To find the area between the curves \( y = 4x \) and \( y = x^2 \) from \( x = 0 \) to \( x = 4 \), we need to evaluate the integral:

\[
\text{Area} = \int_{0}^{4} \left( 4x - x^2 \right) \, dx
\]

### Step-by-Step Calculation:
1. **Integrate \( 4x - x^2 \):**
   \[
   \int \left( 4x - x^2 \right) \, dx = \int 4x \, dx - \int x^2 \, dx
   \]
   - For \( \int 4x \, dx \):
     \[
     \int 4x \, dx = 2x^2
     \]
   - For \( \int x^2 \, dx \):
     \[
     \int x^2 \, dx = \frac{x^3}{3}
     \]
   - So, the indefinite integral is:
     \[
     \int \left( 4x - x^2 \right) \, dx = 2x^2 - \frac{x^3}{3}
     \]

2. **Evaluate the definite integral from 0 to 4:**
   \[
   \text{Area} = \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4}
   \]
   - Plug in \( x = 4 \):
     \[
     2(4)^2 - \frac{(4)^3}{3} = 2(16) - \frac{64}{3} = 32 - \frac{64}{3}
     \]
     Convert 32 to a fraction:
     \[
     32 = \frac{96}{3}
     \]
     So:
     \[
     32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3}
     \]

   - Plug in \( x = 0 \):
     \[
     2(0)^2 - \frac{(0)^3}{3} = 0
     \]

   - Therefore, the area is:
     \[
     \text{Area} = \frac{32}{3} \text{ square units}
     \]

### Final Answer:
The area between the curves \( y = 4x \) and \( y = x^2 \) from \( x = 0 \) to \( x = 4 \) is \( \frac{32}{3} \) square units.

Would you like to explore any related topics or ask additional questions?

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**Follow-up Questions:**
1. How do you determine whether a curve is above or below another within a specific interval?
2. What is the significance of the limits of integration in determining the area?
3. Can the area between two curves ever be negative?
4. How would you approach finding the area if the functions were more complex, like trigonometric functions?
5. How does symmetry of the curves affect the calculation of the area?

**Tip:** When evaluating definite integrals, always check for opportunities to simplify the function before integrating to make the process easier.

Learn how to calculate the area between the curves y = 4x and y = x^2 using definite integrals. This problem involves integrating polynomial functions and applying the Fundamental Theorem of Calculus. Suitable for advanced high school and college students.

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