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Confused about limits in an improper integral

Hi, would somebody kindly explain how in this question

the limits on the integral with the substitution have become 1 and 0? And how do I answer the last question, given that the answer to part (b)(ii) is -1/2?

Thanks

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Reply 1

TeeEm

Original post by foorganders

Hi, would somebody kindly explain how in this question

the limits on the integral with the substitution have become 1 and 0? And how do I answer the last question, given that the answer to part (b)(ii) is -1/2?

Thanks

the reciprocal of 1 is 1
the reciprocal of infinity is zero (loosely before I get lectured...)

Reply 2

foorganders

Original post by TeeEm

the reciprocal of 1 is 1
the reciprocal of infinity is zero (loosely before I get lectured...)

Ok, why have the upper and lower limits swapped?

Reply 3

TeeEm

Original post by foorganders

Ok, why have the upper and lower limits swapped?

they used the minus from the substitution to reverse the limits

Reply 4

foorganders

Original post by TeeEm

they used the minus from the substitution to reverse the limits

I'm not sure what you mean, when I substituted x I didn't have a minus, or rather I had two.

(edited 9 years ago)

Reply 5

TeeEm

Original post by foorganders

I'm not sure what you mean, when I substituted x I didn't have a minus, or rather I had two.

the integral of f(x) with limits x=a to x=b

is the same as

minus the integral of f(x) with limits x=b to x=a

Reply 6

foorganders

Original post by TeeEm

the integral of f(x) with limits x=a to x=b

is the same as

minus the integral of f(x) with limits x=b to x=a

Ok. Is that relevant to the fact that it's a reciprocal? I still can't see why they have done that

Reply 7

TeeEm

Original post by foorganders

Ok. Is that relevant to the fact that it's a reciprocal? I still can't see why they have done that

nothing to do with reciprocals/it is always true

when find dy/dx to do the substitution you generate a minus which you can "throw" it away by reversing the limits

Reply 8

foorganders

Original post by TeeEm

nothing to do with reciprocals/it is always true

when find dy/dx to do the substitution you generate a minus which you can "throw" it away by reversing the limits

Right ok I get that but in the substitution you don't get a minus because when you substitute 1/y into the ln part you get ln(y^-2) or -ln(y^2) so the two minuses cancel

Reply 9

TeeEm

Original post by foorganders

Right ok I get that but in the substitution you don't get a minus because when you substitute 1/y into the ln part you get ln(y^-2) or -ln(y^2) so the two minuses cancel

y = 1/x

dy/dx = - 1/x²

there is a minus and nothing cancels it

Reply 10

foorganders

Original post by TeeEm

y = 1/x

dy/dx = - 1/x²

there is a minus and nothing cancels it

I know that there's a minus there, what I'm saying is that

\ln x^2 = \ln \frac{1}{y^2} = - \ln y^2

Reply 11

TeeEm

Original post by foorganders

I know that there's a minus there, what I'm saying is that

\ln x^2 = \ln \frac{1}{y^2} = - \ln y^2

nowhere does it say in the question that the (i) is equal to (ii)

Reply 12

Zacken

Original post by foorganders

I know that there's a minus there, what I'm saying is that

\ln x^2 = \ln \frac{1}{y^2} = - \ln y^2

\displaystyle \int_0^1 2y \ln y \, \mathrm{d}y = -\int_{1}^{\infty}\frac{\ln x^2}{x^3} \, \mathrm{d}x

Reply 13

foorganders

Original post by Zacken

\displaystyle \int_0^1 2y \ln y \, \mathrm{d}y = -\int_{1}^{\infty}\frac{\ln x^2}{x^3} \, \mathrm{d}x

Sorry but I'm not convinced that this is correct. Unless I've got something wrong, but my working is:

x=\dfrac{1}{y} [br][br]dx=-\dfrac {1}{y^2} dy[br]\int\dfrac{\ln x^2}{x^3}dx[br]=\int -\dfrac{\ln y^{-2}}{y^{-1}}dy[br]=\int 2y\ln ydy[br]

Reply 14

Zacken

Original post by foorganders

Sorry but I'm not convinced that this is correct. Unless I've got something wrong, but my working is:

x=\dfrac{1}{y} [br][br]dx=-\dfrac {1}{y^2} dy[br]\int\dfrac{\ln x^2}{x^3}dx[br]=\int -\dfrac{\ln y^{-2}}{y^{-1}}dy[br]=\int 2y\ln ydy[br]

Yes, I agree. I'm saying the integral from 0 to 1 of blah is NOT equal to the original integral. It never said so in the question. It is in fact equal to negative the original integral.

Reply 15

foorganders

Original post by Zacken

Yes, I agree. I'm saying the integral from 0 to 1 of blah is NOT equal to the original integral. It never said so in the question. It is in fact equal to negative the original integral.