Question

Anyone know how to determine all values of p for which the integral is improper. integral dx/[(x+1)^p] from -2 to 1 ?????????

Answer 1

it is always improper since the integrand is undefined at \(x=-1\)

Answer 2

it's discontinuous at x=-1

Answer 3

yes, so it is always improper
question might be for what \(p\) is
\[\int_{-2}^1\frac{dx}{(x+1)^p}\] finite

Answer 4

so p could be all real #s?

Answer 5

oh no

Answer 6

you have to find
\[\lim_{t\to -1}\int_{-2}^t\frac{dx}{(x+1)^p}\] to see if you get a finite answer
a hint is it will depend on whether \(p>1\) or \(p<1\) try it for \(p=2\) and for \(p=\frac{1}{2}\)

Answer 7

anti derivative of \(\frac{1}{(x+1)^p}=(x+1)^{-p}\) is
\[\frac{(x+1)^{-p+1}}{-p+1}=\frac{1-p}{(x+1)^{p-1}}\]

Answer 8

wow the last row there, r the 2 sides equal? did u just inverse that? i guess p cannot be 1.

Answer 9

well i made a mistake

Answer 10

it is
\[\frac{1}{(1-p)(x+1)^{p-1}}\]

Answer 11

and the question comes down to whether, when you replace \(x\) by -1 you get zero in the denominator (doesn't exist) or a zero in the numerator

Answer 12

which comes down to whether \(1-p <0\) or \(1-p>0\)