Question

Evaluate the double integral.

∫∫ (y/(x^3)+2)dA, R ={(x , y) | 1≤ x≤2, 0≤y ≤2x}
R

Explain please.

Answer 1

is this \[\frac{y}{x^3+2}\]or\[\frac{y}{x^3}+2\]??

Answer 2

the first one

Answer 3

\[y \div ( (x^3) + 2)\]

Answer 4

\[\int\limits_{1}^{2}\int\limits_{0}^{2x}{\frac{y}{x^3+2}dydx}\]

Answer 5

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

Answer 6

would i bring out the 4x^2 to get 1/u du?

Answer 7

what do u mean bring out?

Answer 8

\[4x ^{2\int\limits_{1}^{2}} 1/u du\]

Answer 9

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\[\frac{2}{3}\int\limits{\frac{du}{u}}\]

Answer 10

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

Answer 11

CORRECTION in my 5th post
du=3x^2dx
so x^2dx=du/3

Answer 12

how did you know to sub u = x^3 and not u = x^3+2?

Answer 13

sorry it was supposed to be x^3+2

Answer 14

why cant i write the answer as 2/3 ln (10-3) = 2/3ln 7 instead of 2/3 [ln 10 - ln 3]

Answer 15

oh nevermind i got it xD
thank you so much! i understood everything!