Given that:
$x=f(t)$ and $y=g(t)$
$\int_\alpha^\beta g(t) f'(t)$
$x=8+e^t$ and $y=t-t^2$ bound by the curve given and above the x-axis
I assume I am trying to find values of t? So far I have this setup:
$\int_\alpha^\beta (t-t^2)e^tdt$
Setting y=0 I get values of $t=0$ and $t=1$. Are these correct?
Setting up the integral I get $\int_0^1 (t-t^2)e^t$ Integration by parts gives:
$u=t-t^2$
$du=1-2tdt$
solving for dt gives $-\frac{1}{2}+du=dt$
$dv=e^t$ $v=e^t$
so we have $\int_0^1 uv-\int vdu \to \int_0^1 e^t(t-t^2)-\int -\frac{1}{2}e^t$
which gives:
$\frac{1}{2}e-\frac{1}{2}$ Can someone offer more hints and point out where I am correct to? I get that setting y=0 is the bound of the x axis but how do I find my other bounds?