1. We start with the problem: find the definite integral (ondersom) of the function $f(x) = x^3$ given that the integral of $f(x) = x^2$ is $\frac{1}{3}$. 2. Recall the formula for the integral of a power function: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$ where $n \neq -1$. 3. For $f(x) = x^2$, the integral is: $$\int x^2 \, dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}{3} + C$$ which matches the given result $\frac{1}{3}$ when evaluated at the appropriate limits. 4. Now, for $f(x) = x^3$, apply the same formula: $$\int x^3 \, dx = \frac{x^{3+1}}{3+1} + C = \frac{x^4}{4} + C$$ 5. This is the indefinite integral. If you want the definite integral over an interval, you substitute the limits into $\frac{x^4}{4}$. 6. Therefore, the integral (ondersom) of $f(x) = x^3$ is $$\boxed{\frac{x^4}{4} + C}$$