1. **Problem statement:** Find the value of $a > 0$ such that the area bounded by the curve $y = x^3$, the line $y = 0$, and the vertical line $x = a$ is 4 square units. 2. **Formula and approach:** The area under the curve $y = x^3$ from $x=0$ to $x=a$ above the $x$-axis is given by the definite integral: $$\text{Area} = \int_0^a x^3 \, dx$$ 3. **Calculate the integral:** $$\int_0^a x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^a = \frac{a^4}{4} - 0 = \frac{a^4}{4}$$ 4. **Set the area equal to 4 and solve for $a$:** $$\frac{a^4}{4} = 4$$ Multiply both sides by 4: $$a^4 = 16$$ 5. **Solve for $a$:** Since $a > 0$, take the positive fourth root: $$a = \sqrt[4]{16} = 2$$ 6. **Answer:** The value of $a$ is 2. Therefore, the correct choice is (c) 2.