来源：学生作业帮 编辑：搜狗做题网作业帮 分类：数学作业 时间：2024/08/08 20:35:30
换元积分法和分部积分法求定积分
∫xf''(2x)dx = (1/2)∫xf''(2x)d(2x)
= (1/2)∫xdf'(2x)
= (1/2)[xf'(2x)] - (1/2)∫f'(2x)dx
= (1/2)f'(2) - (1/4)∫f'(2x)d(2x)
= 1 - (1/4)∫df(2x)
= 1 - (1/4)[f(2x)]
= 1 - (1/4)[f(2)-f(0)]
= 1-(1/4)(4-1) = 1/4
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