Q64 / 150 · MHT CET 2022 (07 Aug Shift 2) · Mathematics

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

  • A
    $\frac{2}{3}(4-x)^{3 / 2}+\frac{2}{5}(4-x)^{5 / 2}+C$
  • B
    $-\frac{2}{5}(4-x)^{5 / 2}+\frac{2}{3}(4-x)^{3 / 2}+C$
  • Correct answer
    C
    $\frac{2}{3}(4-x)^{3 / 2}-\frac{2}{5}(4-x)^{5 / 2}+C$
  • D
    $\frac{2}{5}(4-x)^{5 / 2}-\frac{2}{5}(4-x)^{3 / 2}+C$

Solution

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

Hence, the correct answer is option: 'C'.

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