Find the area of the region in the first quadrant bounded by the line \( y=2 x \), the line \( x=5 \), the curve \( y=\frac{2}{x} \), and the \( x \)-axis. The total area of the region is (Type an exact answer, using radicals as needed.)

To find the area of the region bounded by the lines and the curve in the first quadrant, we first need to identify the points of intersection of the lines and the curve. 1. **Intersection of \( y = 2x \) and \( y = \frac{2}{x} \)**: \[ 2x = \frac{2}{x} \implies 2x^2 = 2 \implies x^2 = 1 \implies x = 1 \] (We only take the positive solution since we are in the first quadrant.) Substituting \( x = 1 \) into \( y = 2x \): \[ y = 2(1) = 2 \] So, the intersection point is \( (1, 2) \). 2. **Intersection of \( y = \frac{2}{x} \) and the line \( x = 5 \)**: Substitute \( x = 5 \) into \( y = \frac{2}{x} \): \[ y = \frac{2}{5} \] So, the point is \( (5, \frac{2}{5}) \). 3. **Now we will compute the area between the curves from \( x = 1 \) to \( x = 5 \)**. We need to find the area between \( y = \frac{2}{x} \) and the line \( y = 2x \) until their intersection at \( x = 1 \), then the area bounded by \( y = \frac{2}{x} \) and the line \( x = 5\). 4. **Area from \( x = 1 \) to \( x = 5 \)**: - From \( x = 1 \) to \( x = 5 \), we have the area under \( y = 2x \) and \( y = \frac{2}{x} \). - Area \(A_1\) below \(y = 2\) from \(x = 1\) to \(x = 5\): \[ A_1 = \int_1^5 2x \, dx = [x^2]_1^5 = 5^2 - 1^2 = 25 - 1 = 24. \] - Area \(A_2\) below \(y = \frac{2}{x}\) from \(x = 1\) to \(x = 5\): \[ A_2 = \int_1^5 \frac{2}{x} \, dx = [2\ln|x|]_1^5 = 2\ln(5) - 2\ln(1) = 2\ln(5) - 0 = 2\ln(5). \] 5. **Total area**: The total area \(A\) between these curves is given by: \[ A = A_1 - A_2 = 24 - 2\ln(5). \] Thus, the total area of the region is: \[ \boxed{24 - 2\ln(5)}. \]