1. **Stating the problem:** Calculate the definite integral $$\int_0^2 x^3 \, dx$$ using the Fundamental Theorem of Calculus. 2. **Formula and rules:** The Fundamental Theorem of Calculus states that if $$F(x)$$ is an antiderivative of $$f(x)$$, then $$$\int_a^b f(x) \, dx = F(b) - F(a)$$$ 3. **Find the antiderivative:** For $$f(x) = x^3$$, use the power rule for integration: $$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$$ So, $$$F(x) = \frac{x^{4}}{4}$$$ 4. **Evaluate the definite integral:** $$$\int_0^2 x^3 \, dx = F(2) - F(0) = \frac{2^{4}}{4} - \frac{0^{4}}{4} = \frac{16}{4} - 0 = 4$$$ 5. **Explanation:** We found the antiderivative of $$x^3$$, then substituted the upper and lower limits and subtracted to get the area under the curve from 0 to 2. **Final answer:** $$4$$