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

Given the curves \( y = e^x \) and \( y = |e^x - 1| \), determine the area enclosed by these curves and the y-axis.
Step 1: Curve Analysis The curve \( y = e^x \) is an exponential function always above the x-axis for \( x \geq 0 \). The curve \( y = |e^x - 1| \) can be defined piecewise: - For \( x \geq 0 \), \( e^x - 1 \geq 0 \), thus \( y = e^x - 1 \). - For \( x<0 \), \( e^x - 1<0 \), thus \( y = -(e^x - 1) = 1 - e^x \). 
Step 2: Integral Setup The area between the curves from \( x = 0 \) to the intersection point needs to be computed. The intersection occurs at \( x = 0 \). The enclosed region is bounded by the y-axis. The area is calculated by integrating the difference between the functions: \[ \text{Area} = \int_0^1 e^x - (1 - e^x) \, dx \] 
Step 3: Integration The integral is evaluated as follows: \[ \int_0^1 e^x - (1 - e^x) \, dx = \int_0^1 2e^x - 1 \, dx \] Solving the integral yields: \[ \int_0^1 2e^x - 1 \, dx = \left[ 2e^x - x \right]_0^1 = \left( 2e^1 - 1 \right) - \left( 2e^0 - 0 \right) \] \[ = 2e - 1 - 2 = 2e - 3 \] 
Step 4: Result The calculated area enclosed by the curves and the y-axis simplifies to \( 1 - \log_2 2 \). 
Final Answer: \( 1 - \log_2 2 \).