Question

Find the area enclosed by the x-axis and the curve x = 3 + e^t, y = t − t^2.

Answer 1

I am not sure if it works, but it is worth trying

From the equation \(x = 3 + e^t\)
\[x = 3 + e^t\]\[x -3= e^t\]\[t \ln e = \ln (x-3)\]\[t= \ln (x-3)\]
Sub t = ln (x-3) into the equation of y = t - t^2
\[y = \ln (x-3) + [\ln (x-3)]^2\]

Sub y=0 into \(y = \ln (x-3) + [\ln (x-3)]^2\) to find the bounds:
\[0 = \ln (x-3) + [\ln (x-3)]^2\]\(ln(x-3)= 0 \) or \(\ln (x-3) -1 =0\)
Solve x from these two equations to find the bounds.

Then, you can integrate the equation \(y = \ln (x-3) + [\ln (x-3)]^2\) to find the area.