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

Evaluate the integral.
$$\displaystyle{\int_{{{3}}}^{{{4}}}}{\left({4}{x}-{\frac{{{2}}}{{{x}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}$$

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Arham Warner
$$\displaystyle{\int_{{{3}}}^{{{4}}}}{\left({4}{x}-{\frac{{{2}}}{{{x}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}$$
$$\displaystyle={{\left[{\frac{{{4}{x}^{{{2}}}}}{{{2}}}}-{2}{\left(-{\frac{{{1}}}{{{x}}}}\right)}\right]}_{{{3}}}^{{{4}}}}$$
$$\displaystyle={{\left[{2}{x}^{{{2}}}+{\frac{{{2}}}{{{x}}}}\right]}_{{{3}}}^{{{4}}}}$$
$$\displaystyle={2}{\left({10}\right)}+{\frac{{{2}}}{{{4}}}}-{2}{\left({9}\right)}-{\frac{{{2}}}{{{3}}}}\Rightarrow{32}-{18}+{\frac{{{2}}}{{{4}}}}-{\frac{{{2}}}{{{3}}}}$$
$$\displaystyle={\frac{{{168}+{6}-{8}}}{{{12}}}}={\frac{{{166}}}{{{12}}}}={\frac{{{83}}}{{{6}}}}$$
$$\displaystyle\therefore{\int_{{{3}}}^{{{4}}}}{\left({4}{x}-{\frac{{{2}}}{{{x}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}={\frac{{{83}}}{{{6}}}}$$