We are asked to find the area of the region bounded by the curves:

1. $y = e^x$
2. $y = 3e^{-x} + 2$
3. $x = 0$ (the vertical line at $x = 0$).

To find the area between the curves, we need to follow these steps:

Step 1: Find the points of intersection

The region is bounded between the two curves, so we need to find where the curves intersect. Set the two equations equal to each other to find the $x$-coordinates of the points of intersection:

$e^x = 3e^{-x} + 2$

Multiply both sides of the equation by $e^x$ to eliminate the negative exponent:

$e^{2x} = 3 + 2e^x$

This is a quadratic equation in $e^x$. Let $u = e^x$, so the equation becomes:

$u^2 = 3 + 2u$

Rearranging:

$u^2 - 2u - 3 = 0$

We can solve this quadratic equation using the quadratic formula:

$u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)}$

$u = \frac{2 \pm \sqrt{4 + 12}}{2}$

$u = \frac{2 \pm \sqrt{16}}{2}$

$u = \frac{2 \pm 4}{2}$

Thus, the two solutions are:

$u = \frac{2 + 4}{2} = 3 \quad \text{or} \quad u = \frac{2 - 4}{2} = -1$

Since $u = e^x$ and $e^x$ is always positive, we discard $u = -1$. Therefore, $u = 3$, which means $e^x = 3$, and thus:

$x = \ln 3$

Step 2: Set up the integral

Now we know that the curves intersect at $x = 0$ and $x = \ln 3$. To find the area between the curves, we integrate the difference between the two functions from $x = 0$ to $x = \ln 3$:

$\text{Area} = \int_0^{\ln 3} \left[ (3e^{-x} + 2) - e^x \right] \, dx$

Step 3: Simplify the integrand

The integrand is:

$(3e^{-x} + 2) - e^x = 3e^{-x} + 2 - e^x$

Thus, the area is:

$\text{Area} = \int_0^{\ln 3} (3e^{-x} - e^x + 2) \, dx$

Step 4: Solve the integral

Now, we integrate each term individually:

$\int 3e^{-x} \, dx = -3e^{-x}$ $\int -e^x \, dx = -e^x$ $\int 2 \, dx = 2x$

Therefore:

$\text{Area} = \left[ -3e^{-x} - e^x + 2x \right]_0^{\ln 3}$

Step 5: Evaluate the integral

Substitute the limits of integration:

At $x = \ln 3$:

$-3e^{-\ln 3} - e^{\ln 3} + 2(\ln 3)$ Since $e^{\ln 3} = 3$ and $e^{-\ln 3} = \frac{1}{3}$:

$-3 \times \frac{1}{3} - 3 + 2\ln 3 = -1 - 3 + 2\ln 3 = -4 + 2\ln 3$

At $x = 0$:

$-3e^{0} - e^{0} + 2(0) = -3 - 1 = -4$

Now, subtract the value at $x = 0$ from the value at $x = \ln 3$:

$\text{Area} = \left( -4 + 2\ln 3 \right) - (-4) = 2\ln 3$

Final Answer:

The area of the region is $\boxed{2\ln 3}$.

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Would you like a more detailed explanation or any clarification? Also, here are some related questions:

1. How would the area change if the curves were shifted vertically or horizontally?
2. How do you handle integrals involving more complex functions, such as logarithms or trigonometric functions?
3. What other methods are available to find the area between curves (e.g., numerical methods)?
4. How would you approach finding the volume of a region bounded by similar curves in three dimensions?
5. How do you use the method of integration by parts in these types of problems?

Tip: When solving for the area between curves, always check the points of intersection to ensure your integral limits are correct!