Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises - Page 393: 99

$$\frac{{2{{\left( {1 + a} \right)}^{5/2}}}}{5} - \frac{{2a{{\left( {1 + a} \right)}^{3/2}}}}{3} + \frac{4}{{15}}{a^{5/2}}$$

$$\eqalign{ & \int_0^1 {x\sqrt {x + a} dx,{\text{ }}a > 0} \cr & {\text{taking the substitution }}u = \sqrt {x + a} ,\,\,\,\,\,x = {u^2} - a \cr & du = \frac{1}{{2\sqrt {x + a} }}dx,\,\,\,\,\,\,2\sqrt {x + a} du = dx,\,\,\,\,\,\,\,2udu = dx \cr & {\text{switch the limits of integration using }}u = \sqrt {x + a} \cr & x = 0,\,\,\,\,u = \sqrt a \cr & x = 1,\,\,\,\,u = \sqrt {1 + a} \cr & {\text{use the change of variable}} \cr & \int_0^1 {x\sqrt {x + a} dx = } \int_{\sqrt a }^{\sqrt {1 + a} } {\left( {{u^2} - a} \right)u} \left( {2udu} \right) \cr & = 2\int_{\sqrt a }^{\sqrt {1 + a} } {\left( {{u^2} - a} \right){u^2}} du \cr & = 2\int_{\sqrt a }^{\sqrt {1 + a} } {\left( {{u^4} - a{u^2}} \right)} du \cr & {\text{integrating}} \cr & = 2\left[ {\frac{{{u^5}}}{5} - \frac{{a{u^3}}}{3}} \right]_{\sqrt a }^{\sqrt {1 + a} } \cr & {\text{evaluate the limits}} \cr & = 2\left[ {\frac{{{{\left( {\sqrt {1 + a} } \right)}^5}}}{5} - \frac{{a{{\left( {\sqrt {1 + a} } \right)}^3}}}{3}} \right] - 2\left[ {\frac{{{{\left( {\sqrt a } \right)}^5}}}{5} - \frac{{a{{\left( {\sqrt a } \right)}^3}}}{3}} \right] \cr & {\text{simplifying}} \cr & = 2\left[ {\frac{{{{\left( {1 + a} \right)}^{5/2}}}}{5} - \frac{{a{{\left( {1 + a} \right)}^{3/2}}}}{3}} \right] - 2\left[ {\frac{{{a^{5/2}}}}{5} - \frac{{{a^{5/2}}}}{3}} \right] \cr & = \frac{{2{{\left( {1 + a} \right)}^{5/2}}}}{5} - \frac{{2a{{\left( {1 + a} \right)}^{3/2}}}}{3} + \frac{4}{{15}}{a^{5/2}} \cr & \cr & {\text{taking the substitution }}u = x + a,\,\,\,\,\,x = u - a \cr & du = dx,\,\,\,\,\,\, \cr & {\text{switch the limits of integration using }}u = x + a \cr & x = 0,\,\,\,\,u = a \cr & x = 1,\,\,\,\,u = 1 + a \cr & {\text{use the change of variable}} \cr & \int_0^1 {x\sqrt {x + a} dx = } \int_a^{1 + a} {\left( {u - a} \right){u^{1/2}}} du \cr & = \int_a^{1 + a} {\left( {{u^{3/2}} - a{u^{1/2}}} \right)} du \cr & {\text{integrating}} \cr & = \left[ {\frac{{{u^{5/2}}}}{{5/2}} - \frac{{a{u^{3/2}}}}{{3/2}}} \right]_a^{1 + a} \cr & = \left[ {\frac{{2{u^{5/2}}}}{5} - \frac{{2a{u^{3/2}}}}{3}} \right]_a^{1 + a} \cr & {\text{evaluate the limits}} \cr & = \left[ {\frac{{2{{\left( {1 + a} \right)}^{5/2}}}}{5} - \frac{{2a{{\left( {1 + a} \right)}^{3/2}}}}{3}} \right] - \left[ {\frac{{2{a^{5/2}}}}{5} - \frac{{2a{{\left( a \right)}^{3/2}}}}{3}} \right] \cr & {\text{simplifying}} \cr & = \frac{{2{{\left( {1 + a} \right)}^{5/2}}}}{5} - \frac{{2a{{\left( {1 + a} \right)}^{3/2}}}}{3} - \left[ {\frac{{2{a^{5/2}}}}{5} - \frac{{2{a^{5/2}}}}{3}} \right] \cr & = \frac{{2{{\left( {1 + a} \right)}^{5/2}}}}{5} - \frac{{2a{{\left( {1 + a} \right)}^{3/2}}}}{3} + \frac{4}{{15}}{a^{5/2}} \cr & \cr & {\text{The results are the same for both substitutions}} \cr} $$