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k0ka [10]
3 years ago
11

g find all the values of p for which the following functions are improperly integrable on the indicated domain. be sure to prove

your claims, ie show that it is improperly integrable for those numbers p, but is not for other numbers 1.f(x)=1/x^p on i =(0,1) 2. f(x) = 1/(x(lnx)^p) on I = (e,inf)
Mathematics
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

1. In order to make the integral improper  p  must be 1.

2. In order to make the integral improper  p  must be 1.  

Step-by-step explanation:

Using the rules of integration we get that for

f(x)= \frac{1}{x^p}

\int\limits_{0}^{1}  \frac{1}{x^p}  \, dx  = \frac{x^{1-p}}{(1-p)}  \, |\limits_{0}^{1}   = \frac{1}{1-p} - 0 = \frac{1}{1-p}

Therefore in order to make that integral improper  p  must be 1.

If  p = 1     then you would have a 1/0  indeterminate form.

2.   Using the of integration, specifically substitution we get that for

f(x) = \frac{1}{x(ln(x)^p)}

\int\limits_{e}^{\infty} \frac{1}{x(ln(x))^p} \, dx  = \frac{(ln(x))^{1-p}}{1-p} \, |\limits_{e}^{\infty}

For   p \geq 1  we would have

\int\limits_{e}^{\infty} \frac{1}{x(ln(x))^p} \, dx  = \frac{1}{p-1}

And the problem is the same.  If  p=1   we would have a 1/0 indeterminate form.

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Answer it and you get points
fgiga [73]

Answer:

79 (first blank)

131 (second blank)

157 (last blank)

Step-by-step explanation:

92-13 to get the first blank, and you get 79.

118+13 to get the second blank, and you get 131.

144+13 to get the last blank and you get 157.

----------------------------------------------------------------------------------------------------------

Check your work:

79+13=92

92+13=105

105+13=118

118+13=131

131+13=144

144+13=157

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I hope this helps!

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This is the correct solution.

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