To calculate the area under the curve of $f(x) = 4x - x^2$ from $x = 0$ to $x = 4$, we use the definite integral:

$\text{Area} = \int_{0}^{4} (4x - x^2) \, dx$

Step 1: Integrate $f(x)$

The integral of $4x - x^2$ is:

$\int (4x - x^2) \, dx = 2x^2 - \frac{x^3}{3} + C$

Step 2: Apply the limits of integration

Evaluate the definite integral from $0$ to $4$:

$\int_{0}^{4} (4x - x^2) \, dx = \left[ 2x^2 - \frac{x^3}{3} \right]_0^4$

Step 3: Substitute the limits

Substitute $x = 4$ and $x = 0$:

1. When $x = 4$: $2(4^2) - \frac{4^3}{3} = 2(16) - \frac{64}{3} = 32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}$

2. When $x = 0$: $2(0^2) - \frac{0^3}{3} = 0$

Step 4: Compute the final result

$\text{Area} = \frac{32}{3} - 0 = \frac{32}{3}$

Final Answer:

$\text{Area} = \frac{32}{3} \, \text{square units.}$

Would you like a detailed explanation of any step?

Here are 5 related questions for practice:

1. How do you compute the area under $f(x) = x^2$ from $x = 1$ to $x = 3$?
2. What is the geometric interpretation of a definite integral?
3. How does the Fundamental Theorem of Calculus apply to definite integrals?
4. What happens if the graph dips below the x-axis in an integration problem?
5. Can definite integrals be solved numerically if an antiderivative is unavailable?

Tip: Always sketch the graph of the function to confirm that the area computed is consistent with the problem's geometry!