I'm struggling with this one for a week:
There is a range $R$ that it's points $(x,y)$ are defined as:
For each $0 \le x \le 32$, all the values of $y$ are $\sqrt[5]{x} \le y \le 2$. We need to find:
$$\iint_{R} \sin(y^3) dA$$
Sorry for my english and math writing, it's my first question.
$$\int_0^{32} \int_{x^{1/5}}^2 \sin(y^3)\,dy\,dx=\int_0^2 \int_0^{y^5} \sin(y^3)\,dx\,dy=\int_0^{2}y^5\sin(y^3)\,dy$$ Use the substitution $y^3=t \Rightarrow 3y^2dy=dt$ to obtain: $$\int_0^{2}y^5\sin(y^3)\,dy=\frac{1}{3}\int_0^8 t\sin t\,dt$$ Solve the final integral using integration by parts to obtain: $$\boxed{\dfrac{1}{3}\left(\sin(8)-8\cos(8)\right)}$$