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

Question

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

anonymous · Answer

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

anonymous · Answer

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 $$ I = 16/3$$

anonymous · Answer

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