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Calculus
One, Two, Three, Floor Integral (
Posted on 20071112
)
Evaluate:
_{4}
∫
[t + [t + [t]]] dt
^{4}
where [p] is the greatest integer ≤ p.
Submitted by
K Sengupta
No Rating
Solution:
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Let I = integral [t + [t + [t]]] dt; t = 4 to 4
Since [t + [t + [t]]] = 3*[t] in the given interval, we must have:
I = 3* integral [t] dt; t = 4 to 4
= 3(4321+0+1+2+3)
= 12
Thus, the required value of the definite integral is 12
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Subject
Author
Date
Step by Step Solution
Steve Herman
20071112 18:14:03
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