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91Ó°ÊÓ

Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.

Ifp≥1,then∫011xpdxdiverges.

Short Answer

Expert verified

The given statement is proved.

Step by step solution

01

Step 1. Given Information.

The given integral is∫011xpdx.

02

Step 2. Prove. 

To prove if p≥1,then ∫011xpdx diverges, let aand bbe any real numbers and f(x) be a function, and if fcontinuous on role="math" localid="1649068637899" (a,b],but not x = a, then ∫abf(x)dx=limA→a+∫Abf(x)dx.

So,

∫011xpdx=limA→0+∫A1x-pdx=limA→0+x-p+1-p+1A1=limA→0+11-p-11-pA1-p=11-p+11-p0+=∞p≥1

Thus, p≥1and integral diverges on [0, 1].

If p≥1,then 1xp≥1xforx∈[0,1]and∫011xpdxdiverges.

Hence, the integral is proved.

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