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To solve the problem of finding the area of the shaded region bounded below by the function \( p(x) \) and above by the function \( q(x) \), we can use the following steps:

1. **Identify the functions**:
   \[
   p(x) = \frac{4}{49}x^2 - \frac{2}{7}x + 1
   \]
   \[
   q(x) = -\frac{2}{7}x + 1
   \]

2. **Set up the integral**:
   The area \( A \) of the region between the two curves from \( x = a \) to \( x = b \) is given by:
   \[
   A = \int_{a}^{b} [q(x) - p(x)] \, dx
   \]
   First, we need to find the points of intersection of the two functions, which occur where \( p(x) = q(x) \).

3. **Find the intersection points**:
   Set \( p(x) = q(x) \):
   \[
   \frac{4}{49}x^2 - \frac{2}{7}x + 1 = -\frac{2}{7}x + 1
   \]
   Simplify by subtracting \( -\frac{2}{7}x + 1 \) from both sides:
   \[
   \frac{4}{49}x^2 = 0
   \]
   This simplifies to:
   \[
   x^2 = 0
   \]
   Thus, \( x = 0 \).

4. **Calculate the area**:
   Since the intersection is at \( x = 0 \), and the curves intersect at a single point, we need to check the limits of integration over a reasonable interval. Let's calculate the area from \( x = 0 \) to \( x = 1 \), assuming these are the appropriate bounds based on the graph:

   \[
   A = \int_{0}^{1} \left[-\frac{2}{7}x + 1 - \left(\frac{4}{49}x^2 - \frac{2}{7}x + 1\right)\right] dx
   \]
   Simplify the integrand:
   \[
   A = \int_{0}^{1} \left[-\frac{4}{49}x^2\right] dx
   \]
   Evaluate the integral:
   \[
   A = -\frac{4}{49} \int_{0}^{1} x^2 \, dx
   \]
   \[
   \int x^2 \, dx = \frac{x^3}{3}
   \]
   So:
   \[
   A = -\frac{4}{49} \left[\frac{1^3}{3} - \frac{0^3}{3}\right]
   \]
   \[
   A = -\frac{4}{49} \times \frac{1}{3} = -\frac{4}{147}
   \]
   Since area cannot be negative, take the absolute value:
   \[
   A = \frac{4}{147} \approx 0.0272
   \]

So, the area of the shaded region is approximately **0.0272 square units**.

Do you have any questions or would you like a detailed breakdown of the steps? 

Here are some related questions for further exploration:

1. What happens if you integrate from \( x = 0 \) to another upper bound?
2. How do you find the intersection points if there are two distinct points?
3. What is the general form of finding the area between any two curves?
4. Can you calculate the area using a different method such as subtracting definite integrals?
5. What if \( p(x) \) and \( q(x) \) had higher powers of \( x \)?
6. How does symmetry affect the calculation of the area?
7. How do you find the centroid of the shaded area?
8. How does this area change if \( q(x) \) is a parabola instead of a linear function?

**Tip:** Always check the points of intersection first when dealing with areas bounded by curves.

Learn how to calculate the area between two curves using integral calculus with a step-by-step example problem and solution. This problem involves finding the area of the shaded region bounded below by the function \( p(x) = \frac{4}{49}x^2 - \frac{2}{7}x + 1 \) and above by the function \( q(x) = -\frac{2}{7}x + 1 \), from \( x = 0 \) to \( x = 1 \). The area is approximately 0.0272 square units.

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