# Evaluate the integral. \int_{3}^{4}(4x-\frac{2}{x^{2}})dx

Evaluate the integral.
${\int }_{3}^{4}\left(4x-\frac{2}{{x}^{2}}\right)dx$
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Arham Warner
${\int }_{3}^{4}\left(4x-\frac{2}{{x}^{2}}\right)dx$
$={\left[\frac{4{x}^{2}}{2}-2\left(-\frac{1}{x}\right)\right]}_{3}^{4}$
$={\left[2{x}^{2}+\frac{2}{x}\right]}_{3}^{4}$
$=2\left(10\right)+\frac{2}{4}-2\left(9\right)-\frac{2}{3}⇒32-18+\frac{2}{4}-\frac{2}{3}$
$=\frac{168+6-8}{12}=\frac{166}{12}=\frac{83}{6}$
$\therefore {\int }_{3}^{4}\left(4x-\frac{2}{{x}^{2}}\right)dx=\frac{83}{6}$