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Iterated integral

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,

\(\displaystyle \int_4^1\int_1^2 \frac{x}{y}+\frac{y}{x}dydx\)
My progress:
\(\displaystyle \int_1^4[x\ln(y)+\frac{y^2}{2x}]_1^2\)
\(\displaystyle \int_1^4 x\ln(2)+\frac{2}{x}-\frac{1}{2x}dx\)
\(\displaystyle [\frac{x^2\ln(2)}{2}+2\ln(2)-\ln(2x)]_1^4\)
What I am doing wrong?

Regards,
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: iterated integral

Hello MHB,

\(\displaystyle \int_4^1\int_1^2 \frac{x}{y}+\frac{y}{x}dydx\)
My progress:
\(\displaystyle \int_1^4[x\ln(y)+\frac{y^2}{2x}]_1^2\)
\(\displaystyle \int_1^4 x\ln(2)+\frac{2}{x}-\frac{1}{2x}dx\)
\(\displaystyle [\frac{x^2\ln(2)}{2}+2\ln(2)-\ln(2x)]_1^4\)
What I am doing wrong?

Regards,

Edit: Can't change subject, I did misspelled 'Iterated integral'
How did you integrate \(\displaystyle \frac {1}{2x}\) ?
 

Petrus

Well-known member
Feb 21, 2013
739
Re: iterated integral

How did you integrate \(\displaystyle \frac {1}{2x}\) ?
Typo should be \(\displaystyle \frac{ln(2x)}{2}\)

- - - Updated - - -

Ohh I see..... I antiderivated it wrong.... Should be \(\displaystyle \frac{ln(x)}{2}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: iterated integral

Typo should be \(\displaystyle \frac{ln(2x)}{2}\)

- - - Updated - - -

Ohh I see..... I antiderivated it wrong.... Should be \(\displaystyle \frac{ln(x)}{2}\)
They are actually the same , can you see why ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: iterated integral

\(\displaystyle [\frac{x^2\ln(2)}{2}+2\ln(2)-\ln(2x)]_1^4\)
Also you should have got \(\displaystyle 2\ln(x) \) not \(\displaystyle 2\ln (2) \)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: iterated integral

They are actually the same , can you see why ?
I think I need tea.. Yeah I see they are same
\(\displaystyle \frac{ln(2x)}{2}\) if we derivate we get
\(\displaystyle \frac{\frac{2}{2x}}{2}<=>\frac{1}{2x}\)
\(\displaystyle \frac{ln(x)}{2}\) if we derivate we get \(\displaystyle \frac{\frac{1}{x}}{2} <=> \frac{1}{2x}\)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: iterated integral

Do I got more misstake that I can't see cause I get the answer and facit says \(\displaystyle \frac{21}{2}\ln(2)\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: iterated integral

This is related to the constant of integration

\(\displaystyle \int \frac{1}{2x} dx = \frac { \ln (2x) }{2} + c_1 = \frac { \ln (2) +\ln (x) }{2} + c_1\)

Now assume that \(\displaystyle c= \frac { \ln (2) }{2} + c_1\)

Hence

\(\displaystyle \int \frac{1}{2x} dx = \frac { \ln (x) }{2} + c\)
 

Petrus

Well-known member
Feb 21, 2013
739
Re: iterated integral

This is related to the constant of integration

\(\displaystyle \int \frac{1}{2x} dx = \frac { \ln (2x) }{2} + c_1 = \frac { \ln (2) +\ln (x) }{2} + c_1\)

Now assume that \(\displaystyle c= \frac { \ln (2) }{2} + c_1\)

Hence

\(\displaystyle \int \frac{1}{2x} dx = \frac { \ln (x) }{2} + c\)
Thanks!But have I antiderivated correct:)?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: iterated integral

Do I got more misstake that I can't see cause I get the answer and facit says \(\displaystyle \frac{21}{2}\ln(2)\)
Try evaluating it by hand , it should be the same .