To finish the resolution of this problem, we would continue from the graph shown by Mark McClure : the bounding curve is traced out from $ \ ( \ \cos \ 0 \ , \ e^0 \ ) \ = \ ( \ 1 \ , \ 1 \ ) \ $ to $ \ ( \ \cos \frac{\pi}{2} \ , \ e^{\pi / 2} \ ) \ = \ ( \ 0 \ , \ e^{\pi / 2} \ ) \ $ . The region for which we wish to find the area is thus perched atop a unit square:

We would set up the area integral as $ \ \int_0^1 \ y(x) \ - \ 1 \ \ dx \ $ , but to carry this out parametrically , we need to account for the curve being traced "backwards" (from right to left):
$ \int_0^1 \ y(x) \ - \ 1 \ \ dx \ \ \rightarrow \ \ \left[ \ \int^0_{\pi / 2} \ y(t) \ \frac{dx}{dt} \ \ dt \ \right] \ - \ 1 \ \ = \ \ \left[ \ \int^0_{\pi / 2} \ e^t \ ( - \sin \ t \ \ dt ) \ \right] \ - \ 1 $
$ = \ \ \left[ \ \int_0^{\pi / 2} e^t \ \sin \ t \ \ dt \ \right] \ - \ 1 \ = \ \ \frac{1}{2} e^t \ ( \ \sin \ t \ - \ \cos \ t \ ) \vert_0^{\pi / 2} \ - \ 1 $
[having integrated by parts twice]
$ = \ \ \frac{1}{2} \left[ \ e^{\pi / 2} \ ( \ \sin \frac{\pi}{2} \ - \ \cos \frac{\pi}{2} \ ) \ - \ e^0 \ ( \ \sin \ 0 \ - \ \cos \ 0 \ ) \ \right] \ - \ 1 $
$ = \ \frac{1}{2} \left[ \ e^{\pi / 2} \ ( \ 1 \ - \ 0 \ ) \ - \ 1 \ ( \ 0 \ - \ 1\ ) \ \right] \ - \ 1 \ = \ \frac{1}{2} \left( \ e^{\pi / 2} \ + \ 1 \ \right) \ - \ 1 $
$ = \ \frac{1}{2} \left( \ e^{\pi / 2} \ - \ 1 \ \right) \ \approx \ 1.905 \ \ . $
Despite the curve not (quite) being a straight line, the area is estimated very well by a triangle of appropriate dimensions: $ \ \frac{1}{2} \ \cdot \ ( \ 1 - 0 \ ) \ \cdot \ ( \ e^{\pi / 2} \ - \ 1 \ ) \ $ .