f(x)=∫01一cosxsirit2dt,当x→0时,f(x)是x的n阶无穷小,则n=________.

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问题 f(x)=∫01一cosxsirit2dt,当x→0时,f(x)是x的n阶无穷小,则n=________.

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答案6.

解析 可用下述结论观察求出,也可利用n阶无穷小定义求出.当f(x)连续且x→a时,f(x)是x→a的n阶无穷小量,g(x)是x→a的m阶无穷小量,则当x→a时,∫axf(t)出必为x一a的n+1阶无穷小量,∫ag(x)f(t)出必为x一a的(n+1)m阶无穷小量.因sinx2是x→0=x的2阶无穷小量,1一cosx~x2/2为x的2阶无穷小量,则x→0时,∫01一cosxsint2dt为x的(2+1)×2=6阶无穷小量,即n=6.
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