƒ(0)} + (1/2)∫[0→1] ƒ(x)(2) dx
= (- 1/2){ƒ(1) + ƒ(0)} + ∫[0→1] ƒ(x) dx
= (- 1/2){0 + 0} + ∫[0→1] ƒ(x) dx
= ∫[0→1] ƒ(x) dx
上面共用了两个分部积分法
∫ udv = uv - ∫ vdu
= (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
= (- 1/2)∫[0→1] (2x - 1) d[ƒ(x)]
= (- 1/2)(2x - 1)ƒ(x) |[0→1] + (1/2)∫[0→1] ƒ(x) d(2x - 1)
= (- 1/2){[2(1) - 1]ƒ(1) - [2(0) - 1]ƒ(0)} + (1/2)∫[0→1] ƒ(x)(2) dx
= (- 1/2){ƒ(1) + ƒ(0)} + ∫[0→1] ƒ(x) dx
= (- 1/2){0 + 0} + ∫[0→1] ƒ(x) dx
= ∫[0→1] ƒ(x) dx
上面共用了两个分部积分法
∫ udv = uv - ∫ vdu
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