Key Concepts and Formulas

• Area Under a Curve: The area enclosed between a curve $y = f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$ is given by $A = \int_a^b |f(x)| \, dx$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.
• Logarithm Properties: $\log_a a = 1$ and $\log_a 1 = 0$.

Step-by-Step Solution

Step 1: Understanding the Curve and Finding Intercepts

The goal is to find the area enclosed by the curve $y = \log_e(x+e)$ and the coordinate axes. We need to find the x and y intercepts to define the region of integration.

1. Finding the X-intercept: The x-intercept occurs when $y=0$. $0 = \log_e(x+e)$ $e^0 = x+e$ $1 = x+e$ $x = 1-e$ So, the x-intercept is at $(1-e, 0)$.

2. Finding the Y-intercept: The y-intercept occurs when $x=0$. $y = \log_e(0+e)$ $y = \log_e(e)$ $y = 1$ So, the y-intercept is at $(0, 1)$.

3. Determining the Region of Integration: The region is bounded by the curve $y = \log_e(x+e)$, the x-axis, the y-axis, and the line $x = 1-e$. The interval of integration for $x$ is therefore $[1-e, 0]$. Since $x > -e$, $x+e > 0$ and $\log_e(x+e)$ is well-defined. Also, on the interval $[1-e, 0]$, $1 \le x+e \le e$, so $\log_e(x+e) \ge 0$. Therefore, the function is non-negative on the interval of integration.

Step 2: Setting up the Definite Integral

The area $A$ is given by the integral of the function from $x=1-e$ to $x=0$. $A = \int_{1-e}^0 \log_e(x+e) \, dx$

Step 3: Evaluating the Integral using Substitution

We use substitution to simplify the integral.

1. Substitution: Let $u = x+e$. Then, $du = dx$.

2. Changing the Limits of Integration: When $x = 1-e$, $u = (1-e) + e = 1$. When $x = 0$, $u = 0 + e = e$.

3. Rewriting the Integral: The integral becomes: $A = \int_1^e \log_e(u) \, du$

Step 4: Integrating the Logarithmic Function using Integration by Parts

We use integration by parts to solve the integral.

1. Integration by Parts: Let $v = \log_e(u)$ and $dw = du$. Then, $dv = \frac{1}{u} \, du$ and $w = u$. Using the integration by parts formula $\int v \, dw = vw - \int w \, dv$: $\int \log_e(u) \, du = u \log_e(u) - \int u \cdot \frac{1}{u} \, du$ $= u \log_e(u) - \int 1 \, du$ $= u \log_e(u) - u + C$

Step 5: Applying the Limits of Integration and Final Calculation

We evaluate the definite integral using the antiderivative and the limits of integration. $A = \left[ u \log_e(u) - u \right]_1^e$ $A = (e \log_e(e) - e) - (1 \log_e(1) - 1)$ $A = (e \cdot 1 - e) - (1 \cdot 0 - 1)$ $A = (e - e) - (0 - 1)$ $A = 0 - (-1)$ $A = 1$

Common Mistakes & Tips

• Sketching the Graph: Always sketch the graph to visualize the region and determine the correct limits of integration.
• Changing Limits in Substitution: When using substitution for definite integrals, remember to change the limits of integration to match the new variable.
• Integrating by Parts: Correctly identify $u$ and $dv$ when applying integration by parts.

Summary

We found the area enclosed between the curve $y = \log_e(x+e)$ and the coordinate axes by first determining the x and y intercepts. We then set up a definite integral, used substitution to simplify the integral, and finally used integration by parts to solve it. The area is 1 square unit.

Final Answer

The final answer is \boxed{1}, which corresponds to option (A).