设f(x)=sinx,x∈[0,π/2) f(x)= 1,x∈[π/2,2].则f(x)dx在0到2上的积分为
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/17 09:08:50
![设f(x)=sinx,x∈[0,π/2) f(x)= 1,x∈[π/2,2].则f(x)dx在0到2上的积分为](/uploads/image/z/9325139-59-9.jpg?t=%E8%AE%BEf%28x%29%3Dsinx%2Cx%E2%88%88%5B0%2C%CF%80%2F2%EF%BC%89+f%28x%29%3D+1%2Cx%E2%88%88%5B%CF%80%2F2%2C2%5D.%E5%88%99f%28x%29dx%E5%9C%A80%E5%88%B02%E4%B8%8A%E7%9A%84%E7%A7%AF%E5%88%86%E4%B8%BA)
x){n_Fmqf^Nţh
Ft*%! AX3A)O0xڱɎZ/_Ɏ]6IE40UΆV}Ա(6xg6D=X!*
`:*MiBҢ_i{H_] @Ȃ%leO HS
设f(x)=sinx,x∈[0,π/2) f(x)= 1,x∈[π/2,2].则f(x)dx在0到2上的积分为
设f(x)=sinx,x∈[0,π/2) f(x)= 1,x∈[π/2,2].则f(x)dx在0到2上的积分为
设f(x)=sinx,x∈[0,π/2) f(x)= 1,x∈[π/2,2].则f(x)dx在0到2上的积分为
∫上2下0;f(x)dx
=∫上2下π/2;f(x)dx+∫上π/2下0;f(x)dx
=∫上2下π/2;1dx+∫上π/2下0;sinxdx
=x│上2下π/2-cosx│上π/2下0
=(2-π/2)-(0-1)
=3-π/2
∫上2下0;f(x)dx
=∫上2下π/2;f(x)dx+∫上π/2下0;f(x)dx
=∫上2下π/2;1dx+∫上π/2下0;sinxdx
=x│上2下π/2-cosx│上π/2下0
=(2-π/2)-(0-1)
=3-π/2
设f(x)=(cosx+sinx)sinx,且x∈{0,π/2},则函数f(x)的最大值
设f(x)=sinx f(x+2π)=
设f(x)=x*sinx,求f'' (π/2)
设f(x)∈C[0,1],证明∫(π,0)*x*f(sinx)dx =π/2*∫(π,0)*f(sinx)dx
设函数f(x)=sinx,则[f(π/2)]'
三角函数:设f(x)={sinx(0
设f(x)满足f(-sinx)+3f(sinx)=4sinx*cosx(x的绝对值小于等于π/2)
设x∈(0,π),则f(x)=cos²x+sinx的最大值是多少?
设f(x-1)={-sinx/x,x>0;2,x=0;x-1,x
设f(x-1)={-sinx/x,x>0;2,x=0;x-1,x
设函数f(x)=(x+sinx)/x,g(x)=xcosx-sinx. (1).求证:当x∈(0,π】时,g(x)<0 (2).存在x
已知函数f(x)=3-2cosx-2sinx,其中x∈[0,2π],设g(x)=(sinx-1)/f(x),求函数g(x)的值域如题
设f(x)=∫(x,x+2π)e^sinx*sinxdx,则f(x)=
设函数f(x)=(2sinxcosx+5/2)/(sinx+cosx),[0 ,π/2] 求f(x)的最小值
设x∈(0,π),则f(x)=cos²x+sinx的最大值是
f(x)=4sinx(sinx-根号3cosx)-3,x∈[0,π/2],求f(x)值域
设y=sinx/x^2,求f'(π/3)
设f(sinx)=(cosx)^2+sinx+5,求f(x)