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

Yasmin 2021-10-03 Answered
Evaluate the integral.
\(\displaystyle{\int_{{{3}}}^{{{4}}}}{\left({4}{x}-{\frac{{{2}}}{{{x}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}\)

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Expert Answer

Arham Warner
Answered 2021-10-04 Author has 23063 answers
\(\displaystyle{\int_{{{3}}}^{{{4}}}}{\left({4}{x}-{\frac{{{2}}}{{{x}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[{\frac{{{4}{x}^{{{2}}}}}{{{2}}}}-{2}{\left(-{\frac{{{1}}}{{{x}}}}\right)}\right]}_{{{3}}}^{{{4}}}}\)
\(\displaystyle={{\left[{2}{x}^{{{2}}}+{\frac{{{2}}}{{{x}}}}\right]}_{{{3}}}^{{{4}}}}\)
\(\displaystyle={2}{\left({10}\right)}+{\frac{{{2}}}{{{4}}}}-{2}{\left({9}\right)}-{\frac{{{2}}}{{{3}}}}\Rightarrow{32}-{18}+{\frac{{{2}}}{{{4}}}}-{\frac{{{2}}}{{{3}}}}\)
\(\displaystyle={\frac{{{168}+{6}-{8}}}{{{12}}}}={\frac{{{166}}}{{{12}}}}={\frac{{{83}}}{{{6}}}}\)
\(\displaystyle\therefore{\int_{{{3}}}^{{{4}}}}{\left({4}{x}-{\frac{{{2}}}{{{x}^{{{2}}}}}}\right)}{\left.{d}{x}\right.}={\frac{{{83}}}{{{6}}}}\)
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