Getting area between curve and line? Suppose we have an arbitrary continuous parametric function $f(t)$ differentiable at any point $t$.
Further suppose we have a point at position $f(t)$ and we add $h$ units to $t$. Between time $t$ and $t+h$ the point has moved and has created a well-defined path.
I'm looking for the absolute total area that is created between the path and a straight line that passes from $f(t)$ and $f(t+h)$.
Here is an image to help clarify:

EDIT:
$f(t)$ is a 2 dimensional vector and the picture shows a two dimensional space.
 A: @Garmekain , regards. Hope this will be useful. So in the problem, we have
$$ f(t) = (x(t), y(t)) $$
I may not give full detailed answer here. In a point of view : to calculate the area, you may transform the straight line (from $f(t_{0})$ to $f(t_{0}+h)$) and all the points on the curve so that $$(x(t_{0}),y(t_{0})) \rightarrow (0, 0)$$ and $$(x(t_{0}+h),y(t_{0}+h)) \rightarrow (|l|, 0)$$ $|l|$ is the length of the straight line. 
This can be done by translating all the points by referring translation $(x(t_{0}),y(t_{0}))$ to $(0, 0)$, and then make a rotation of $ -\theta $. (*The rotation $-\theta$ is clockwise)

$$ \left[ \begin{matrix} X(t) \\ Y(t) \end{matrix} \right] = \left[ \begin{matrix} \cos(-\theta) & -\sin(-\theta) \\ \sin(-\theta) & \cos(-\theta) \end{matrix} \right] \left[ \begin{matrix} x(t)-x(t_{0}) \\ y(t)-y(t_{0}) \end{matrix} \right]  $$
The 
$$ \left[ \begin{matrix} x(t)-x(t_{0}) \\ y(t)-y(t_{0}) \end{matrix} \right]  $$
is the result of the translation.
Then the area can be calculated by the integral below :
$$ A = \int_{t_{0}}^{t_{0} + h} |Y(t)| dt $$
Thanks. All the best.
