MCQ
Area bounded by curve $y = k\sin x$ between $x = \pi $ and $x = 2\pi ,$ is
  • $2k$ sq. unit
  • B
    $0$
  • C
    $\frac{{{k^2}}}{2}\,\, sq. \,unit$
  • D
    $k\,\, sq. \,unit$

Answer

Correct option: A.
$2k$ sq. unit
a
(a) Required area $ = k\int_\pi ^{2\pi } {\sin x\,\,dx = k} [ - \cos x]_\pi ^{2\pi } = - 2k$

Hence, area $= 2k\,\, sq. \,unit$

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