## Thomas' Calculus 13th Edition

$12$
Here,we have $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ This implies that $S=\int_{0}^{4} \sqrt{t^2+(\sqrt{2t+1})^2} dt=\int_{0}^{4} (t+1) dt$ Therefore, $S=[\dfrac{t^{2}}{2}+t]_{0}^{4}$ or, $S=\dfrac{16}{2}+4=8+4=12$