I am assuming that this is a parametric curve.
We see that the curve intersects the x-axis when:
t - t^2 = 0 ==> t = 0 and t = 1.
Then, since x = 1 + e^t is an increasing function, the curve is being traced exactly once on the interval (0, 1).
Using the fact that the area under the curve given by the parametric equations x = f(t) and y = g(t) on (a, b) is:
A = ∫ f'(t)g(t) dt (from t=a to b),
and that f(t) = 1 + e^t ==> f'(t) = e^t, the area under the curve is:
A = ∫ e^t(t - t^2) dt (from t=0 to 1)
= e^t(-t^2 + 3t - 3) (evaluated from t=0 to 1), by integrating by parts
= e(-1 + 3 - 3) - (0 + 0 - 3)
= 3 - e.
