Question

The area of the figure bounded by the curve \( y = \log_e x \), the x-axis and the straight line \( x = e \) is

• A. \( 5 - e \)
• B. \( 3 + e \)
• C. \( 1 \)
• D. None of the above

Hint:

The area bounded by a curve \( y = f(x) \) (\( f(x) \ge 0 \)), the x-axis, and the lines \( x = a \) and \( x = b \) is given by the definite integral \( \int_{a}^{b} f(x) \, dx \). Remember the integral of \( \ln x \) is \( x \ln x - x \).

Answer 1

The Correct Option is C

Step 1: Identify the boundaries of the region. The region is bounded by the curve \( y = \log_e x \) (which is \( y = \ln x \)), the x-axis (\( y = 0 \)), and the vertical line \( x = e \). We need to find the point where the curve intersects the x-axis to determine the lower limit of integration. Set \( y = 0 \): \[ \ln x = 0 \implies x = e^0 = 1 \] So, the region is bounded by \( y = \ln x \), \( y = 0 \), \( x = 1 \), and \( x = e \). Step 2: Set up the definite integral for the area. The area \( A \) under the curve \( y = f(x) \) from \( x = a \) to \( x = b \) above the x-axis is given by \( A = \int_{a}^{b} f(x) \, dx \). Here, \( f(x) = \ln x \), \( a = 1 \), and \( b = e \). Since \( \ln x \ge 0 \) for \( x \ge 1 \), the area is: \[ A = \int_{1}^{e} \ln x \, dx \] Step 3: Evaluate the indefinite integral \( \int \ln x \, dx \) using integration by parts. Let \( u = \ln x \) and \( dv = dx \). Then \( du = \frac{1}{x} dx \) and \( v = x \). Using the formula \( \int u \, dv = uv - \int v \, du \): \[ \int \ln x \, dx = (\ln x)(x) - \int x \left(\frac{1}{x}\right) dx \] \[ \int \ln x \, dx = x \ln x - \int 1 \, dx \] \[ \int \ln x \, dx = x \ln x - x + C \] Step 4: Evaluate the definite integral using the limits of integration. \[ A = [x \ln x - x]_{1}^{e} \] \[ A = (e \ln e - e) - (1 \ln 1 - 1) \] Substitute \( \ln e = 1 \) and \( \ln 1 = 0 \): \[ A = (e \cdot 1 - e) - (1 \cdot 0 - 1) \] \[ A = (e - e) - (0 - 1) \] \[ A = 0 - (-1) \] \[ A = 1 \] Step 5: Compare the result with the given options. The calculated area is 1, which matches option (C).