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Calculus
One, Two, Three, Floor Integral (
Posted on 2007-11-12
)
Evaluate:
4
∫
[t + [t + [t]]] dt
-4
where [p] is the greatest integer ≤ p.
Submitted by
K Sengupta
Rating:
4.0000
(1 votes)
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(-4-3-2-1+0+1+2+3)
= -12
Thus, the required value of the definite integral is -12
Comments: (
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Subject
Author
Date
Step by Step Solution
Steve Herman
2007-11-12 18:14:03
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