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Evaluate integral of floor(log₂(log₃(x))) between 3 and 81.

User LBarret
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Final answer:

To evaluate the integral of floor(log₂(log₃(x))) between 3 and 81, we find the values of log₃(x) for the given range and evaluate floor(log₂(log₃(x))) based on those values.

Step-by-step explanation:

To evaluate the integral of floor(log₂(log₃(x))) between 3 and 81, we need to first find the values of log₃(x) for x in the given range. Since log₃(x) > 0 for x > 1, we can evaluate floor(log₂(log₃(x))) as 0 if log₂(log₃(x)) < 0. Otherwise, we round down log₂(log₃(x)) to the nearest integer.

For x = 3, log₃(3) = 1 and log₂(1) = 0, so the value of floor(log₂(log₃(3))) is 0. For x = 81, log₃(81) = 4 and log₂(4) = 2, so the value of floor(log₂(log₃(81))) is 2.

Therefore, the integral of floor(log₂(log₃(x))) between 3 and 81 is equal to 0 + 0 + 0 + 0 + 2 = 2.

User Kemar
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