## Calculus: Early Transcendentals 8th Edition

$f(t) = t^{3/2}$ $a = 9$
Consider the function $f(t) = t^{3/2}$ $6+\int_{a}^{x}\frac{f(t)}{t^2}~dt$ $= 6+\int_{a}^{x}\frac{t^{3/2}}{t^2}~dt$ $= 6+\int_{a}^{x}t^{-1/2}~dt$ $= 6+2t^{1/2}~\vert_{a}^{x}$ $= 6+2\sqrt{x}- 2\sqrt{a}$ To find the required value of $a$, we can set this expression equal to $2\sqrt{x}$: $6+2\sqrt{x}- 2\sqrt{a} = 2\sqrt{x}$ $2\sqrt{a} = 6$ $\sqrt{a} = 3$ $a = 9$ Therefore: $f(t) = t^{3/2}$ $a = 9$