Q.
If the value of the definite integral A=∫010π[sinx]dx is equal to kπ, then the absolute value of k is equal to (where, [⋅] is the greatest integer function)
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NTA AbhyasNTA Abhyas 2020Integrals
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Answer: 5
Solution:
As period of sinx is 2π, I=(5)∫02π[sinx]dx I=(5)(∫0π0dx+∫π2π(−1)dx) =5[−x]π2π =−5π ∴k=−5⇒∣k∣=5