A curve C has parametric equations $x = e^{t},\ \ y = \sin t$ where $0 \le t \le \pi$.

1. Apply integration by parts to the given integral.

Let $I = \int e^t \sin t \, dt$.

Using the integration by parts formula $\int u \frac{dv}{dt} \, dt = uv - \int v \frac{du}{dt} \, dt$:
Let $u = \sin t \implies \frac{du}{dt} = \cos t$
Let $\frac{dv}{dt} = e^t \implies v = e^t$

$I = e^t \sin t - \int e^t \cos t \, dt$

2. Apply integration by parts a second time to the new integral.

For $\int e^t \cos t \, dt$:
Let $u = \cos t \implies \frac{du}{dt} = -\sin t$
Let $\frac{dv}{dt} = e^t \implies v = e^t$

$\int e^t \cos t \, dt = e^t \cos t - \int e^t (-\sin t) \, dt$
$\int e^t \cos t \, dt = e^t \cos t + \int e^t \sin t \, dt$

3. Substitute the result back and solve algebraically for the original integral $I$.

Substitute the result from the second integration by parts back into the equation for $I$:
$I = e^t \sin t - \left( e^t \cos t + \int e^t \sin t \, dt \right)$
$I = e^t \sin t - e^t \cos t - I$

Rearranging to solve for $I$:
$2I = e^t (\sin t - \cos t)$
$I = \frac{1}{2}e^t (\sin t - \cos t) + C$

1. Find the coordinates of the endpoints by substituting the boundary values of $t$.

When $t = 0$:
$x = e^0 = 1$
$y = \sin 0 = 0$
Start point is $(1, 0)$.

When $t = \pi$:
$x = e^{\pi}$
$y = \sin \pi = 0$
End point is $(e^{\pi}, 0)$.

2. Find the coordinates of the maximum point.

The maximum value of $y = \sin t$ for $0 \le t \le \pi$ occurs at $t = \frac{\pi}{2}$, where $y = 1$.
When $t = \frac{\pi}{2}$:
$x = e^{\pi/2}$
Turning point (maximum) is $(e^{\pi/2}, 1)$.

3. Sketch the curve noting its domain and shape.

Since $x = e^t$ is strictly increasing, the curve progresses continuously from left to right. The sketch is a bell-like curve in the positive $x, y$ quadrant, starting at $(1,0)$, peaking at $(e^{\pi/2}, 1)$, and returning to the x-axis at $(e^{\pi}, 0)$. The curve is skewed such that the peak is horizontally closer to the starting point.

1. Determine the limits for the parameter $t$ and set up the parametric integral for the area.

When $x = 1$, $e^t = 1 \implies t = 0$.
When $x = e^a$, $e^t = e^a \implies t = a$.

The area is given by $\int_{x=1}^{x=e^a} y \, dx$.
Using $x = e^t \implies \frac{dx}{dt} = e^t$:
$\text{Area} = \int_{t=0}^{t=a} y \left(\frac{dx}{dt}\right) dt = \int_{0}^{a} \sin t \, (e^t) \, dt$

2. Find the antiderivative $\int e^t \sin t \, dt$ (concisely re-deriving the result required from part (i)).

Let $I = \int e^t \sin t \, dt$.
Using integration by parts ($u = \sin t$, $dv = e^t \, dt$):
$I = e^t \sin t - \int e^t \cos t \, dt$
Applying integration by parts again on the second integral ($u = \cos t$, $dv = e^t \, dt$):
$I = e^t \sin t - \left( e^t \cos t - \int e^t (-\sin t) \, dt \right)$
$I = e^t \sin t - e^t \cos t - I$
$2I = e^t(\sin t - \cos t)$
$\int e^t \sin t \, dt = \frac{1}{2} e^t (\sin t - \cos t)$

3. Evaluate the definite integral to find the area in terms of $a$.

$\text{Area} = \left[ \frac{1}{2} e^t (\sin t - \cos t) \right]_0^a$
$= \frac{1}{2} e^a (\sin a - \cos a) - \frac{1}{2} e^0 (\sin 0 - \cos 0)$
$= \frac{1}{2} e^a (\sin a - \cos a) - \frac{1}{2}(1)(0 - 1)$
$= \frac{1}{2} e^a (\sin a - \cos a) + \frac{1}{2}$