Answer
$A= \frac{24}{5}$
Work Step by Step
The curve
$x$ = $t^{3}+1$
$y$ = $2t-t^{2}$
intersects the $x$-axis when $y$ = $0$, that is when $2t-t^2=t(2-t)=0$, so $t$ = $0$ and $t$ = $2$.
$A$ = $\int_0^2{[y(t)-0]x'(t)}dt$
$A$ = $\int_0^2{(2t-t^{2})(3t^{2})}dt$
$A$ = $3\int_0^2{(2t^{3}-t^{4})}dt$
$A$ = $3\left[\frac{t^{4}}{2}-\frac{t^{5}}{5}\right]_0^2$ = $\frac{24}{5}$