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To find the area of the shaded region in this problem, we need to set up an integral that represents the area between the two curves over the given interval. Here’s a step-by-step solution:

### Step 1: Identify the Equations of the Curves
The curves are:
1. \( y = 4x - x^2 \)
2. \( y = -4x^2 \)

The shaded region lies between these curves.

### Step 2: Determine the Points of Intersection
To find where these curves intersect, set them equal to each other:

\[
4x - x^2 = -4x^2
\]

Rearrange terms:

\[
5x^2 - 4x = 0
\]

Factor out \( x \):

\[
x(5x - 4) = 0
\]

This gives \( x = 0 \) and \( x = \frac{4}{5} \) as points of intersection.

### Step 3: Set Up the Integral
The area of the shaded region is given by the integral of the difference between the two functions over the interval from \( x = 0 \) to \( x = \frac{4}{5} \):

\[
\text{Area} = \int_0^{\frac{4}{5}} \left((4x - x^2) - (-4x^2)\right) \, dx
\]

Simplify the integrand:

\[
= \int_0^{\frac{4}{5}} (4x - x^2 + 4x^2) \, dx
\]

\[
= \int_0^{\frac{4}{5}} (4x + 3x^2) \, dx
\]

### Step 4: Integrate
Now, integrate term by term:

\[
\int_0^{\frac{4}{5}} (4x + 3x^2) \, dx = \int_0^{\frac{4}{5}} 4x \, dx + \int_0^{\frac{4}{5}} 3x^2 \, dx
\]

1. **Integrate \( 4x \):**

   \[
   \int 4x \, dx = 2x^2
   \]

2. **Integrate \( 3x^2 \):**

   \[
   \int 3x^2 \, dx = x^3
   \]

Thus,

\[
= \left[ 2x^2 + x^3 \right]_0^{\frac{4}{5}}
\]

### Step 5: Evaluate the Integral
Substitute \( x = \frac{4}{5} \):

\[
= 2\left(\frac{4}{5}\right)^2 + \left(\frac{4}{5}\right)^3
\]

\[
= 2 \cdot \frac{16}{25} + \frac{64}{125}
\]

Convert to a common denominator:

\[
= \frac{32}{25} + \frac{64}{125} = \frac{160 + 64}{125} = \frac{224}{125} = \frac{10}{3}
\]

### Final Answer
The area of the shaded region is:

\[
\frac{10}{3} \text{ square units}
\]

Thus, the correct answer is:

**(d) \( \frac{10}{3} \)**.

---

Would you like more details or have any questions about this solution?

### Related Questions:
1. How do you determine the points of intersection between two curves?
2. What is the general process for finding the area between two curves?
3. How do you integrate polynomials?
4. How do you handle integrals with absolute values in areas between curves?
5. What are some applications of finding areas between curves in real-world problems?

### Tip:
Always check for symmetry in graphs, as it can simplify the calculation of areas by allowing you to compute over half the region and then double the result if needed.

The area of the shaded region = ............. square units in the figure where y = -4x^2 and y = 4x - x^2.

This problem guides you in finding the area of the shaded region between two curves, y = -4x^2 and y = 4x - x^2, using integral calculus. The step-by-step solution includes setting up the integral, finding points of intersection, and evaluating the area. Suitable for Grades 11-12.

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