Question

Find the area under the curve. x=t^2+2. y=t^2+3t+2. from t=0 to t=1.

Answer 1

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

Answer 2

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 \[ I = 16/3\]

Answer 3

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