∫(上限π/2 下限0) [(sint)^4-(sint)^6] dt
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/08/06 04:04:15
∫(上限π/2 下限0) [(sint)^4-(sint)^6] dt
当n为正整偶数时,即n=2m,m=1,2...
∫(0→π/2)(sinx)^ndx=[(2m-1)!/(2m)!](π/2)
当n为正整奇数时,即n=2m+1,m=0,1,2...
∫(0→π/2)(sinx)^ndx=[(2m)!/(2m+1)!]
∫(0→π/2)(sinx)^4dx
=(3/4)×(1/2)×(π/2)
=3π/16
∫(0→π/2)(sinx)^6dx
=(5/6)×(3/4)×(1/2)×(π/2)
=5π/32
3π/16-5π/32=π/32
∫(0→π/2)(sinx)^ndx=[(2m-1)!/(2m)!](π/2)
当n为正整奇数时,即n=2m+1,m=0,1,2...
∫(0→π/2)(sinx)^ndx=[(2m)!/(2m+1)!]
∫(0→π/2)(sinx)^4dx
=(3/4)×(1/2)×(π/2)
=3π/16
∫(0→π/2)(sinx)^6dx
=(5/6)×(3/4)×(1/2)×(π/2)
=5π/32
3π/16-5π/32=π/32
d/dx∫(sint/t)dt上限π下限x
d[∫f(sint)dt]/dx,上限x,下限0
d/dx[∫(上限x^2 下限0 )sint^2dt]=?
d/dx[∫(上限x^3 下限0 )sint^2dt]=?
d[∫f(sint)dt]/dx,上限x^2 下限0
x→0时∫(上限x,下限-x)sint+sint^2dt 与ax^k 等价无穷小 求a与k
126.设F(x)=∫x (积分上限) 0 (积分下限) sint / t dt ,求 F’(0)
求(d/dx)∫(sint/t)dt=?上限为x 下限为0
设f(x)=∫(上限x 下限pain) sint/t dt , 计算 ∫(上限π 下限0) f(x) dx
设f(x)=∫(下限x上限1)sint²dt,则∫(下限0上限1)f(x)dx=__.
变限积分f(x)=∫sint^2 dt 积分下限x,上限x^2,求f(x)导数
f(x)=∫(sint/t)dt,积分上限是π/2,积分下限是x^2,求这个函数的定义域.