Given curve C defined by the equations $x=3t^2,y=2t^3$ where $t$ is a parameter and line e which is tangent to C at point P where $t=2$, find the area $R$ between $C$ and $e$.
For this question I have worked out that $y=2x-8$ is the line $e$ by taking the derivative of $C$ and that the line $e$ intersects $C$ at a second point $Q(3,-2)$. However I can't work out how to find the area of $R$. I assume that it is with integrals, but I am stuck. Could you please explain to me how to solve this question?
$$x(2)=3(2)^2=12$$ $$y(2)=2(2)^3=16$$ so: $P=(12,16)$
Now: $$x'(t)=6t,y'(t)=6t^2$$ so: $$\frac{dy}{dx}=\frac{6(2)^2}{6(2)}=2$$ then you have: $$(y-16)=2(x-12)$$ $$(e):y=2x-8$$ so that part is all good, and the second point is in fact $Q=(3,-2)$ now for $C$ try and get $y$ in terms of $x$