Answer
$0 \leq \int_{-2}^{3}sin~\sqrt[3] {x} \leq 1$
Work Step by Step
For all real numbers $x$:
$\sqrt[3] {-x} = -\sqrt[3] x$
Also:
$sin~(-x) = -sin~x$
Therefore:
$sin~\sqrt[3] {-x} = sin~(-\sqrt[3] {x}) = -sin~\sqrt[3] {x}$
We can see that $~~f(x) = sin \sqrt[3] {x}~~$ is an odd function.
Therefore:
$\int_{-2}^{2} sin \sqrt[3] {x} = 0$
Then:
$\int_{-2}^{3} sin \sqrt[3] {x} = \int_{-2}^{2} sin \sqrt[3] {x}+\int_{2}^{3} sin \sqrt[3] {x} = 0+\int_{2}^{3} sin \sqrt[3] {x}$
On the interval $2 \leq x \leq 3$:
$0 \leq \sqrt[3] {x} \leq \frac{\pi}{2}$
$0 \leq sin~\sqrt[3] {x} \leq 1$
Therefore:
$(0)(3-2) \leq \int_{2}^{3}sin~\sqrt[3] {x} \leq (1)(3-2)$
$0 \leq \int_{2}^{3}sin~\sqrt[3] {x} \leq 1$
$0 \leq \int_{-2}^{3}sin~\sqrt[3] {x} \leq 1$
