Answer
$F(x)=\displaystyle \frac{1}{x^{2}}-\frac{1}{4}$
$F(2)=0$
$F(5)=-0.21$
$F(8)=-\displaystyle \frac{15}{64}$
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,
$\displaystyle \frac{d}{dx}[\int_{a}^{x}f(t)dt]=f(x)$.
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$F(x)=\displaystyle \int_{2}^{x}-\frac{2}{t^{3}}dt=-\int_{2}^{x}2t^{-3}dt$
$=-2[\displaystyle \frac{t^{-2}}{-2}]_{2}^{x}=[\frac{1}{t^{2}}]_{2}^{x}=\frac{1}{x^{2}}-\frac{1}{4}$
$F(x)=\displaystyle \frac{1}{x^{2}}-\frac{1}{4}$
$F(2)=\displaystyle \frac{1}{4}-\frac{1}{4}=0$
$F(5)=\displaystyle \frac{1}{25}-\frac{1}{4}=-\frac{21}{100}=-0.21$
$F(8)=\displaystyle \frac{1}{64}-\frac{1}{4}=-\frac{15}{64}$
