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设曲线x1n+y1n=1(正整数n≥1)在第一象限与坐标轴围成图形的面积为I(n),证明:(1)I(n)=2n∫10(1−t2)nt2n−1dt(2)I(n)=2n4n∫π20sin2n−1tdt且I(n)<2n4n(n>1)(3)∞n=1I(n)<1.
题目详情
设曲线x
+y
=1(正整数n≥1)在第一象限与坐标轴围成图形的面积为I(n),证明:
(1)I(n)=2n
(1−t2)nt2n−1dt
(2)I(n)=
sin2n−1tdt且I(n)<
(n>1)
(3)
I(n)<1.
1 |
n |
1 |
n |
(1)I(n)=2n
∫ | 1 0 |
(2)I(n)=
2n |
4n |
∫ |
0 |
2n |
4n |
(3)
∞ |
n=1 |
▼优质解答
答案和解析
由已知条件,I(n)=
y(x)dx=
(1−x
)ndx.
(1)令x=t2n(0≤t≤1),则dx=2nt2n-1dt,从而,
I(n)=
(1−x
)ndx
=
(1−t2)n•(2nt2n−1)dt
=2n
(1−t2)nt2n−1dt.
(2)令t=sinu(0≤u≤
),则dt=cosudu,
从而,
I(n)=2n
(1−t2)nt2n−1dt
=2n
cos2n+1usin2n−1udu.①
令t=
−u,则
cos2n+1usin2n−1udu
=
sinn+1ucos2n−1udu
=
(sin2n+1ucosn−1u+cos2n+1usin2n−1u)du
=
∫ | 1 0 |
∫ | 1 0 |
1 |
n |
(1)令x=t2n(0≤t≤1),则dx=2nt2n-1dt,从而,
I(n)=
∫ | 1 0 |
1 |
n |
=
∫ | 1 0 |
=2n
∫ | 1 0 |
(2)令t=sinu(0≤u≤
π |
2 |
从而,
I(n)=2n
∫ | 1 0 |
=2n
∫ |
0 |
令t=
π |
2 |
∫ |
0 |
=
∫ |
0 |
=
1 |
2 |
∫ |
0 |
=
1 |
2 |
∫ |
0
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