Calculus: Early Transcendentals 8th Edition

$6\sqrt 6$ - $2\sqrt 3$
$\int^{18}_{1} \sqrt \frac{3}{z} dz$ The first step is to simplify the integral and bring the constant outside the integral: $= \int^{18}_{1} \frac{\sqrt 3}{\sqrt z} dz$ $= \sqrt 3 \int^{18}_{1} \frac{1}{\sqrt z} dz$ $= \sqrt 3 \int^{18}_{1} z^{-1/2} dz$ Next step is to solve for the definite integral using power rule. [ $\int^b_a x^{n} dx = (\frac{x^{n+1}}{n+1})|^b_a$ ] $= \sqrt{3} (\frac{z^{\frac{1}{2}}}{\frac{1}{2}})|^{18}_1$ $= \sqrt 3 (2\sqrt{z})|^{18}_1$ Plug in the values from 18 to 1. $= \sqrt{3}( 2\sqrt{18} - 2\sqrt{1} )$ Simplify until final answer: $= \sqrt{3} (6\sqrt{2} - 2)$ $= 6\sqrt{6} - 2\sqrt{3}$