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Area under Simple Curves
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the area bounded by the curve y x between x 0 and
Question:
The area bounded by the curve
$y = \sin x$
between
$x = 0$
and
$x = 2\pi $
given by
COMEDK UGET - 2010
COMEDK UGET
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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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:
\[\text{Area}=\int_a^bydx=\int_a^bf(x)dx\]
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