Question

The area of the region enclosed by the curves \(y = x\), \(x = e\), \(y = \frac{1}{x}\) and the positive X-axis, is

• A. \(\frac{1}{2}\) sq unit
• B. 1 sq unit
• C. \(\frac{3}{2}\) sq units
• D. \(\frac{5}{2}\) sq units

Hint:

Draw the curves to identify the correct region boundaries.

Answer 1

The Correct Option is C

 To find the area of the region enclosed by the curves \(y = x\), \(x = e\), \(y = \frac{1}{x}\), and the positive X-axis, we will follow these steps:

1. Identify the points of intersection for the curves \(y = x\) and \(y = \frac{1}{x}\) within the given domain.
2. Calculate the area under each curve from their respective intersection points by integrating.
3. Subtract the smaller area from the larger area to find the enclosed area.

Step 1: Find the points of intersection:

We solve the equation \(y = x\) and \(y = \frac{1}{x}\) to find the point of intersection:

\(x = \frac{1}{x}\)

\(x^2 = 1\)

Thus, \(x = 1\) and \(x = -1\). Since we are only considering the positive X-axis and \(x = e\), we are concerned with \(x = 1\).

The region of interest is bounded by \(x = 1\) and \(x = e\).

Step 2: Compute the area between \(x = 1\) and \(x = e\).

Now, we compute the area between these curves:

The desired area \(A\) can be expressed as:

\(A = \int_{1}^{e} (x - \frac{1}{x}) \, dx\)

Split the integral into two separate integrals:

\(A = \int_{1}^{e} x \, dx - \int_{1}^{e} \frac{1}{x} \, dx\)

Evaluate each integral separately:

• \(\int x \, dx = \frac{x^2}{2} \quad \text{so,} \quad \left[\frac{x^2}{2}\right]_{1}^{e} = \frac{e^2}{2} - \frac{1}{2}\)
• \(\int \frac{1}{x} \, dx = \ln|x| \quad \text{so,} \quad [\ln x]_{1}^{e} = \ln e - \ln 1 = 1 - 0 = 1\)

Step 3: Calculate the enclosed area:

Substitute the values back into our equation for \(A\):

\(A = \left(\frac{e^2}{2} - \frac{1}{2}\right) - 1 = \frac{e^2}{2} - \frac{3}{2}\)

Since \(e\) is approximately 2.718, the enclosed area simplifies to:

\(A = \frac{3}{2} \, \text{sq units}\)

Thus, the area of the region enclosed by the given curves is \(\frac{3}{2}\) sq units.