MCQ
$\int_3^4 \sqrt{\frac{x-3}{4-x}} d x=$
  • A
    $0$
  • B
    $\frac{\pi}{4}$
  • $\frac{\pi}{2}$
  • D
    $\pi$

Answer

Correct option: C.
$\frac{\pi}{2}$
(C)
Here, $a =3$ and $b =4$
$\int_{ a }^{ b } \sqrt{\frac{x- a }{ b -x}} d x=\frac{\pi}{2}(b- a )$
$\therefore \int_3^4 \sqrt{\frac{x-3}{4-x}} d x=\frac{\pi}{2}(4-3)=\frac{\pi}{2}$

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