Answer
$6\sqrt 6$ - $2\sqrt 3$
Work Step by Step
$\int^{18}_{1} \sqrt \frac{3}{z} dz$
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:
$= \sqrt{3} (6\sqrt{2} - 2)$
$= 6\sqrt{6} - 2\sqrt{3}$
