## anonymous 5 years ago Find the area under the curve. x=t^2+2. y=t^2+3t+2. from t=0 to t=1.

1. anonymous

if possible keep the parameters and solve the integral with respect to t

2. anonymous

I can't I need to put it in terms of x. so: $t = \sqrt(x-2)$ and $y = (x-2) + 3\sqrt(x-2) + 2$ $x= (0 + 2) = 2$ and $x = (1 + 2) = 3$ $dt = 1/(2\sqrt(x-2)$ $I = \int\limits_{x=2}^{x=3} (x + 3\sqrt(x-2))(1/(2\sqrt(x-2))) = 1/2\int\limits_{x=2}^{x=3} (x/sqrt(x-2) +3)dx = 1/3(x-2)^{3/2} + 6\sqrt(x-2) +9x$(integration by parts) which is from x=2 to x=3 $I = 16/3$

3. anonymous

dammit my stuff disappeared. $I = 1/3(x-2) + 6(\sqrt(x-2) + 9x)$ from 2 to 3