Question

$\int (1-x)\sqrt{x}dx$ is equal to(where $C$ is constant of integration)

• A. $\frac{2}{3}{x}^{\frac{3}{2}}+\frac{2}{5}{x}^{\frac{5}{2}}+C$
• B. $\frac{4}{3}{x}^{\frac{3}{4}}-\frac{2}{5}{x}^{\frac{5}{2}}+C$
• C. $\frac{2}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}+C$
• D. $\frac{4}{3}{x}^{\frac{3}{4}}-\frac{3}{5}{x}^{\frac{5}{3}}+C$

Answer 1

The correct option is C $\frac{2}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}+C$

$\int (1-x)\sqrt{x}dx=\int (\sqrt{x}-x\sqrt{x})dx=\int x^{\frac{1}{2}}dx-\int x^{\frac{3}{2}}dx=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}+C=\frac{2}{3}{x}^{\frac{3}{2}}-\frac{2}{5}{x}^{\frac{5}{2}}+C$