Answer
$L = e^{2}+1$
Work Step by Step
Remember the arc length formula for parametrics:
$L=\int^{b}_{a}\sqrt((\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2})dt$
$x= e^{t}-t$
$\frac{dx}{dt}= e^{t}-1$
$y= 4e^{\frac{t}{2}}$
$\frac{dy}{dt}= 2e^{\frac{t}{2}}$
Now that we have the derivatives, we can now find the arc length across $0\leq t\leq2$
$L=\int^{2}_{0}\sqrt(( e^{t}-1)^{2}+(2e^{\frac{t}{2}})^{2})dt$
$=\int^{2}_{0}\sqrt( e^{2t}-2e^{t}+1+4e^{t})dt$
$=\int^{2}_{0}\sqrt( e^{2t}+2e^{t}+1)dt$
$=\int^{2}_{0}\sqrt( (e^{t}+1)^{2})dt$
$=\int^{2}_{0}e^{t}+1 dt$
$= e^{t}+t]^{2}_{0}$
$=(e^{2}+2)-(e^{0}+0)$
$=e^2+2-1$
$=e^2+1$
