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 \(\boxed{6 - 8e^{-1}}\).