Answer
$\frac{24}{5}$
Work Step by Step
$x=t^3+1$ and $y=2t-t^2$
Find the intersection points between the curve and the x-axis:
$y=0$
$2t-t^2=0$
$t(2-t)=0$
$t=0\vee t=2$
For $t=0$, $x=0^3+1=1$
For $t=2$, $x=2^3+1=9$
Then, the curve above the x-axis is given by $g(t)=2t-t^2$ and $f(t)=t^3+1$ where $0\leq t\leq 2$.
Find the area:
$A=\int_0^2g(t)f'(t)dt$
$A=\int_0^2(2t-t^2)(3t^2)dt$
$A=\int_0^26t^3-3t^4dt$
$A=[\frac{3t^4}{2}-\frac{3t^5}{5}]_0^2$
$A=(\frac{3\cdot 2^4}{2}-\frac{3\cdot 2^5}{5})-(\frac{3\cdot 0^4}{2}-\frac{3\cdot 0^5}{5})$
$A=(24-\frac{96}{5})-0$
$A=\frac{24}{5}$
Thus, the area is $\frac{24}{5}$.
