Answer
$6\sqrt6 -2\sqrt3$
Work Step by Step
Evaluate the Integral: $\int^{18}_{1}\sqrt{\frac{3}{z}} dz $
Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$
Find $F(x)$:
$F(x) = \int\sqrt{\frac{3}{x}} dx$
$F(x) = \int\frac{\sqrt3}{\sqrt x}dx$
$F(x) = \sqrt3 \int\frac{1}{\sqrt x}dx$
$F(x) = \sqrt3 \int x^{-\frac{1}{2}}dx$
$F(x) = \sqrt3 \frac{x^{(-1/2) +1}}{(-1/2) +1}$
$F(x) = \sqrt3(2\sqrt x)$
Now Evaluate: $F(b) - F(a)$
$F(18) - F(1)$
$\sqrt3(2\sqrt{18}) - \sqrt3(2\sqrt1)$
$\sqrt3(6\sqrt2 - 2)$
$6\sqrt6 -2\sqrt3$
