To find the total area of the region bounded by the given boundaries, we need to divide the problem into parts based on the functions and integrate accordingly.

Step 1: Define the region

The boundaries of the region are:

1. $y = 4x$ (a line passing through the origin with slope 4).
2. $x = 4$ (a vertical line at $x = 4$).
3. $y = \frac{4}{x}$ (a hyperbolic curve).
4. The $x$-axis.

Step 2: Identify intersection points

The region is in the first quadrant. We'll find the intersections of the curves:

1. Intersection of $y = 4x$ and $x = 4$: Substituting $x = 4$ into $y = 4x$: $y = 4(4) = 16.$ So, the intersection point is $(4, 16)$.

2. Intersection of $y = 4x$ and $y = \frac{4}{x}$: Set $4x = \frac{4}{x}$: $4x^2 = 4 \implies x^2 = 1 \implies x = 1 \quad (\text{since } x > 0).$ Substituting $x = 1$ into $y = 4x$: $y = 4(1) = 4.$ So, the intersection point is $(1, 4)$.

3. Intersection of $y = \frac{4}{x}$ and $x = 4$: Substituting $x = 4$ into $y = \frac{4}{x}$: $y = \frac{4}{4} = 1.$ So, the intersection point is $(4, 1)$.

Step 3: Divide the region

The region is split into two parts:

1. From $x = 1$ to $x = 4$, bounded by $y = 4x$ (upper curve) and $y = \frac{4}{x}$ (lower curve).
2. A triangular region formed by $y = 4x$, $x = 4$, and the $x$-axis.

Part 1: Area between $y = 4x$ and $y = \frac{4}{x}$ for $x \in [1, 4]$

The area is given by: $A_1 = \int_{1}^{4} \left[ (4x) - \left(\frac{4}{x}\right) \right] \, dx.$ Simplify the integral: $A_1 = \int_{1}^{4} 4x \, dx - \int_{1}^{4} \frac{4}{x} \, dx.$ Compute each term:

1. $\int_{1}^{4} 4x \, dx = 4 \int_{1}^{4} x \, dx = 4 \left[\frac{x^2}{2}\right]_1^4 = 4 \left(\frac{16}{2} - \frac{1}{2}\right) = 4 \cdot \frac{15}{2} = 30.$
2. $\int_{1}^{4} \frac{4}{x} \, dx = 4 \int_{1}^{4} \frac{1}{x} \, dx = 4 [\ln x]_1^4 = 4 (\ln 4 - \ln 1) = 4 \ln 4.$

So, $A_1 = 30 - 4 \ln 4.$

Part 2: Area of the triangle

The triangle is formed by the line $y = 4x$, the line $x = 4$, and the $x$-axis. The base is along $x = 4$, and the height is the $y$-coordinate of the point $(4, 16)$, which is 16.

The area of the triangle is: $A_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 16 = 32.$

Step 4: Total area

The total area is: $\text{Total Area} = A_1 + A_2 = (30 - 4 \ln 4) + 32 = 62 - 4 \ln 4.$

Final Answer:

$\text{Total Area} = 62 - 4 \ln 4.$

Let me know if you want further details or explanations!

---

Relative Questions:

1. What are the steps for setting up integrals for bounded regions?
2. How do you find intersection points of curves?
3. What is the significance of logarithmic terms in integration?
4. How do you determine the area of regions involving a line and a hyperbola?
5. Can we use numerical methods to approximate areas for more complex regions?

Tip: Always sketch the region to visualize and confirm the boundaries before setting up integrals.