Answer
$F(x)=20(1-\displaystyle \frac{1}{x})$
$F(2)=10$
$F(5)=16$
$F(8)=17.5$
Work Step by Step
Apply The Second Fundamental Theorem of Calculus (Th.4.11 )
If $f$ is continuous on an open interval I containing $a$, then, for every $x$ in the
interval,
$\displaystyle \frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right]=f(x)$.
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$F(x)=\displaystyle \int_{1}^{x}\frac{20}{v^{2}}dv=\int_{1}^{x}20v^{-2}dv$
$F(x)=20\left[\displaystyle \frac{v^{-1}}{-1}\right]_{1}^{x} = 20[-\frac{1}{x}-(-1)]$
$F(x)=20(1-\displaystyle \frac{1}{x})$
$F(2)=20\left(\displaystyle \frac{1}{2}\right)=10$
$F(5)=20\left(1-\displaystyle \frac{1}{5}\right)=20\left(\frac{4}{5}\right)=16$
$F(8)=20\left(1-\displaystyle \frac{1}{8}\right)=20\left(\frac{7}{8}\right)=\frac{35}{2}$
