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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_0^\pi \sqrt {1 + 4{{\sin }^2}\frac{x}{2} - 4\sin \frac{x}{2}} \;dx = $
  • A
    $4\sqrt 3 - 4$
  • ✓
    $\;4\sqrt 3 - 4 - \frac{\pi }{3}$
  • C
    $\pi - 4\;$
  • D
    $\frac{{2\pi }}{3} - 4\sqrt 3 - 4$
✓

Answer

Correct option: B.
$\;4\sqrt 3 - 4 - \frac{\pi }{3}$
b
$\int\limits_0^\pi  {\left| {\left( {1 - 2\sin \frac{x}{2}} \right)} \right|} dx$

$ = \int\limits_0^{\frac{\pi }{3}} {\left| {\left( {1 - 2\sin \frac{x}{2}} \right)} \right|} dx - \int\limits_{\frac{\pi }{3}}^\pi  {\left| {\left( {1 - 2\sin \frac{x}{2}} \right)} \right|} dx$

$=\left(x+4 \cos \frac{x}{2}\right)_{0}^{\frac{\pi}{3}}-\left(x+4 \cos \frac{x}{2}\right)_{\frac{\pi}{3}}^{\pi}$

$=\frac{\pi}{3}+4 \cos \frac{\pi}{6}-0-4-\left(\pi+4 \cos \frac{\pi}{2}-\frac{\pi}{3}-4 \cos \frac{\pi}{6}\right)$

$=-\frac{\pi}{3}+4 \sqrt{3}-4$

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