# Evaluate the following integral. \int_{1}^{32}(x^{-\frac{6}{5}})dx

Question
Applications of integrals
Evaluate the following integral.
$$\displaystyle{\int_{{{1}}}^{{{32}}}}{\left({x}^{{-{\frac{{{6}}}{{{5}}}}}}\right)}{\left.{d}{x}\right.}$$

2020-11-09
$$\displaystyle{\int_{{{1}}}^{{{32}}}}{\left({x}^{{-{\frac{{{6}}}{{{5}}}}}}\right)}{\left.{d}{x}\right.}={{\left[{\frac{{{x}^{{-{\frac{{{6}}}{{{5}}}}+{1}}}}}{{-{\frac{{{6}}}{{{5}}}}+{1}}}}\right]}_{{{1}}}^{{{32}}}}={{\left[{\frac{{{x}^{{{\frac{{-{1}}}{{{5}}}}}}}}{{{\frac{{-{1}}}{{{5}}}}}}}\right]}_{{{1}}}^{{{32}}}}$$
$$\displaystyle={\left(-{5}\right)}{\left[{\left({32}\right)}^{{{\frac{{-{1}}}{{{5}}}}}}-{\left({1}\right)}^{{{\frac{{-{1}}}{{{5}}}}}}\right]}$$
$$\displaystyle=-{5}{\left[{\left({2}^{{{5}}}\right)}^{{{\frac{{-{1}}}{{{5}}}}}}-{1}\right]}=-{5}{\left({2}^{{-{1}}}-{1}\right)}$$
Hence, $$\displaystyle{\int_{{{1}}}^{{{32}}}}{\left({x}^{{{\frac{{-{6}}}{{{5}}}}}}\right)}{\left.{d}{x}\right.}=-{5}{\left({\frac{{{1}}}{{{2}}}}-{1}\right)}=-{5}{\left({\frac{{-{1}}}{{{2}}}}\right)}={\frac{{{5}}}{{{2}}}}$$

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