## University Calculus: Early Transcendentals (3rd Edition)

$4t^{5/2}+4t^{3/2}-8t$
Calculate the anti-derivative. Given: $\dfrac{d^2 r}{dt^2}=15 \sqrt t+\dfrac{3}{\sqrt t}$ We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$ and where $C$ is a constant of proportionality. Now, $r'=15 (\dfrac{t^{1/2+1}}{1/2+1})+3 (\dfrac{t^{-1/2+1}}{-1/2+1})+C=10t^{3/2}+6^{1/2}+C$ Here, $r'(1)=8 \implies C=-8$ 'Then, $r'=10t^{3/2}+6^{1/2}-8$ Now, $r=10(\dfrac{t^{3/2+1}}{3/2+1})+6(\dfrac{t^{1/2+1}}{1/2+1})-8t+C=4t^{5/2}+4t^{3/2}-8t+C$ and $r(1)=0 \implies C=0$ Hence, $r=4t^{5/2}+4t^{3/2}-8t$