## Calculus: Early Transcendentals 8th Edition

The area is $\frac{24}{5}$
$x = t^3+1$ $dx = 3t^2~dt$ 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 $\frac{24}{5}$