To find the area of the region in the first quadrant bounded by the lines $y = 4x$, $x = 5$, the curve $y = \frac{4}{x}$, and the $x$-axis, we need to follow these steps:

Step 1: Identify the points of intersection

First, find the points where the line $y = 4x$ intersects the curve $y = \frac{4}{x}$.

Set $4x = \frac{4}{x}$.

Multiply both sides by $x$ to get: $4x^2 = 4$ $x^2 = 1$ x = 1 \quad \text{(since we are in the first quadrant, x > 0)} So, the point of intersection is $(1, 4)$.

Step 2: Set up the integrals

The region is bounded between $x = 1$ and $x = 5$. The total area is the sum of the areas under the curve $y = \frac{4}{x}$ from $x = 1$ to $x = 5$, minus the area under the line $y = 4x$ over the same interval.

Area under the curve $y = \frac{4}{x}$: $\text{Area}_1 = \int_{1}^{5} \frac{4}{x} \, dx$

Area under the line $y = 4x$: $\text{Area}_2 = \int_{1}^{5} 4x \, dx$

Step 3: Compute the integrals

1. Integral of $\frac{4}{x}$: $\text{Area}_1 = 4 \int_{1}^{5} \frac{1}{x} \, dx = 4[\ln(x)]_{1}^{5} = 4(\ln(5) - \ln(1)) = 4 \ln(5)$ (Since $\ln(1) = 0$)

2. Integral of $4x$: [ \text{Area}2 = 4 \int{1}^{5} x , dx = 4\left[\frac{x^2}{2}\right]_{1}^{5} = 4\left[\frac{25}{2} - \frac{1}{2}\right] = 4 \times \frac{24}{2} = 48 ]

Step 4: Subtract the areas

The total area is: $\text{Total Area} = \text{Area}_1 - \text{Area}_2 = 4 \ln(5) - 48$

Final Answer:

The total area of the region is $4 \ln(5) - 48$.

Would you like any further details or have any questions?

Here are some related questions for further practice:

1. How do you find the area between two curves?
2. How do you interpret the result when the area comes out negative?
3. Can you calculate the area if the curve or line is in the second quadrant?
4. How do you compute definite integrals involving logarithms?
5. What happens if the limits of integration include a point where the function is undefined?

Tip: Always visualize the region before setting up integrals, as it helps in correctly identifying the limits and the functions to integrate.