$\int(3-x) \sqrt{4-x} d x=$ (Where $C$ is a constant of integration.)

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int(3-x) \sqrt{4-x} d x \ & =\int\{(4-x)-1\} \sqrt{4-x} d x=\int\left\{(4-x)^{\frac{3}{2}}-(4-x)^{\frac{1}{2}}ight\} d x \ & =\frac{2}{5}(4-x)^{5 / 2}+\frac{2}{3}(4-x)^{3 / 2}+C \ & =\frac{2}{3}(4-x)^{3 / 2}-\frac{2}{5}(4-x)^{5 / 2}+C\end{aligned}$

$\int(3-x) \sqrt{4-x} d x=$ (Where $C$ is a constant of integration.)

Solution