## Thomas' Calculus 13th Edition

$$\frac{3}{5}\left( {\root 3 \of 7 - \root 3 \of 2 } \right)$$
\eqalign{ & \int_0^1 {\frac{{dr}}{{\root 3 \of {{{\left( {7 - 5r} \right)}^2}} }}} \cr & = \int_0^1 {{{\left( {7 - 5r} \right)}^{ - 2/3}}dr} \cr & {\text{use the substitution method }} \cr & u = 7 - 5r,\,\,\,\,du = - 5dr,\,\,\,\,\,dr = - \frac{{du}}{5} \cr & {\text{Then}}{\text{,}} \cr & \int_{}^{} {{{\left( {7 - 5r} \right)}^{ - 2/3}}dr} = \int_{}^{} {{u^{ - 2/3}}\left( { - \frac{{du}}{5}} \right)} \cr & = - \frac{1}{5}\int {{u^{ - 2/3}}} du \cr & {\text{integrate}} \cr & = - \frac{1}{5}\left( {\frac{{{u^{1/3}}}}{{1/3}}} \right) + C \cr & = - \frac{3}{5}{u^{1/3}} + C \cr & {\text{write in terms of }}r;{\text{ replace }}u = 7 - 5r \cr & = - \frac{3}{5}{\left( {7 - 5r} \right)^{1/3}} + C \cr & {\text{Then}}{\text{,}} \cr & \int_0^1 {\frac{{dr}}{{\root 3 \of {{{\left( {7 - 5r} \right)}^2}} }}} = \left( { - \frac{3}{5}{{\left( {7 - 5r} \right)}^{1/3}}} \right)_0^1 \cr & {\text{Evaluating, we get:}} \cr & = - \frac{3}{5}{\left( {7 - 5\left( 1 \right)} \right)^{1/3}} + \frac{3}{5}{\left( {7 - 5\left( 0 \right)} \right)^{1/3}} \cr & = - \frac{3}{5}{\left( 2 \right)^{1/3}} + \frac{3}{5}{\left( 7 \right)^{1/3}} \cr & = \frac{3}{5}\left( {\root 3 \of 7 - \root 3 \of 2 } \right) \cr}