# Where is the mistake?

1. Jul 17, 2007

### engin

(Double integral on D) sin[((x^3)/3) - x] dxdy = ? where
D={(x,y): 1<=y<=4 , sqrt(y)<=x<=2 }.

Okay, we change the order of integration and then we get

(Double integral on D') sin [((x^3)/3) - x] dxdy where
D'={(x,y): 1<=x<=2 , 1<=y<=(x^2). Thus, we get the one variable integral

(Integral from x=1 to x=2) (x^2 - 1)* sin[((x^3)/3) - x]dx.

Letting u = ((x^3)/3) converts the integral to

(Integral from u=-2/3 to u=2/3) sin(u)du = 0. So where is the mistake?

Last edited: Jul 17, 2007
2. Jul 17, 2007

### CompuChip

Isn't this correct?

3. Jul 17, 2007

### nicktacik

Shouldn't the substitution be u = x^3/3 - x? Which would make the bounds in u space be -2/3 to 5/3?

Edit: Nevermind, -2/3 to 2/3 is the correct bounds.

Last edited: Jul 17, 2007
4. Jul 17, 2007

### engin

Pardon, the substitution is of course u = x^3/3 - x but i have found the integral 0. The answer in the worksheet is different though.

Last edited: Jul 17, 2007
5. Jul 17, 2007

### nicktacik

Woops... I see no problem with your work, and it looks like the integral should indeed be 0.