
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
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 β
$\int {\sin xdx = - \cos x + c} $
Trigonometric table values β
$\cos \pi = - 1$, $\cos {0^ \circ } = 1$
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
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 βCβ 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.
Formula Used:Integration formula β
$\int {\sin xdx = - \cos x + c} $
Trigonometric table values β
$\cos \pi = - 1$, $\cos {0^ \circ } = 1$
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
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 βCβ 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.
Recently Updated Pages
Geometry of Complex Numbers β Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Instantaneous Velocity - Formula based Examples for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges

NEET Total Marks 2025
