anonymous
  • anonymous
find the double integral the integral from 5 to 4 the integral from 5 to 4 (4x+y)^-1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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reemii
  • reemii
- integrating \(\frac1{4x+y}\) with respect to \(y\) gives \(\ln(4x+y)\). - a primitive of \(\ln(x)\) is \(x\ln(x)- x \). This might help, I didn't do it but im confident it's what you need.
anonymous
  • anonymous
im not really sure what you did im just had my first lecture on double integrals today so im not sure how to solve it
anonymous
  • anonymous
Marry me, I'm an Asian.

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reemii
  • reemii
you can choose the order of integration. I'll fix it to this: \(\int_4^5(\int_4^5 \dots\,dy)dx\) You will first evaluate what's inside the parenthesis.
reemii
  • reemii
since it's an integration of the variable \(y\) you consider \(x\) as constant. A primitive of the function is ln(4x+y). -> the parenthesis is equal to \(\ln(4x+5) - \ln(4x+4)\). Then you can split the integrals and use "normal" integration.
anonymous
  • anonymous
so the equation you typed is my function and i first have to find the integral with respect to y and whatever i get from there i take the integral with respect to x
reemii
  • reemii
yes. in general, the parenthesis , after integrating once, still contains one variable. Here \(dy\) is in the parenthesis. After comupting this, all \(y\)'s have disappeared. only \(x\)'s remain.
anonymous
  • anonymous
ok so the integral for the above functions is 1/4x+5 - 1/4x+4 right
reemii
  • reemii
which integral? I don't think this function appears in the answer. One part of the answer is \[\int_4^5 \ln(4x+5) \,dx = \int_{21}^{25} \ln(u) \frac14\,du\\ = \left[ u\ln u - u \right]_{21}^{25} \]i think.
dan815
  • dan815
hi im here!!
reemii
  • reemii
i forgot \(\frac14\)
dan815
  • dan815
my favourite questions
dan815
  • dan815
|dw:1370581533606:dw|
anonymous
  • anonymous
where did you get the points 21 and 25 for the second integral
dan815
  • dan815
now u kindly go to www.wolframalpha.com and type this in
dan815
  • dan815
hahhahh
reemii
  • reemii
u = 4x+5. then du = 4dx. -> dx = du/4. and , when x=4, u = 21, when x=5, u=25.
anonymous
  • anonymous
ok so i substitute the equation back to u and that where i take the integral with respect to x
dan815
  • dan815
|dw:1370581743812:dw|
dan815
  • dan815
do u know what integral of lnx is ?
anonymous
  • anonymous
the integral of ln x is 1/x right
dan815
  • dan815
int(lnx) = x (-1+ln(x))
dan815
  • dan815
do by parts
dan815
  • dan815
|dw:1370582013728:dw|
reemii
  • reemii
I didnt use the fact that \(\ln(4x+5) = \ln(4x)\ln5\), instead I made a change of variables. \(u = 4x+5\). After doing that, Idon't need to revert to the variable \(x\). because I'm supposed to fund a real number as answer. After, you take the other integral (integral1 - integral2, only "dx"), and do the same work.
reemii
  • reemii
you're supposed to find a real number as answer *
anonymous
  • anonymous
ok, so then i would plug in 5 and 4 for x(ln x-1 ) for what dan815 put
dan815
  • dan815
http://www.wolframalpha.com/input/?i=int%28ln%284x%2B5%29-ln%284x%2B4%29%29dx#_=_
dan815
  • dan815
thats ur answer, check after u finish
dan815
  • dan815
no im just show u how to integrate ln x by parts u arent integrating that u have a ln(f(x))
reemii
  • reemii
@91004775 not so easily. he showed the integration for \(\ln(x)\). But you have a coefficient '\(4\)' in front of \(x\). So you must take this into account. That is why a change of variable is necessary.
dan815
  • dan815
yes this is why u must take f(x) as u and do (lnu) and sub for du in terms of dx
anonymous
  • anonymous
ok i understand i tried working it out and i got 0.044 is that correct
dan815
  • dan815
check it on wolfram lol xD
dan815
  • dan815
that is our integrate and evaluate all tool right there
reemii
  • reemii
I think it's correct.
dan815
  • dan815
yep u are right
dan815
  • dan815
http://www.wolframalpha.com/input/?i=int+from+4+to+5+%28ln%284x%2B5%29-ln%284x%2B4%29%29dx
dan815
  • dan815
see
dan815
  • dan815
0.04456...
anonymous
  • anonymous
thanks that was helpful

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