Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises - Page 393: 109

$$\frac{{32}}{{105}}$$

$$\eqalign{ & \int_0^1 {\sqrt {x - x\sqrt x } dx} \cr & {\text{Factoring}} \cr & = \int_0^1 {\sqrt {x\left( {1 - \sqrt x } \right)} dx} \cr & = \int_0^1 {\sqrt x \sqrt {1 - \sqrt x } dx} \cr & {\text{Let }}u = \sqrt x {\text{ }} \Rightarrow {\text{ }}du = \frac{1}{{2\sqrt x }}dx \cr & {\text{ }} \Rightarrow {\text{ }}dx = 2udu \cr & {\text{Limits of integration}} \cr & x = 0{\text{ }} \Rightarrow {\text{ }}u = 0 \cr & x = 1{\text{ }} \Rightarrow {\text{ }}u = 1 \cr & {\text{Use the substitution}} \cr & \int_0^1 {\sqrt x \sqrt {1 - \sqrt x } dx} = \int_0^1 {u\sqrt {1 - u} \left( {2u} \right)du} \cr & {\text{ }} = 2\int_0^1 {{u^2}\sqrt {1 - u} du} \cr & {\text{Let }}z = 1 - u{\text{ }} \Rightarrow {\text{ }}u = 1 - z,{\text{ }}du = - dz \cr & {\text{Limits of integration}} \cr & u = 0{\text{ }} \Rightarrow {\text{ }}z = 1 \cr & u = 1{\text{ }} \Rightarrow {\text{ }}z = 0 \cr & 2\int_0^1 {{u^2}\sqrt {1 - u} du} = - 2\int_1^0 {{{\left( {1 - z} \right)}^2}\sqrt z dz} \cr & {\text{ }} = 2\int_0^1 {{{\left( {1 - z} \right)}^2}{z^{1/2}}dz} \cr & {\text{ }} = 2\int_0^1 {\left( {1 - 2z + {z^2}} \right){z^{1/2}}dz} \cr & {\text{ }} = 2\int_0^1 {\left( {{z^{1/2}} - 2{z^{3/2}} + {z^{5/2}}} \right)dz} \cr & {\text{Integrating}} \cr & {\text{ }} = 2\left[ {\frac{{{z^{3/2}}}}{{3/2}} - 2\left( {\frac{{{z^{5/2}}}}{{5/2}}} \right) + \frac{{{z^{7/2}}}}{{7/2}}} \right]_0^1 \cr & {\text{ }} = 2\left[ {\frac{{2{z^{3/2}}}}{3} - \frac{{4{z^{5/2}}}}{5} + \frac{{2{z^{7/2}}}}{7}} \right]_0^1 \cr & {\text{Evaluating}} \cr & = 2\left[ {\frac{2}{3} - \frac{4}{5} + \frac{2}{7}} \right] - 2\left[ 0 \right] \cr & = 2\left( {\frac{{16}}{{105}}} \right) \cr & = \frac{{32}}{{105}} \cr} $$