The integral taken over R, the area bounded by curves y = 0, x = 1, and y = x. The function we are integrating is f(x,y) = $sin(x^{1000}y)$.
So the integral is $\int_0^1 \int_0^x sin(x^{1000}y)dydx$
$= \int_0^1 \frac{1-cos(x^{1001})}{x^{1000}}$
But I have been having trouble getting any further than that. I know that the integrand here is less than or equal to 1-cos(1) but that does not help in proving either of the inequalities I have been asked to prove.
Which reminds me, I am asked to prove that the integral is less than or equal to both $\frac{sin(1)}{2}$ and $\frac{1}{2004}$.
I would greatly appreciate any help or hints, even if it is just in proving one of the two inequalities.
Start with simple estimates instead of evaluating complex explicit integral. We have that $$ 0\le \sin(x^{1000}y)\le \sin(1),\quad \sin(x^{1000}y)\le x^{1000}y\tag{*} $$ by monotonicity of $t\mapsto \sin t$ on $[0,1]$ and the fact that $\sin t\le t$ for all $t\ge 0$. From $(*)$, it follows that $$ 0\le I\le \int_0^1\int_0^x \sin(1)\mathrm dy\mathrm dx =\frac{\sin(1)}{2} $$ and $$ 0\le I\le \int_0^1\int_0^x x^{1000}y \ \mathrm dy\mathrm dx =\frac1{2006}. $$ This gives the estimate as wanted.