Question:

The area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi $ given by

Updated On: May 12, 2024
  • $4\, s\, units$
  • $3\, s\, units$
  • $0\, s\, units$
  • $2\, s \, units$
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The Correct Option is A

Solution and Explanation

We have, $y = \sin x, x = 0$ and $x = 2\pi$
$ \therefore$ Required area = Area of shaded region
$\int\limits_{0}^{2\pi} \left|\sin x\right|dx = \int\limits_{0}^{\pi} \sin x dx + \int\limits_{\pi}^{2\pi} \left(-\sin x\right)dx $
$=- \left[\cos x\right]^{\pi}_{0} + \left[\cos x\right]^{2\pi}_{\pi} $
$= -\left[\cos \pi -\cos 0\right]+\left[\cos 2\pi -\cos \pi\right] $
$=- \left[-1 -1 \right]+\left[1+1\right]=4$
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Concepts Used:

Area under Simple Curves

  • The area of the region bounded by the curve y = f (x), x-axis and the lines x = a and x = b (b > a) - given by the formula:
\[\text{Area}=\int_a^bydx=\int_a^bf(x)dx\]
  • The area of the region bounded by the curve x = φ (y), y-axis and the lines y = c, y = d - given by the formula:
\[\text{Area}=\int_c^dxdy=\int_c^d\phi(y)dy\]

Read More: Area under the curve formula