1. \( y=\frac{1}{2 x-1} \quad y=0 \quad x=1 \quad x=5 \quad s=? \)

We need to find the area \( s \) of the region bounded by the curve \[ y = \frac{1}{2x-1}, \] the line \[ y=0, \] and the vertical lines \[ x=1 \quad \text{and} \quad x=5. \] **Step 1. Set Up the Integral** The area \( s \) is given by the integral \[ s=\int_{x=1}^{x=5} \frac{1}{2x-1}\,dx. \] **Step 2. Use a Substitution** Let \[ u=2x-1. \] Then the differential \( du \) is \[ du=2\,dx \quad\Longrightarrow\quad dx=\frac{du}{2}. \] We also need to change the limits: - When \( x=1 \): \[ u=2(1)-1=1. \] - When \( x=5 \): \[ u=2(5)-1=9. \] **Step 3. Rewrite and Evaluate the Integral** Substitute into the integral: \[ s=\int_{u=1}^{u=9} \frac{1}{u}\cdot\frac{du}{2} = \frac{1}{2}\int_{u=1}^{u=9}\frac{du}{u}. \] The integral \( \int \frac{du}{u} \) is the natural logarithm: \[ \int_{u=1}^{u=9}\frac{du}{u} = \ln|u|\Big|_{1}^{9} = \ln(9) - \ln(1). \] Since \( \ln(1)=0 \), we have: \[ \int_{u=1}^{u=9}\frac{du}{u} = \ln 9. \] Thus, the area is \[ s=\frac{1}{2}\ln 9. \] **Step 4. Simplify the Result** We can simplify the result by recognizing that \[ \frac{1}{2}\ln 9 = \ln\left(9^{\frac{1}{2}}\right)=\ln 3. \] **Final Answer** \[ s = \ln 3. \]