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What is the area of the portion of the curve $y = \sin x$, lying $x = 0$, $y = 0$ and $x = 2\pi $?
A) $1$ square unit
B) $2$ square units
C) $4$ square units
D) $8$ square units


Answer
VerifiedVerified
162.9k+ views
Hint: In this question, we are given the equation of the curve and the lines. We have to find the area bounded by them. Firstly, plot the graph of the following. Then, for the area integrate the equation of the curve with respect to $dx$ from $x = 0$ to $x = 2\pi $. Also, the area from $0$ to $\pi $ and $\pi $ to $2\pi $ is same. So, you can solve any of the integration part and then double it. Solve it further using integration and trigonometric formulas.



Formula Used:Integration formula –
$\int {\sin xdx = - \cos x + c} $
Trigonometric table values –
$\cos \pi = - 1$, $\cos {0^ \circ } = 1$



Complete step by step solution:Given that,
Equation of the curve is $y = \sin x$ and the line is $x = 2\pi $
Graph of the given equations is attached below (figure 1);


Figure 1: A graph containing the curve y=sin x and the line x=2πœ‹
Now, as you can see the area of bounded region lies between $x = 0$ to $x = 2\pi $
So, integrate $y = \sin x$ with respect to $dx$ from $0$ to $2\pi $
Therefore, area will be $A = \int\limits_0^{2\pi } {\sin xdx} $
Now, the area is divided into two parts
So, $A = \int\limits_0^\pi {\sin xdx} + \int\limits_\pi ^{2\pi } {\sin xdx} $
As we know that, $\int\limits_0^\pi {\sin xdx} = \int\limits_\pi ^{2\pi } {\sin xdx} $
We can write the area as $A = 2\int\limits_0^\pi {\sin xdx} $
As we know that, integration of the function $\sin x$ is $ - \cos x + c$
It implies that,
$A = 2\left[ { - \cos x} \right]_0^\pi $
Now, resolving the limits we get
$A = 2\left[ { - \cos \pi + \cos 0} \right]$
$A = 2\left[ { - \left( { - 1} \right) + 1} \right]$
$A = 4$ square units



Option β€˜C’ is correct

Note: Different methods are used to determine the area under the curve, with the antiderivative method being the most prevalent. The area under the curve can be calculated by knowing the curve's equation, borders, and the axis surrounding the curve. There exist formulas for obtaining the areas of regular shapes such as squares, rectangles, quadrilaterals, polygons, and circles, but no formula for finding the area under a curve. The integration procedure aids in solving the equation and determining the required area. Antiderivative methods are highly useful for determining the areas of irregular planar surfaces.