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Why does it make sense that 011xpdxdiverges when p1? Consider how 1xpcompares with 1xin this case.

Short Answer

Expert verified

For p1,1xpis greater than 1xin the interval [0,1], whose improper integral on[0,1] is known to diverge.

Step by step solution

01

Step 1. Given information.    

We are given,

011xpdx

02

Step 2. Comparing the integral

When the integral 011xpdxis solved in the interval p1, it gives an result of . Consider 1x, since p=1.

Then,

011xdx=[lnx]01=ln1-ln0=

So, the integral is diverges.

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