$\displaystyle x=2-t^2$, $\displaystyle y= t(t+1)$
Find the exact area of the region bounded by the curve and the coordinate axes.
To find the x-intercepts we set y =0 and solve for t.
$\displaystyle 0=t(t+1) \\\ t=0 \\\ t=-1$
The area under the curve can be found using
$\displaystyle \int_{x_0}^{x_1}ydx$
Now we need to calculate dx so
$\displaystyle dx=-2t dt$
so we get
$\displaystyle \int_{1}^{2}ydx=\int_{-1}^{0}-2t^3-2t^2dt=-\frac{1}{6}$
Note that the area is negative because it is below the x-axis
Here is a graph. Good luck.