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STD 12 - 7.2 definite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.2 definite integral1 Mark
MCQ
$\mathop \smallint \limits_3^6 \frac{{\sqrt x }}{{\sqrt {9 - x} + \sqrt x }}\;dx = $
  • A
    $\frac{1}{2}$
  • ✓
    $\frac{3}{2}$
  • C
    $2$
  • D
    $1$
✓

Answer

Correct option: B.
$\frac{3}{2}$
b
Let $I=\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} d x \ldots(i)$

$=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{9-9+x}+\sqrt{9-x}} d x$

$\Rightarrow I=\int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x \ldots(i i)$

On adding Eqs. (i) and (ii), we get

$2 I=\int_{3}^{6} \frac{\sqrt{x}+\sqrt{9-x}}{\sqrt{x}+\sqrt{9-x}} d x$

$=\int_{3}^{6} 1 d x=[x]_{3}^{6}$

$=6-3=3$

$\Rightarrow I=\frac{3}{2}$

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