Question

Evaluate: ${\displaystyle {\int }_0^1{\text{t}}^2\sqrt{1-\text{t}} \text{dt}}$

Answer 1

Let I = ${\displaystyle {\int }_0^1{\text{t}}^2\sqrt{1-\text{t}} \text{dt}}$

= ${\displaystyle {\int }_0^1{(1-\text{t})}^2\sqrt{1-(1-\text{t})} \text{dt}}$   ......${\displaystyle [\because {\int }_0^{\text{a}}\text{f}\left(x\right)\text{d}x={\int }_0^{\text{a}}\text{f}(\text{a}-x)\text{d}x]}$

= ${\displaystyle {\int }_0^1(1-2\text{t}+{\text{t}}^2)\sqrt{\text{t}} \text{dt}}$

= ${\displaystyle {\int }_0^1({\text{t}}^{\frac{1}{2}}-2{\text{t}}^{\frac{3}{2}}+{\text{t}}^{\frac{5}{2}})\text{dt}}$

= ${\displaystyle {\int }_0^1{\text{t}}^{\frac{1}{2}} \text{dt}-2{\int }_0^1{\text{t}}^{\frac{3}{2}} \text{dt}+{\int }_0^1{\text{t}}^{\frac{5}{2}} \text{dt}}$

=  ${\displaystyle {\left[\frac{{\text{t}}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_0^1-2{\left[\frac{{\text{t}}^{\frac{5}{2}}}{\frac{5}{2}}\right]}_0^1+{\left[\frac{{\text{t}}^{\frac{7}{2}}}{\frac{7}{2}}\right]}_0^1}$

= ${\displaystyle \frac{2}{3}(1^{\frac{3}{2}}-0)-\frac{4}{5}(1^{\frac{5}{2}}-0)+\frac{2}{7}(1^{\frac{7}{2}}-0)}$

= ${\displaystyle \frac{2}{3}-\frac{4}{5}+\frac{2}{7}}$

= ${\displaystyle \frac{70-84+30}{105}}$

∴ I = ${\displaystyle \frac{16}{105}}$