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Question

For what values of p, is the integral 1dxxp is convergent?

A
p>1
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B
p>1
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C
p<1
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D
None of these
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Solution

The correct option is D p>1
Given : 1dxxp
So to find its convergent value
1dxxp=limtt1xpdx=[x1p1p]t1=t1p1p=1(1p)
Sign of 1p is positive when p<1
And the scene will then be divergent
Sign of 1p is negative when p>1
And limtt1p1p=1(1p)=0 Convergent for (p>1)
Hence the correct answer is p>1

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