Answer
$\displaystyle \frac{40}{3}$
Work Step by Step
FTC: Let $f$ be a continuous function defined on the interval $[a, b]$,
and let $F$ be any antiderivative of f defined on $[a, b]$. Then
$\displaystyle \int_{a}^{b}f(x)dx=\left[F(x)\right]_{a}^{b}=F(b)-F(a)$.
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$f(x)=2x^{-2}+3x$, which is continuous on [1,3].
First, find an antiderivative $F(x)$:
$\displaystyle \int f(x)dx=\frac{2x^{-1}}{-1}+\frac{3x^{2}}{2}+C$
$=-\displaystyle \frac{2}{x}+\frac{3x^{2}}{2}+C=$
(we take C=0, since by FTC we need ANY antiderivative)
$F(x)=-\displaystyle \frac{2}{x}+\frac{3x^{2}}{2}$
Next, evaluate:
$F(3)=-\displaystyle \frac{2}{3}+\frac{3(9)}{2}=\frac{-4+81}{6}=\frac{77}{6}$
$F(1)=-\displaystyle \frac{2}{1}+\frac{3(1)}{2}=\frac{-4+3}{2}=-\frac{1}{2}$
Finally, apply the theorem:
$\displaystyle \int_{1}^{3}f(x)dx=F(3)-F(1)$
$=\displaystyle \frac{77}{6}-(-\frac{1}{2}) =\frac{80}{6}=\frac{40}{3}$
