(Double integral on D) sin[((x^3)/3) - x] dxdy = ? where(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Where is the mistake?

**Physics Forums | Science Articles, Homework Help, Discussion**