anonymous
  • anonymous
Evaluate the double integral. ∫∫ (y/(x^3)+2)dA, R ={(x , y) | 1≤ x≤2, 0≤y ≤2x} R Explain please.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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lalaly
  • lalaly
is this \[\frac{y}{x^3+2}\]or\[\frac{y}{x^3}+2\]??
anonymous
  • anonymous
the first one
anonymous
  • anonymous
\[y \div ( (x^3) + 2)\]

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lalaly
  • lalaly
\[\int\limits_{1}^{2}\int\limits_{0}^{2x}{\frac{y}{x^3+2}dydx}\]
lalaly
  • lalaly
first u start integrating the y and keep the x constant,, so u can make it easier for urself to integrate the x so first we start with the integral with the y limits\[\int\limits_{0}^{2x}\frac{y}{x^3+2}dy\]\[\int\limits{\frac{y}{x^3+2}dy}=\frac{1}{x^3+2}\int\limits{ydy}= \frac{1}{x^3+2}\frac{y^2}{2}\]now evaluate from 0-2x\[\int\limits_{0}^{2x}{\frac{y}{x^3+2}dy}=\frac{4x^2}{2x^3+4}\]now u find the outer integral\[\int\limits_{1}^{2}{\frac{4x^2}{2x^3+4}dx}\]just substitute u=2x^3+4 ... du=6x^2 and then find the integral at the given limits
anonymous
  • anonymous
would i bring out the 4x^2 to get 1/u du?
lalaly
  • lalaly
what do u mean bring out?
anonymous
  • anonymous
\[4x ^{2\int\limits_{1}^{2}} 1/u du\]
lalaly
  • lalaly
no u can easily divide top and bottom by 2 it will be >> 2x^2/(x^3+2) then u can factor out the 2\[\large{2\int\limits_{1}^{2}\frac{x^2}{x^3+2}dx}\]let u=x^3 du=3x^2 so dx=du/3 so the integra; becomes\[2\int\limits{\frac{1}{u}\frac{du}{3}}\]factpr pit the 1/3\[\frac{2}{3}\int\limits{\frac{du}{u}}\]
lalaly
  • lalaly
so u will get\[\frac{2}{3}\ln|u|\]substituting the x back\[\large{2\int\limits{\frac{x^2}{x^3+2}dx}=\frac{2}{3}\ln(x^3+2)}\]now just evaluate at the limits from 1 to 2
lalaly
  • lalaly
CORRECTION in my 5th post du=3x^2dx so x^2dx=du/3
anonymous
  • anonymous
how did you know to sub u = x^3 and not u = x^3+2?
lalaly
  • lalaly
sorry it was supposed to be x^3+2
anonymous
  • anonymous
why cant i write the answer as 2/3 ln (10-3) = 2/3ln 7 instead of 2/3 [ln 10 - ln 3]
anonymous
  • anonymous
oh nevermind i got it xD thank you so much! i understood everything!

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