Question

The area of the region enclosed between the curve \[ y=\log_e(x+e) \] and the coordinate axes is

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\) \bigskip

Hint:

For regions bounded by a logarithmic curve and coordinate axes, first find the intercepts carefully. A substitution of the form \[ u=x+a \] usually converts the integral into the standard form \[ \int \ln u\,du. \]

Answer 1

The Correct Option is D

Concept: The required area is bounded by: \[ y=\ln(x+e), \] the \(x\)-axis \((y=0)\), and the \(y\)-axis \((x=0)\). First find the intercept with the \(x\)-axis.

Step 1: Find the point where the curve meets the \(x\)-axis. For the \(x\)-axis, \[ y=0. \] Hence \[ \ln(x+e)=0. \] \[ x+e=1. \] \[ x=1-e. \] Thus the enclosed region lies between \[ x=1-e \quad \text{and} \quad x=0. \]

Step 2: Set up the area integral. \[ A = \int_{1-e}^{0}\ln(x+e)\,dx. \] Let \[ u=x+e. \] Then \[ du=dx. \] When \[ x=1-e, \quad u=1, \] and when \[ x=0, \quad u=e. \] Therefore, \[ A = \int_{1}^{e}\ln u\,du. \]

Step 3: Evaluate the integral. \[ \int \ln u\,du = u\ln u-u. \] Hence \[ A = \Big[u\ln u-u\Big]_{1}^{e}. \] \[ = (e\ln e-e)-(1\cdot \ln1-1). \] Using \[ \ln e=1, \qquad \ln1=0, \] \[ A=(e-e)-(0-1). \] \[ A=1. \]

Step 4: Write the final answer. \[ \boxed{1} \]