1. **State the problem:** Find the area of the region $R$ bounded by the y-axis, the line $y=1$, and the curve $y=\sqrt[3]{x}$ by writing $x$ as a function of $y$ and integrating with respect to $y$. 2. **Rewrite the curve:** Given $y=\sqrt[3]{x}$, express $x$ in terms of $y$: $$y=\sqrt[3]{x} \implies x=y^3$$ 3. **Set up the integral:** The region $R$ is bounded by $y=0$ (y-axis), $y=1$, and $x=y^3$. The area is: $$\text{Area} = \int_0^1 x \, dy = \int_0^1 y^3 \, dy$$ 4. **Integrate:** $$\int_0^1 y^3 \, dy = \left[ \frac{y^4}{4} \right]_0^1 = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$$ 5. **Answer:** The area of the region $R$ is $\boxed{\frac{1}{4}}$. **Second question:** Determine if the statement is true or false: "If $f$ and $g$ are differentiable and $f(x) \ge g(x)$ for $a < x < b$, then $f'(x) \ge g'(x)$ for $a < x < b$." This statement is **False** because the inequality of functions does not imply the inequality of their derivatives. For example, two functions can satisfy $f(x) \ge g(x)$ but have $f'(x) < g'(x)$ at some points. Final answers: - Area of region $R$: $\frac{1}{4}$ - Statement: False