Find the area of the region that lies above the $x$-axis, below the curve $x=t^{2}+4 t+3, y=e^{-t}$ with $0 \leq t \leq 1$. Give your answer exactly or round to four decimal places.
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Step 1 ：Given the parametric equations \(x = t^{2} + 4t + 3\) and \(y = e^{-t}\) where \(0 \leq t \leq 1\), we are asked to find the area of the region that lies above the x-axis and below the curve.

Step 2 ：The area under a curve from a to b is given by the definite integral from a to b of the function. In this case, we are given a parametric equation, so we need to find the derivative of x with respect to t, \(dx/dt\), and then integrate \(y \cdot dx/dt\) from 0 to 1.

Step 3 ：First, we find the derivative of x with respect to t, \(dx/dt\). Given \(x = t^{2} + 4t + 3\), we find that \(dx/dt = 2t + 4\).

Step 4 ：Next, we integrate \(y \cdot dx/dt\) from 0 to 1. Given \(y = e^{-t}\) and \(dx/dt = 2t + 4\), we find that the integral is \(6 - 8e^{-1}\).

Step 5 ：Thus, the area of the region that lies above the x-axis, below the curve \(x = t^{2} + 4t + 3, y = e^{-t}\) with \(0 \leq t \leq 1\) is \(\boxed{6 - 8e^{-1}}\).