Sketch the region bounded above by the curve y = 2e
−x + x, below by the
curve y = (e^x)/2, on the left by x = 0, and on the right by x = 1 and find the area of the region

We are tasked with finding the area of the region bounded by the following curves:

- **Upper curve:** \( y = 2e^{-x} + x \)
- **Lower curve:** \( y = \frac{e^x}{2} \)
- **Left boundary:** \( x = 0 \)
- **Right boundary:** \( x = 1 \)

### Step 1: Set up the integral

The area between two curves over a given interval \([a, b]\) is found using the formula:

\[
A = \int_a^b [f(x) - g(x)] \, dx
\]

Where:
- \( f(x) \) is the upper curve
- \( g(x) \) is the lower curve
- \( a \) and \( b \) are the boundaries along the \(x\)-axis

For our problem:
- \( f(x) = 2e^{-x} + x \) (the upper curve)
- \( g(x) = \frac{e^x}{2} \) (the lower curve)
- \( a = 0 \)
- \( b = 1 \)

So the area is:

\[
A = \int_0^1 \left[ (2e^{-x} + x) - \frac{e^x}{2} \right] \, dx
\]

### Step 2: Simplify the integrand

Simplifying the expression inside the integral:

\[
A = \int_0^1 \left( 2e^{-x} + x - \frac{e^x}{2} \right) \, dx
\]

### Step 3: Solve the integral

Now, we integrate each term separately:

1. \(\int 2e^{-x} \, dx = -2e^{-x}\)
2. \(\int x \, dx = \frac{x^2}{2}\)
3. \(\int \frac{e^x}{2} \, dx = \frac{e^x}{2}\)

Now, applying the limits \( x = 0 \) to \( x = 1 \):

\[
A = \left[ -2e^{-x} + \frac{x^2}{2} - \frac{e^x}{2} \right]_0^1
\]

### Step 4: Evaluate at the bounds

First, evaluate at \( x = 1 \):

\[
-2e^{-1} + \frac{1^2}{2} - \frac{e^1}{2} = -\frac{2}{e} + \frac{1}{2} - \frac{e}{2}
\]

Then, evaluate at \( x = 0 \):

\[
-2e^{0} + \frac{0^2}{2} - \frac{e^0}{2} = -2 + 0 - \frac{1}{2} = -\frac{5}{2}
\]

### Step 5: Subtract to find the area

Now subtract the values at \( x = 1 \) and \( x = 0 \):

\[
A = \left( -\frac{2}{e} + \frac{1}{2} - \frac{e}{2} \right) - \left( -\frac{5}{2} \right)
\]
\[
A = -\frac{2}{e} + \frac{1}{2} - \frac{e}{2} + \frac{5}{2}
\]
\[
A = \frac{5}{2} + \frac{1}{2} - \frac{e}{2} - \frac{2}{e}
\]
\[
A = 3 - \frac{e}{2} - \frac{2}{e}
\]

### Final Result:

The area of the region is:

\[
A = 3 - \frac{e}{2} - \frac{2}{e}
\]

Let me know if you'd like a sketch of this region or further explanation!

---

### Related Questions:
1. How do you find the area between two curves when they intersect?
2. What are the steps to solve definite integrals involving exponential functions?
3. Can you explain how the boundary conditions affect the area of a region?
4. What would change if the limits of integration were different (e.g., \(x = 2\))?
5. How would this problem change if both functions were reversed (upper and lower)?

### Tip:
When setting up integrals, always carefully check which curve is the upper and which is the lower in the given interval.

Sketch the region bounded above by the curve y = 2e^(-x) + x, below by the curve y = (e^x)/2, on the left by x = 0, and on the right by x = 1 and find the area of the region.

Learn how to calculate the area of the region bounded by the curves y = 2e^(-x) + x and y = (e^x)/2 using definite integrals. This detailed step-by-step solution applies exponential functions and the Fundamental Theorem of Calculus. Ideal for undergraduate calculus students.

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