Someone could help me or explain how to solve this exercise. I understand that the equations are in parametric form but I have not been able to go further and find the enclosed area that they ask me.
I share the statement:
- Find the area enclosed by the curves $x=t-t^2$, $y=1+e^{-t}$ and the $y$-axis.
I would appreciate your help and comments.
First find where the curve intersects the $y$-axis, by setting $x=0$
We get $t = 0 $ and $t = 1$.
Secondly, the area is given by
$A = \displaystyle \int x dy = \int_{t=0}^{t=1} x \dfrac{dy}{dt} dt = \int_{t=0}^{t=1} (t - t^2) (- e^{-t} ) dt $
Since we want the positive area, we negate the above expression to obtain
$A = \displaystyle \int_{0}^{1} (t - t^2) e^{-t} dt $
Integrating each term by parts, yields,
$A = 3 e^{-1} - 1 $