To find the area of the region bounded by the curve $y = 9x e^{-x}$, the x-axis, and the vertical line $x = 3$, we need to follow these steps:

1. Sketching the Region:

• The function is $y = 9x e^{-x}$, which is a product of a linear function and an exponential decay function.
• The curve intersects the x-axis when $y = 0$, which happens when $x = 0$.
• The line $x = 3$ defines a vertical boundary on the right side.
• So, the region of interest is between $x = 0$ and $x = 3$, under the curve and above the x-axis.

2. Finding the Area:

The area under the curve can be found by integrating the function from $x = 0$ to $x = 3$:

$A = \int_0^3 9x e^{-x} \, dx$

3. Performing Integration:

To integrate $9x e^{-x}$, we will use integration by parts. Let:

• $u = 9x$, so $du = 9dx$,
• $dv = e^{-x}dx$, so $v = -e^{-x}$.

The formula for integration by parts is:

$\int u \, dv = uv - \int v \, du$

Substitute the values:

$\int 9x e^{-x} \, dx = -9x e^{-x} - \int (-9 e^{-x}) \, dx$

$= -9x e^{-x} + 9 \int e^{-x} \, dx$

$= -9x e^{-x} - 9e^{-x} + C$

Now, evaluate this from $x = 0$ to $x = 3$:

$A = \left[ -9x e^{-x} - 9e^{-x} \right]_0^3$

4. Evaluating the Definite Integral:

At $x = 3$:

$-9(3)e^{-3} - 9e^{-3} = -27e^{-3} - 9e^{-3} = -36e^{-3}$

At $x = 0$:

$-9(0)e^{-0} - 9e^{-0} = -9$

So, the area is:

$A = \left( -36e^{-3} \right) - (-9) = 9 - 36e^{-3}$

Now, calculate the numerical value of $e^{-3} \approx 0.0498$:

$A \approx 9 - 36 \times 0.0498 = 9 - 1.7928 = 7.2072$

Final Answer:

The area of the region is approximately:

$A \approx 7.2072 \, \text{square units}$

Would you like more details on any step or a follow-up explanation? Here are some related questions you might find useful:

1. How do we apply integration by parts in general?
2. What is the significance of the exponential function in modeling real-world problems?
3. How can you calculate areas between curves for non-exponential functions?
4. How do you interpret the meaning of negative values in an integral evaluation?
5. What are other methods for finding the area under a curve?

Tip: If you encounter an integral involving a product of polynomials and exponentials, integration by parts is often a useful method to simplify the process.