Answer
$12$
Work Step by Step
Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$
Here, $\dfrac{dx}{dt}=t$
and $\dfrac{dy}{dt}=(2t+1)^{1/2}$
Thus,
$S=\int_{0}^{4} \sqrt{(t)^2+((2t+1)^{1/2})^2} dt=\int_{0}^{4} (t+1) dt$
Thus,
$S=[\dfrac{t^{2}}{2}+t]_{0}^{4} =\dfrac{16}{2}+4=12$
