MHB MHBIntegral Calculation Help: \int_4^1\int_1^2

[Petrus]

Hello MHB,

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

Regards,

 

[alyafey22]

Re: iterated integral

Hello MHB,

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

Regards,

Edit: Can't change subject, I did misspelled 'Iterated integral'

How did you integrate $$\frac {1}{2x}$$ ?

 

[Petrus]

Re: iterated integral

How did you integrate $$\frac {1}{2x}$$ ?

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

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

 

[alyafey22]

Re: iterated integral

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

- - - Updated - - -

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

They are actually the same , can you see why ?

 

[alyafey22]

Re: iterated integral

$$[\frac{x^2\ln(2)}{2}+2\ln(2)-\ln(2x)]_1^4$$

Also you should have got $$2\ln(x) $$ not $$2\ln (2) $$

 

[Petrus]

Re: iterated integral

They are actually the same , can you see why ?

I think I need tea.. Yeah I see they are same
$$\frac{ln(2x)}{2}$$ if we derivate we get
$$\frac{\frac{2}{2x}}{2}<=>\frac{1}{2x}$$
$$\frac{ln(x)}{2}$$ if we derivate we get $$\frac{\frac{1}{x}}{2} <=> \frac{1}{2x}$$

 

[Petrus]

Re: iterated integral

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

 

[alyafey22]

Re: iterated integral

This is related to the constant of integration

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

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

Hence

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

 

[Petrus]

Re: iterated integral

This is related to the constant of integration

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

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

Hence

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

Thanks!But have I antiderivated correct:)?

 

[alyafey22]

Re: iterated integral

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

Try evaluating it by hand , it should be the same .