解ï¼4ãï¼1ï¼å½æ°f(x)å¨xâï¼-âï¼+âï¼ï¼å¨æ´ä¸ªå®æ°åºé´æå®ä¹ï¼f(x)max=lim(x-->1-0)2x=2,0<f(x)<2,æçï¼
ï¼2ï¼å 为lim(x-->0+)2x=0=lim(x-->0-)0,æ以å½æ°å¨åºé´ï¼-âï¼1ï¼è¿ç»å¯å¯¼ï¼èå¨lim(x-->1-0)2x=2â lim(x-->1+0)0=0ï¼äº§çäºä¸å¯å»é´æç¹ï¼å¨xâ[1,+âï¼è¿ç»ï¼ç§¯åå½æ°å¨å¼åºé´xâï¼0,1ï¼ä¸ºå常积åï¼ç§¯åå¼å¨å¼åºé´xâï¼0,1ï¼ç积åçåäºåºé´xâï¼-âï¼+â)ç积åï¼
ï¼3ï¼ç§¯åå¼çæ大å¼æé为â«(0,1)2xdx=x^2](0,1)=lim(x-->1-0)2x-lim(x-->0+)2x=2ã
5ã设平å温度为Tpï¼
Tp=(1/12){â«(9,21)[50+14sin(tÏ/12)]dt=(1/12)[50t-(12/Ï)cos(tÏ/12)](9,21)
=(1/12){12*50-12/Ï[cos(7Ï/4)-cos3Ï/4]=50-(1/Ï)[cos(2Ï-Ï/4)-cos(Ï-Ï/4)]
=50-(1/Ï)2cos(Ï/4)=50-â2/Ïã
温馨提示:答案为网友推荐,仅供参考