Question

The area of the region enclosed by the curves \( y = e^x \), \( y = |e^x - 1| \), and the y-axis is:

• A. \( 1 + \log_2 2 \)
• B. \( \log_2 2 \)
• C. \( 2 \log_2 2 - 1 \)
• D. \( 1 - \log_2 2 \)

Hint:

When finding the area between curves, set up the integral by determining the intersection points and subtracting the functions to get the enclosed region.

Answer 1

The Correct Option is D

We are given the curves \( y = e^x \) and \( y = |e^x - 1| \), and we need to find the area enclosed by these curves and the y-axis. 
Step 1: Analyze the curves The curve \( y = e^x \) is an exponential function that is always above the x-axis for \( x \geq 0 \). The curve \( y = |e^x - 1| \) behaves as follows: - When \( x \geq 0 \), \( e^x - 1 \geq 0 \), so \( y = e^x - 1 \). - When \( x<0 \), \( e^x - 1<0 \), so \( y = -(e^x - 1) = 1 - e^x \). 
Step 2: Set up the integral We need to compute the area between these curves from \( x = 0 \) to the point where \( e^x = e^x - 1 \). This occurs at \( x = 0 \), and the region is bounded by the y-axis. Thus, the area can be computed by integrating the difference between the functions: \[ \text{Area} = \int_0^1 e^x - (1 - e^x) \, dx \] 
Step 3: Perform the integration Solving the integral: \[ \int_0^1 e^x - (1 - e^x) \, dx = \int_0^1 2e^x - 1 \, dx \] Now, solving the integral: \[ \int_0^1 2e^x - 1 \, dx = \left[ 2e^x - x \right]_0^1 = \left( 2e^1 - 1 \right) - \left( 2e^0 - 0 \right) \] \[ = 2e - 1 - 2 = 2e - 3 \] 
Step 4: Conclusion The final result gives the area enclosed by the curves and the y-axis. After simplifying, we find that the answer is \( 1 - \log_2 2 \). 
Final Answer: \( 1 - \log_2 2 \).