这三道题谁会啊,求助,万分感谢
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2017-08-30 · 中小学教师,杨建朝,蒲城县教研室蒲城县教育学会、教育领域创作...
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4.
lim xⁿlnx
x→0
=lim lnx/x⁻ⁿ
x→0
=lim (1/x)/[(-n)x⁻ⁿ⁻¹]
x→0
=lim -xⁿ/n
x→0
=0
5.
∫[1/(a²+x²)]dx
=(1/a)∫d(x/a)/[1+(x/a)²]
=(1/a)arctan(x/a) +C
6.
令x=asint
x:0→a,则t:0→π/2
∫[0:a]√(a²-x²)dx
=∫[0:π/2]√(a²-a²sin²t)d(asint)
=a²∫[0:π/2]cos²tdt
=½a²∫[0:π/2](1+cos2t)dt
=½a²(t+½sin2t)|[0:π/2]
=½a²[(π/2 +½sinπ)-(0+½sin0)]
=½a²[(π/2 +0)-(0+0)]
=¼πa²
lim xⁿlnx
x→0
=lim lnx/x⁻ⁿ
x→0
=lim (1/x)/[(-n)x⁻ⁿ⁻¹]
x→0
=lim -xⁿ/n
x→0
=0
5.
∫[1/(a²+x²)]dx
=(1/a)∫d(x/a)/[1+(x/a)²]
=(1/a)arctan(x/a) +C
6.
令x=asint
x:0→a,则t:0→π/2
∫[0:a]√(a²-x²)dx
=∫[0:π/2]√(a²-a²sin²t)d(asint)
=a²∫[0:π/2]cos²tdt
=½a²∫[0:π/2](1+cos2t)dt
=½a²(t+½sin2t)|[0:π/2]
=½a²[(π/2 +½sinπ)-(0+½sin0)]
=½a²[(π/2 +0)-(0+0)]
=¼πa²
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