求定积分

显示全部楼层 · 2013-2-9 08:01:41

设f(x)=∫[1,x^2] sint/t dt，则定积分∫[1,0]f(x)dx=()A 1-cos1 B -1+cos1 C (1-cos1)/2 D (-1+cos1)/2

千问 · 2013-2-9 08:01:41

设f(x)=∫defds123lt;x^2,1defds123gt; (sint/t) dt,∫defds123lt;0,1defds123gt;xf(x)dx=∫defds123lt;0,1defds123gt;xdx)∫defds123lt;x^2,1defds123gt; (sint/t) dt(交换二次积分次序）=∫defds123lt;0,1defds123gt; (sint/t) dt∫defds123lt;0,t^(1/2)defds123gt;xdx=(1/2)∫defds123lt;0,1defds123gt; sintdt=(1-cos1)/2.

千问 · 2013-2-9 08:01:41

<b class=\\\"o\\\">提问者对答案的评价：<img src=\\\'http://i0.sinaimg.cn/pfp/ask/images/zhishi/ssr50a.gif\\\'><img src=\\\'http://i0.sinaimg.cn/pfp/ask/images/zhishi/ssr50a.gif\\\'><img src=\\\'http://i0.sinaimg.cn/pfp/ask/images/zhishi/ssr50a.gif\\\'><img src=\\\'http://i0.sinaimg.cn/pfp/ask/images/zhishi/ssr50a.gif\\\'><img src=\\\'http://i0.sinaimg.cn/pfp/ask/images/zhishi/ssr50a.gif\\\'></b>

千问 · 2013-2-9 08:01:41

用分部积分法：∫[1,0]f(x)dx=xf(x)[1,0]-∫[1,0]xdf(x)（注意f(1))=∫[1,1]sint/tdt=0, df(x)=2x*sinx^2/x^2dx）=0-∫[1,0]x*2x*sinx^2/x^2dx=-2∫[1,0]sinx^2dx应该是积不出来的。请核查题目？