Find the area bounded by the curves `y = e^(x), y = |x-1|` and `x = 2`.

Answer

To find the area bounded by the curves \( y = e^x \), \( y = |x - 1| \), and \( x = 2 \), we will follow these steps:

### Step 1: Understand the Functions
The function \( y = |x - 1| \) can be expressed in piecewise form:
- For \( x < 1 \), \( |x - 1| = 1 - x \)
- For \( x \geq 1 \), \( |x - 1| = x - 1 \)

### Step 2: Find Intersection Points
We need to find the points where \( y = e^x \) intersects with \( y = |x - 1| \).

1. **For \( x < 1 \)**:
Set \( e^x = 1 - x \):
\[ e^x + x - 1 = 0 \]
This equation can be solved numerically or graphically to find the intersection point.

2. **For \( x \geq 1 \)**:
Set \( e^x = x - 1 \):
\[ e^x - x + 1 = 0 \]
Again, this can be solved numerically or graphically.

### Step 3: Determine the Area
The area is bounded between the curves from \( x = 0 \) to \( x = 2 \). We will split the area into two parts:
- From \( x = 0 \) to \( x = 1 \)
- From \( x = 1 \) to \( x = 2 \)

#### Area Calculation
1. **From \( x = 0 \) to \( x = 1 \)**:
The area \( A_1 \) is given by:
\[ A_1 = \int_0^1 (e^x - (1 - x)) \, dx \]
Simplifying:
\[ A_1 = \int_0^1 (e^x + x - 1) \, dx \]

2. **From \( x = 1 \) to \( x = 2 \)**:
The area \( A_2 \) is given by:
\[ A_2 = \int_1^2 (e^x - (x - 1)) \, dx \]
Simplifying:
\[ A_2 = \int_1^2 (e^x + 1 - x) \, dx \]

### Step 4: Compute the Integrals
1. **Calculate \( A_1 \)**:
\[ A_1 = \left[ e^x + \frac{x^2}{2} - x \right]_0^1 \]
Evaluating at the limits:
\[ A_1 = \left( e^1 + \frac{1^2}{2} - 1 \right) - \left( e^0 + \frac{0^2}{2} - 0 \right) \]
\[ A_1 = \left( e + \frac{1}{2} - 1 \right) - 1 = e - \frac{1}{2} \]

2. **Calculate \( A_2 \)**:
\[ A_2 = \left[ e^x + x - \frac{x^2}{2} \right]_1^2 \]
Evaluating at the limits:
\[ A_2 = \left( e^2 + 2 - \frac{2^2}{2} \right) - \left( e^1 + 1 - \frac{1^2}{2} \right) \]
\[ A_2 = \left( e^2 + 2 - 2 \right) - \left( e + 1 - \frac{1}{2} \right) \]
\[ A_2 = e^2 - e + \frac{1}{2} \]

### Step 5: Total Area
The total area \( A \) is:
\[ A = A_1 + A_2 = \left( e - \frac{1}{2} \right) + \left( e^2 - e + \frac{1}{2} \right) \]
\[ A = e^2 \]

### Final Answer
The area bounded by the curves is \( e^2 \).

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