Find the area bounded by the parametric curve... Find the area bounded by the parametric curve $x = \cos(t)$, $y = e^t, 0 < t < \pi/2$, and the lines $y = 1$ and $x = 0$.
I do not even know where to start with this problem. I know that I need to draw a graph, but that's all I know. Thanks for the help!
 A: If $x=\cos(t)$, then $t=\arccos(x)$, at least when $0\leq t \leq \pi/2$.  Thus, your curve can be expressed as $y=e^{\arccos(x)}$, which you can integrate over $[0,1]$.
I certainly agree that a picture like the one below is useful.  You could plot both the parametric plot and the function plot using a tool like WolframAlpha.  It then becomes apparent that you should subtract 1 from the integral.  Of course, you can also ask WolframAlpha for help with the integral, which might be a bit tricky.

A: 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 \ ) \ $ .
