f(x)=∫sint⼀pi-t dt (上限x,下限0) , ∫f(x)dx (上限pi,下限0)是多少啊,怎么做

是sint⼀(pi-t)
2024-12-26 00:31:51
推荐回答(4个)
回答1:

∫sint/pi-t dt
=1/pi∫sintdt-∫tdt
=1/pi∫-dcost-1/2∫dt^2
=-cost/pi-t^2/2+c
当t=x时
f(x)=-cosx/pi+x^2/2+C
当t=0时
f(0)=-cos0/pi+0^2/2+C=-1/pi+C
f(x)-f(0)=-cosx/pi+x^2/2+C-(-1/pi+C)
=-cosx/pi+x^2/2+C+1/pi-C
=-cosx/pi+x^2/2+1/pi

∫f(x)dx
=∫-cosxdx/pi+∫x^2dx/2+∫dx/pi
=-sinx/pi+x^3/6+x/pi+C
当x=pi时
f(pi)=0/pi+pi^3/6+pi/pi+C=pi^3/6+1+C
当x=0时
f(0)=0/pi+0^3/6+0/pi+C=C
所以
f(pi)-f(0)=pi^3/6+1

回答2:

为什么不适合发表?

回答3:

f(x)=∫sint/(pi-t )dt
=∫-sin(t-pi)/(pi-t )dt
=∫sin(pi-t)/(pi-t )dt
=-∫sin(pi-t)/(pi-t )d(pi-t)(令pi-t=y)
=-∫siny/ydy(积分限为[pi-x,pi])
参考http://zhidao.baidu.com/question/2488735.html
该积分无准确表示式,只能近似表示

回答4:

你那个sint/pi-t最后面的t是在分母里的还是在式子里的?

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