Question
By using the properties of definite integral, evaluate the integral in Exercise:
$\int^{2}_{0}\text{x}\sqrt{2-\text{x}}\ \text{dx}$
$\int^{2}_{0}\text{x}\sqrt{2-\text{x}}\ \text{dx}$
✓
Answer
$\text{Let}\ \text{I}=\int\limits_{0}^{2}\text{x}\sqrt{2-\text{x}}\ \text{dx}=\int\limits_{0}^{2}(2-\text{x})\sqrt{2-(2-\text{x)}}\ \text{dx}\ \bigg[\because\int\limits_{0}^{\text{a}}\text{f}\text{(x)}\text{dx}=\int\limits_{0}^{\text{a}}(\text{a}-\text{x})\text{dx}=\bigg]$
$\Rightarrow\ \text{I}=\int\limits_{0}^{2}(2-\text{x})\sqrt{\text{x}}\ \text{dx}=\int\limits_{0}^{2}\bigg({2\text{x}^{\frac{1}{2}}-\text{x}^{\frac{3}{2}}}\bigg)\text{dx} =\Bigg[2.\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}\Bigg]^{2}_{0}=\bigg(\frac{4}{3}.2^{\frac{3}{2}}-\frac{2}{5}.2^{\frac{5}{2}}\bigg)-(0-0)$
$\Rightarrow\ \text{I}=\frac{4}{3}\times2\sqrt{2}-\frac{2}{5}\times4\sqrt{2}=\bigg(\frac{8}{3}-\frac{8}{5}\bigg)\sqrt{2}={\frac{16\sqrt{2}}{15}}$
$\Rightarrow\ \text{I}=\int\limits_{0}^{2}(2-\text{x})\sqrt{\text{x}}\ \text{dx}=\int\limits_{0}^{2}\bigg({2\text{x}^{\frac{1}{2}}-\text{x}^{\frac{3}{2}}}\bigg)\text{dx} =\Bigg[2.\frac{\text{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{x}^{\frac{5}{2}}}{\frac{5}{2}}\Bigg]^{2}_{0}=\bigg(\frac{4}{3}.2^{\frac{3}{2}}-\frac{2}{5}.2^{\frac{5}{2}}\bigg)-(0-0)$
$\Rightarrow\ \text{I}=\frac{4}{3}\times2\sqrt{2}-\frac{2}{5}\times4\sqrt{2}=\bigg(\frac{8}{3}-\frac{8}{5}\bigg)\sqrt{2}={\frac{16\sqrt{2}}{15}}$
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