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find the values of p for which the integral converges and evaluate the integral for those values of p. 1 ∫ 1/xp dx 0

User CMArg
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Answer:

To find the values of p for which the integral converges, we need to consider the exponent p in the integral. For the integral ∫(1/x^p) dx from 1 to 0, we need to determine the values of p that make the integral converge.

The integral ∫(1/x^p) dx converges when p > 1, as the function 1/x^p approaches 0 as x approaches infinity.

To evaluate the integral for those values of p, we can use the property of integrals:

∫(1/x^p) dx = [(x^(1-p))/(1-p)] + C

Substituting the limits of integration, we have:

∫(1/x^p) dx from 1 to 0 = [(0^(1-p))/(1-p)] - [(1^(1-p))/(1-p)]

Simplifying further, we get:

∫(1/x^p) dx from 1 to 0 = -1/(1-p) - 1/(1-p)

= -2/(1-p)

Therefore, the integral evaluates to -2/(1-p) for values of p > 1.

User Clmccomas
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