Question
Find the area bounded by the curve y = sin x between x = 0 and $x = 2\pi $
✓
Answer
Equation of the curve is y = sin x …(i)
$\therefore y = \sin x \geqslant 0$ for $0 \leqslant x \leqslant \pi $ i.e., graph is in first and second quadrant.
And $y = \sin x \leqslant 0$ for $\pi \leqslant x \leqslant 2\pi $ i.e., graph is in third and fourth quadrant.
If tangent is parallel to x - axis, then $\frac{{dy}}{{dx}} = 0$
$ \Rightarrow \cos x = 0$
$ \Rightarrow x = \frac{\pi }{2},\frac{{3\pi }}{2}$
Table of values for curve y = sin x between x = 0 and $x = 2\pi $
$\therefore y = \sin x \geqslant 0$ for $0 \leqslant x \leqslant \pi $ i.e., graph is in first and second quadrant.
And $y = \sin x \leqslant 0$ for $\pi \leqslant x \leqslant 2\pi $ i.e., graph is in third and fourth quadrant.
If tangent is parallel to x - axis, then $\frac{{dy}}{{dx}} = 0$
$ \Rightarrow \cos x = 0$
$ \Rightarrow x = \frac{\pi }{2},\frac{{3\pi }}{2}$
Table of values for curve y = sin x between x = 0 and $x = 2\pi $
| x | 0 | $\frac{\pi }{2}$ | $\pi $ | $\frac{{3\pi }}{2}$ | $2\pi $ |
| y | 0 | 1 | 0 | - 1 | 0 |
Now Required shaded area = Area OAB + Area BCD
$= \left| {\int\limits_0^\pi {ydx} } \right| + \left| {\int\limits_\pi ^{2\pi } {ydx} } \right|$
$= \left| {\int\limits_0^\pi {\sin xdx} } \right| + \left| {\int\limits_\pi ^{2\pi } {\sin xdx} } \right|$
$= \left| { - \left( {\cos x} \right)_0^\pi } \right| + \left| { - \left( {\cos x} \right)_\pi ^{2\pi }} \right|$
$= \left| { - \left( {\cos \pi - \cos 0} \right)} \right| + \left| { - \left( {\cos 2\pi - \cos \pi } \right)} \right|$
$= \left| { - \left( { - 1 - 1} \right)} \right| + \left| { - \left( {1 + 1} \right)} \right|$
= 2 + 2 = 4 sq. units
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