# Chapter 8 - Further Techniques and Applications of Integration - Chapter Review - Review Exercises - Page 455: 14

$$2{\left( {8 - x} \right)^{5/2}}\left( {\frac{{27}}{{35}} - \frac{{2x}}{7}} \right) + C$$

#### Work Step by Step

\eqalign{ & \int {x{{\left( {8 - x} \right)}^{3/2}}} dx \cr & {\text{set }}u = 8 - x{\text{ then }}\frac{{du}}{{dx}} = - 1,\,\,\,\,\, - du = dx \cr & u = 8 - x \to x = 8 - u \cr & {\text{write the integrand in terms of }}u \cr & \int {x{{\left( {8 - x} \right)}^{3/2}}} dx = \int {\left( {8 - u} \right)} {u^{3/2}}\left( { - du} \right) \cr & = \int {\left( {u - 8} \right)} {u^{3/2}}du \cr & = \int {\left( {{u^{5/2}} - 8{u^{3/2}}} \right)} du \cr & {\text{using }}\int {{u^n}du = \frac{{{u^{n + 1}}}}{{n + 1}} + C} \cr & = \frac{{{u^{7/2}}}}{{7/2}} - 8\left( {\frac{{{u^{5/2}}}}{{5/2}}} \right) + C \cr & = \frac{{2{u^{7/2}}}}{7} - \frac{{16{u^{5/2}}}}{5} + C \cr & {\text{factoring}} \cr & = 2{u^{5/2}}\left( {\frac{{2u}}{7} - \frac{8}{5}} \right) + C \cr & {\text{replace }}8 - x{\text{ for }}u \cr & = 2{\left( {8 - x} \right)^{5/2}}\left( {\frac{{2\left( {8 - x} \right)}}{7} - \frac{8}{5}} \right) + C \cr & = 2{\left( {8 - x} \right)^{5/2}}\left( {\frac{{16}}{7} - \frac{{2x}}{7} - \frac{8}{5}} \right) + C \cr & = 2{\left( {8 - x} \right)^{5/2}}\left( {\frac{{27}}{{35}} - \frac{{2x}}{7}} \right) + C \cr}

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