if x[x[x[x]]]=2013 then x equal to
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MATHS
For a real number x let [x] denote the largest integer less than or equal to x and {x} = x - [x] . The smallest possible integer value of n for which ∫
1
n
[x]{x}dx exceeds 2013 is
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VIDEO EXPLANATION
ANSWER
∫
1
n
[x]{x}dx
=∫
1
2
1(x−1)dx+∫
2
3
2(x−2)dx+...+∫
n−1
n
(n−1)(x−n+1)dx
This can be generalized as
t=1
∑
n−1
∫
t
t+1
t(x−t)dx
=
t=1
∑
n−1
[
2
tx
2
−t
2
x]
t
t+1
=
t=1
∑
n−1
[
2
t(t
2
+2t+1−t
2
)
−t
2
(t+1−t)]
=
t=1
∑
n−1
[t
2
+
2
t
−t
2
]
=
2
1
×
2
(n−1)n
=
4
(n−1)n
We need the smallest n so that (n−1)n>2013×4
i.e. (n−1)n>8052
The smallest square number greater than 8052 is 8100, whose square root is 90.
If n=90,90∗89<8052
∴n=91