To calculate the area of the region above the $x$-axis and below the given parametric curve:

$x = t^2 + 3t + 9, \quad y = e^{-t}, \quad 0 \leq t \leq 1,$

we will use the following formula for the area under a parametric curve:

$\text{Area} = \int_{t=a}^{t=b} y \frac{dx}{dt} \, dt.$

Steps:

1. Compute $\frac{dx}{dt}$ from the given $x = t^2 + 3t + 9$: $\frac{dx}{dt} = 2t + 3.$

2. Substitute $y = e^{-t}$ and $\frac{dx}{dt} = 2t + 3$ into the formula: $\text{Area} = \int_{0}^{1} e^{-t}(2t + 3) \, dt.$

3. Expand the integrand: $\text{Area} = \int_{0}^{1} \left( 2te^{-t} + 3e^{-t} \right) \, dt.$

4. Split the integral into two parts: $\text{Area} = \int_{0}^{1} 2te^{-t} \, dt + \int_{0}^{1} 3e^{-t} \, dt.$

5. Solve each integral:

   • For $\int_{0}^{1} 3e^{-t} \, dt$, the result is: $\int_{0}^{1} 3e^{-t} \, dt = \left[ -3e^{-t} \right]_{0}^{1} = -3e^{-1} + 3.$

   • For $\int_{0}^{1} 2te^{-t} \, dt$, use integration by parts: Let $u = 2t$, $dv = e^{-t} dt$. Then: $u = 2t, \, du = 2 \, dt, \, v = -e^{-t}.$ Apply the formula: $\int 2te^{-t} \, dt = -2te^{-t} - \int -2e^{-t} \, dt = -2te^{-t} + 2e^{-t}.$ Evaluate from $0$ to $1$: $\int_{0}^{1} 2te^{-t} \, dt = \left[ -2te^{-t} + 2e^{-t} \right]_{0}^{1} = \left( -2e^{-1} + 2e^{-1} \right) - \left( 0 + 2 \right) = -2e^{-1} + 2e^{-1} - 2.$

6. Combine the results:

   • First term: $-3e^{-1} + 3$.
   • Second term: $-2$.

   Total: $\text{Area} = (-3e^{-1} + 3) + (-2) = -3e^{-1} + 1.$

7. Approximate the result: Substitute $e^{-1} \approx 0.3679$: $\text{Area} \approx -3(0.3679) + 1 \approx -1.1037 + 1 = -0.1037.$

Would you like more detailed steps or clarifications?