Answer
$e^2+1$
Work Step by Step
Recall: The integral which represents the length of the parametric curve for $a\leq t\leq b$ is given by $L=\int_a^b\sqrt{(dx/dt)^2+(dy/dt)^2}dt$.
Given: $x=e^t-t$ and $y=4e^{t/2}$ where $0\leq t\leq 2$
Find $dx/dt$ and $dy/dt$:
$dx/dt=e^t-1$ and $dy/dt=2e^{t/2}$
Find the length of the given curve:
$L=\int_0^2\sqrt{(e^t-1)^2+(2e^{t/2})^2}dt$
$L=\int_0^2\sqrt{(e^{t})^2-2e^t+1+4e^t}dt$
$L=\int_0^2\sqrt{(e^t)^2+2e^t+1}dt$
$L=\int_0^2\sqrt{(e^t+1)^2}dt$
$L=\int_0^2e^t+1 dt$
$L=[e^t+t]_0^2$
$L=(e^2+2)-(e^0+0)$
$L=e^2+2-1$
$L=e^2+1$
Thus, the length of the curve is $e^2+1$.