#### Answer

The area is $\frac{24}{5}$

#### Work Step by Step

$x = t^3+1$
$dx = 3t^2~dt$
Since the area is enclosed by the x-axis, we need to find the values of $t$ when $y=0$:
$y = 2t-t^2 = 0$
$t(2-t) = 0$
$t = 0, 2$
We can find the area:
$A = \int_{0}^{2} y~dx$
$A = \int_{0}^{2} (2t-t^2)~3t^2~dt$
$A = \int_{0}^{2} (6t^3-3t^4)~dt$
$A = (\frac{3}{2}t^4-\frac{3}{5}t^5)\Big\vert_{0}^{2}$
$A = [\frac{3}{2}(2)^4-\frac{3}{5}(2)^5]- [\frac{3}{2}(0)^4-\frac{3}{5}(0)^5]$
$A = (24-\frac{96}{5})- (0)$
$A = \frac{120}{5}-\frac{96}{5}$
$A = \frac{24}{5}$
The area is $\frac{24}{5}$