哪一位牛人帮忙解一道高数题?
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/29 16:03:38
哪一位牛人帮忙解一道高数题?
本人已做到求最大值的部分,就是那个定积分,(单击图可以放大)
![](http://img.wesiedu.com/upload/1/1d/11d100f407ca376661ac23ca5896b260.jpg)
本人已做到求最大值的部分,就是那个定积分,(单击图可以放大)
![](http://img.wesiedu.com/upload/1/1d/11d100f407ca376661ac23ca5896b260.jpg)
![哪一位牛人帮忙解一道高数题?](/uploads/image/z/3147186-66-6.jpg?t=%E5%93%AA%E4%B8%80%E4%BD%8D%E7%89%9B%E4%BA%BA%E5%B8%AE%E5%BF%99%E8%A7%A3%E4%B8%80%E9%81%93%E9%AB%98%E6%95%B0%E9%A2%98%3F)
Φ'(x)=(3x+1)/(x^2-x+1)>0(在[0,1]内)
故Φ(x)单调上升,最大值为
Φ(1)=∫[0,1](3t+1)/(t^2-t+1)*dt
=∫[0,1]3(t-1/2)/(t^2-t+1)dt+5/2∫[0,1]1/[(t-1/2)^2+3/4]dt
=3/2∫[0,1]1/(t^2-t+1)d(t^2-t+1)+10/3∫[0,1]1/[1+(2/√3*(t-1/2))^2]dt
=3/2ln(t^2-t+1)|[0,1]+5√3/3arctan[2/√3*(t-1/2)]|[0,1]
=5√3/3[π/6-(-π/6)]=5π√3/9
故Φ(x)单调上升,最大值为
Φ(1)=∫[0,1](3t+1)/(t^2-t+1)*dt
=∫[0,1]3(t-1/2)/(t^2-t+1)dt+5/2∫[0,1]1/[(t-1/2)^2+3/4]dt
=3/2∫[0,1]1/(t^2-t+1)d(t^2-t+1)+10/3∫[0,1]1/[1+(2/√3*(t-1/2))^2]dt
=3/2ln(t^2-t+1)|[0,1]+5√3/3arctan[2/√3*(t-1/2)]|[0,1]
=5√3/3[π/6-(-π/6)]=5π√3/9