Question

1. find the area bounded by the curves $x = 3$, $y = 1$, and $y = 1/x$.

Explanation:

Step1: Determine the intersection point

Find where \( y = 1 \) and \( y=\frac{1}{x} \) intersect. Set \( 1=\frac{1}{x} \), so \( x = 1 \).

Step2: Set up the integral for area

The area is bounded on the left by \( x = 1 \) (intersection), on the right by \( x = 3 \), below by \( y=\frac{1}{x} \), and above by \( y = 1 \). The formula for the area between curves is \( A=\int_{a}^{b}(top - bottom)dx \). So \( A=\int_{1}^{3}(1-\frac{1}{x})dx \).

Step3: Integrate the function

Integrate \( 1-\frac{1}{x} \). The integral of \( 1 \) is \( x \), and the integral of \( \frac{1}{x} \) is \( \ln|x| \). So \( \int(1 - \frac{1}{x})dx=x-\ln|x|+C \).

Step4: Evaluate the definite integral

Evaluate from \( 1 \) to \( 3 \): \( [x-\ln x]_{1}^{3}=(3-\ln 3)-(1 - \ln 1) \). Since \( \ln 1 = 0 \), this simplifies to \( 3-\ln 3 - 1=2-\ln 3 \).

Answer:

The area is \( 2 - \ln 3 \) (or approximately \( 2 - 1.0986 = 0.9014 \))