1) The integral (between 4 and 1) of the integral (between 2 and 1) (x/y+y/x)dydx

2)The integral (between pi/3 and 0) of the integral (between pi/6 and 0)

xsin(x+y)dydx

Thanks!

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- Oct 28th 2009, 09:17 AMdaskywalkerMultiple integrals help
1) The integral (between 4 and 1) of the integral (between 2 and 1) (x/y+y/x)dydx

2)The integral (between pi/3 and 0) of the integral (between pi/6 and 0)

xsin(x+y)dydx

Thanks! - Oct 28th 2009, 09:38 AMtonio

1.- $\displaystyle \int\limits_1^4\int\limits_1^2 \frac{x}{y}+\frac{y}{x}\;dy\,dx=\int\limits_1^4dx\ left(\int\limits_1^2\left(\frac{x}{y}+\frac{y}{x}\ right)dy\right)=\int\limits_1^4dx\left(x\ln y+\frac{1}{2x}y^2\right)_1^2$

Well, I did the first one now you do the second one AFTER plugging in the limits (for y!) in the primitive function we got above. You get two nice, not hard integrals in x.

Try to walk now your way with problem 2 and if you get stuck somewhere in the way write back AFTER you do some work on it.

Tonio - Oct 28th 2009, 12:21 PMdaskywalker
ok got 1 now...almost got 2, I am having trouble with a +- sign somewhere...

so for 2 I got [-xcos(x+pi/3)]-[-xcos(x)] (after I evaluated the first integral between pi/3 and 0)

integrate it becomes [-xsin(x+pi/3)+cos(x+pi/3)+cos(x)+xsin(x)]

evaluate it between pi/6 and 0 I get (-pi/12)+0.5[sqrt(3)-3], but the correct answer is (-pi/12)+0.5[sqrt(3)-1]