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Today's Calculation Of Integral
2005 Today's Calculation Of Integral
56
56
Part of
2005 Today's Calculation Of Integral
Problems
(1)
Today's Calculation of Integaral 56
Source: 2000 Osaka University
6/29/2005
Evaluate
lim
n
→
∞
∑
k
=
1
n
[
2
n
2
−
k
2
]
n
2
\lim_{n\to\infty} \sum_{k=1}^n \frac{[\sqrt{2n^2-k^2}\ ]}{n^2}
n
→
∞
lim
k
=
1
∑
n
n
2
[
2
n
2
−
k
2
]
[
x
]
[x]
[
x
]
is the greatest integer
≤
x
\leq x
≤
x
.
limit
integration
calculus
calculus computations