a) \(\int\limits_1^2 {\frac{1}{{{x^3}}}dx} = \int\limits_1^2 {{x^{ - 3}}dx} = \frac{{{x^{ - 2}}}}{{ - 2}}\left| {\begin{array}{*{20}{c}}{^2}\\{_1}\end{array}} \right.\)
\(= \frac{{ - 1}}{{2{x^2}}}\left| {\begin{array}{*{20}{c}}{^2}\\{_1}\end{array}} \right. = \frac{{ - 1}}{{{{2.2}^2}}} - \frac{{ - 1}}{{{{2.1}^2}}} = \frac{3}{8}\).
b) \(\int\limits_1^3 {{x^{\frac{2}{3}}}dx} = \frac{{{x^{\frac{2}{3} + 1}}}}{{\frac{2}{3} + 1}}\left| {\begin{array}{*{20}{c}}{^3}\\{_1}\end{array}} \right. = \frac{{{x^{\frac{5}{3}}}}}{{\frac{5}{3}}}\left| {\begin{array}{*{20}{c}}{^3}\\{_1}\end{array}} \right. = \frac{3}{5}x\sqrt[3]{{{x^2}}}\left| {\begin{array}{*{20}{c}}{^3}\\{_1}\end{array}} \right. \)
\(= \frac{3}{5}.3.\sqrt[3]{{{3^2}}} - \frac{3}{5}.1.\sqrt[3]{{{1^2}}} = \frac{{9\sqrt[3]{9} - 3}}{5}\).
c) \(\int\limits_1^8 {\sqrt[3]{x}dx} = \int\limits_1^8 {{x^{\frac{1}{3}}}dx} = \frac{3}{4}{x^{\frac{4}{3}}}\left| {\begin{array}{*{20}{c}}{^8}\\{_1}\end{array}} \right. = \frac{3}{4}x\sqrt[3]{x}\left| {\begin{array}{*{20}{c}}{^8}\\{_1}\end{array}} \right.\)
\(= \frac{3}{4}.8.\sqrt[3]{8} - \frac{3}{4}.1.\sqrt[3]{1} = \frac{{45}}{4}\).