The problem asks to find the area enclosed by the curve $4x + 5y = 20$, the x-axis ($y = 0$), and the vertical lines $x = 1$ and $x = 4$.

Step-by-Step Solution:

1. Rewriting the curve equation: Solve for $y$: $y = \frac{20 - 4x}{5}$

2. Area under the curve: The enclosed area can be computed as the integral of $y$ with respect to $x$ from $x = 1$ to $x = 4$: $A = \int_{1}^{4} \frac{20 - 4x}{5} \, dx$

3. Simplifying the integral: Factor out $\frac{1}{5}$: $A = \frac{1}{5} \int_{1}^{4} (20 - 4x) \, dx$ Split the integral: $A = \frac{1}{5} \left( \int_{1}^{4} 20 \, dx - \int_{1}^{4} 4x \, dx \right)$

4. Evaluating the integrals:

   • For $\int_{1}^{4} 20 \, dx$: $\int_{1}^{4} 20 \, dx = 20[x]_{1}^{4} = 20(4 - 1) = 60$
   • For $\int_{1}^{4} 4x \, dx$: $\int_{1}^{4} 4x \, dx = 4 \int_{1}^{4} x \, dx = 4\left[\frac{x^2}{2}\right]_{1}^{4} = 4\left(\frac{4^2}{2} - \frac{1^2}{2}\right) = 4\left(\frac{16}{2} - \frac{1}{2}\right) = 4\left(8 - 0.5\right) = 4(7.5) = 30$

5. Combine results: Substitute back into the area formula: $A = \frac{1}{5} \left( 60 - 30 \right) = \frac{1}{5} (30) = 6$

Final Answer:

The enclosed area is: $\boxed{6 \, \text{square units}}$

Would you like further clarification or a visual graph of this problem? Here are related questions to deepen understanding:

1. What is the geometric interpretation of integrating a curve?
2. How do vertical boundaries ($x = a$, $x = b$) affect the area calculation?
3. What changes if the curve intersects the x-axis within the boundaries?
4. Can this area be solved using definite sums instead of integration?
5. How would the area change if one of the limits was modified (e.g., $x = 2$ to $x = 4$)?

Tip: Always sketch the curve and boundaries to visualize the enclosed region before setting up the integral.