已知f(x)具有二阶连续导数,且f(0)=1,f(2)=4,f'(2)=2 求∫xf''(2x)dx
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已知f(x)具有二阶连续导数,且f(0)=1,f(2)=4,f'(2)=2 求∫xf''(2x)dx
∫(0→1) xƒ''(2x) dx
= (1/2)∫(0→1) xƒ''(2x) d(2x)
= (1/2)∫(0→1) x d[ƒ'(2x)]
= (1/2)[xƒ'(2x)] |(0→1) - (1/2)∫(0→1) ƒ'(2x) dx
= (1/2)xƒ'(2x) - (1/4)ƒ(2x) |(0→1)
= [(1/2)ƒ'(2) - (1/4)ƒ(2)] - [(- 1/4)ƒ(0)]
= (1/2)(2) - (1/4)(4) + (1/4)(1)
= 1/4
= (1/2)∫(0→1) xƒ''(2x) d(2x)
= (1/2)∫(0→1) x d[ƒ'(2x)]
= (1/2)[xƒ'(2x)] |(0→1) - (1/2)∫(0→1) ƒ'(2x) dx
= (1/2)xƒ'(2x) - (1/4)ƒ(2x) |(0→1)
= [(1/2)ƒ'(2) - (1/4)ƒ(2)] - [(- 1/4)ƒ(0)]
= (1/2)(2) - (1/4)(4) + (1/4)(1)
= 1/4
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