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: \( 1 - \log_2 2 \).

Answer 2

Step 1: Understand the curves.
We are given the curves \( y = e^x \) and \( y = |e^x - 1| \), and we are asked to find the area of the region enclosed by these curves and the y-axis.
The curve \( y = e^x \) is the exponential function, while \( y = |e^x - 1| \) represents the absolute value of the difference between \( e^x \) and 1. The absolute value function causes a change in the shape of the curve depending on whether \( e^x - 1 \) is positive or negative.

Step 2: Find the intersection points of the curves.
The two curves will intersect where \( e^x = |e^x - 1| \). To solve for this, we consider two cases:
1. If \( e^x - 1 >= 0 \), then \( |e^x - 1| = e^x - 1 \). The equation becomes: \[ e^x = e^x - 1 \] which simplifies to \( 0 = -1 \), a contradiction. Therefore, there are no solutions for \( e^x - 1 >= 0 \).
2. If \( e^x - 1 < 0 \), then \( |e^x - 1| = -(e^x - 1) = 1 - e^x \). The equation becomes: \[ e^x = 1 - e^x \] Adding \( e^x \) to both sides, we get: \[ 2e^x = 1 \] Solving for \( x \): \[ e^x = \frac{1}{2} \quad \Rightarrow \quad x = \log_e \left( \frac{1}{2} \right) = -\log_e 2 \] Therefore, the curves intersect at \( x = -\log_e 2 \).

Step 3: Set up the integral to find the area.
The area between the curves is given by the integral of the difference between the two functions from the y-axis (\( x = 0 \)) to the intersection point (\( x = -\log_e 2 \)): \[ \text{Area} = \int_{0}^{-\log_e 2} \left( e^x - (1 - e^x) \right) dx \] Simplifying the integrand: \[ \text{Area} = \int_{0}^{-\log_e 2} \left( e^x - 1 + e^x \right) dx = \int_{0}^{-\log_e 2} \left( 2e^x - 1 \right) dx \] Step 4: Evaluate the integral.
We now evaluate the integral: \[ \int_{0}^{-\log_e 2} \left( 2e^x - 1 \right) dx = \left[ 2e^x - x \right]_{0}^{-\log_e 2} \] Substituting the limits of integration: \[ = \left( 2e^{-\log_e 2} - (-\log_e 2) \right) - \left( 2e^0 - 0 \right) \] \[ = \left( 2 \cdot \frac{1}{2} + \log_e 2 \right) - 2 \] \[ = \left( 1 + \log_e 2 \right) - 2 = \log_e 2 - 1 \] Since \( \log_e 2 = \log_2 e \), we can rewrite the answer as: \[ 1 - \log_2 2 \] Thus, the area enclosed by the curves is \( \boxed{1 - \log_2 2} \).