Integrals
In mathematics, Integration and Differentiation are the most significant ways to solve complex mathematical problems. Besides mathematics, Integration and Differentiation play key roles in Science, Engineering, and various other facets of our life. We have outlined what's, ifs, and how's related to integrals and their application. But how? Let us take an example.
To determine the area of a rectangle, the formulae 'length × breadth' is used. But what if you are given to calculate the area of the shaded portion, of a shaded rectangle with an unshaded circle within?
Importance of Integrals in Maths
Integration is an important chapter in Maths that needs to be studied well in advance by the students before they attempt tests on the topic. It always comes in use even if the students take up engineering, Science in their future years. Students can read from Application of Integrals on Vedantu to know more. They need to practice sums based on Integrals too to perfect them. It is an important topic that will prove to be quite instrumental later on.
How is the Calculation done when Integrals come into the Play
Integrals
In mathematics, the application of Integrals is applied to find the area under a curve, areas bounded by any curve, and so on.
Definition of Integrals
An integral is a function, of which a given function is a derivative. It is also known as the anti-derivative or reverse of a derivative. Integrals are used to determine the area of 2D objects and the volume of 3D objects in real life.
Types of Integrals
There are two types of Integrals.
Definite Integrals
Indefinite Integrals
Definite Integrals
An integral which has a start and an end value is known as a definite integral. In simple words, the function is restricted within an interval a,b, where a and b are upper limit and lower limit, respectively. It is represented as
y = \[\int_{b}^{a}\] f(x)dx
y = \[\int_{a}^{b}\] f(x)dx
Example: y = \[\int_{2}^{4}\] 6x dx
Sol: y = \[\int_{2}^{4}\] 6x dx
y = \[\int_{4}^{2}\] 6x dx , here a = 4, b = 2
y = 6\[\left [\frac{x^{2}}{2} \right ]_{2}^{4}\]
= 6\[\left [\frac{4^{2}}{2} \right ]\]-\[\left [ \frac{2^{2}}{2} \right ]\]=6\[\left [ 8-6 \right ]\]=36
Indefinite Integrals
An integral which does not have an upper and lower limit is known as a definite integral. It is represented as
y = \[\int_{b}^{a}\] f(x)dx = F(x)+C, where ‘C’ is a constant
Example: y = \[\int_{2}^{4}\] 6x dx
Sol:
y = \[\int\] 6xdx
y = 6 \[\left [\frac{x^{2}}{2} \right ]\] = 3x + C, where C is a constant
Solved Examples
Question 1: Determine the Area Enclosed By a Circle x²+ y² =a²
Sol: It is observed that the area enclosed by the given circle is ‘4 x area of the region AOBA bounded by the curve, x-axis and the ordinates x=0 and x=a’.
As the circle is symmetrical about both x−axis and y−axis=4 \[\int_{o}^{a}\]ydx (taking vertical strips)
= 4 \[\int_{o}^{a}\] \[\sqrt{\left (a^{2}-x^{2} \right )}\]dx
Since x²+ y² = a² gives y = ± \[\sqrt{a^{2}-x^{2}}\]
The quarter AOBA lies in the first quadrant, hence 'y' is taken as positive. On integrating, we get the entire area enclosed by the given circle.
= 4 \[\left [ \left [\frac{x}{2} \right ]\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a} \right ]\]
= 4 \[\left [ \left (\left [\frac{x}{2} \right ]\times 0+\frac{a^{2}}{2}sin^{-1}1 \right )-0 \right ]\]
= 4 \[\left (\frac{a^{2}}{2} \right )\]\[\left ( \frac{\pi }{2} \right )\]
= πa²
Question 2: Determine the Area of the Region Bounded b the Curve y = x² and the line Y = 4
Sol: Since the given curve expressed by the equation y = x² is a parabola symmetric about y-axis only, therefore, the required area of the region AOBA is given bt
= 2 \[\int_{0}^{4}\] xdy
= 2 area of the region BONB bounded by the curve, y-axis and the lines y=0 and y =4
=2 x ( area of the region BONB bounced by the curve, y-axis and the lines y = 0 and y = 4)
= 2 \[\int_{0}^{4}\] ydy
= 2 \[\times \frac{2}{3}\] \[\left [ y^{\frac{3}{2}} \right ]_{0}^{4}\]
= \[\frac{4}{3}\]×8
= \[\frac{32}{3}\]
Application of Integrals
Integrals have their application in both science and maths. In maths, the application of integral is made to determine the area under a curve, the area between two curves, the center of mass of a body, and so on. Whereas in science (Physics in particular), the application of integrals is made to calculate the Centre of Gravity, Mass, Momentum, Work done, Kinetic Energy, Velocity, Trajectory, and Thrust.
Application of Integrals in Engineering Fields
There’s a vast application of integration in the fields of engineering.
In Architecture
To determine the amount of material required in a curved surface. For instance, take the construction of a dome.
In Electrical Engineering
Integrals are used in Electrical Engineering to calculate the length of a power cable required for transmission between two power stations.
Application of Integrals in Different Fields
In Medical Science
Integrals are used to determine the growth of bacterias in the laboratory by keeping variables such as a change in temperature and foodstuff.
In Medicine
To study the rate of spread of infectious disease, the field of epidemiology uses medical seine to determine how fast a disease is spreading, its origin, and how to best treat it.
In Statistics
To estimate survey data to help improve marketing plans for different companies because a survey requires many different questions with a range of possible answers.
Does Vedantu have anything on Integrals?
Vedantu has study material on Integrals that can be used by all students to study from. It has an application of Integrals on its platform which students can read and understand. The material is completely free of cost and can even be downloaded in offline mode and studied.
FAQs on Application of Integrals
1. What is meant by the application of integrals in Class 12 Maths?
The application of integrals refers to using definite and indefinite integrals to determine quantities such as the area under curves, area between curves, volumes of solids, and the center of mass. In the CBSE 2025–26 syllabus for Class 12 Maths, this includes solving area-based problems with definite integrals, interpreting geometric regions, and applying integration in real-life contexts.
2. How do definite integrals help find the area between two curves in board exams?
To find the area between two curves using definite integrals, subtract the lower function from the upper function and integrate over the interval where they intersect:
- Area = ∫ab [f(x) − g(x)] dx, where f(x) is above g(x) between x=a and x=b.
- This method ensures accurate calculation of the bounded region, as required in CBSE Class 12 application questions.
3. What types of area problems are important for the CBSE Class 12 exam from the Application of Integrals chapter?
Important area problems often include:
- Finding the area under simple curves
- Calculating area between two intersecting curves
- Area enclosed by circles, parabolas, and ellipses
- Word problems with real-life applications of integration
4. Why is drawing and interpreting graphs essential when solving integral-based area questions?
Drawing graphs helps to visually represent the region whose area is to be calculated, identify correct limits of integration, determine which curve is upper and lower, and avoid mistakes in setup. It improves accuracy in CBSE board questions, especially for areas between curves.
5. How does the choice between vertical and horizontal strips affect integration problems for bounded areas?
The choice of integrating with respect to x (vertical strips) or y (horizontal strips) depends on how the region is best described:
- Use vertical strips (dx) when curves are functions of x.
- Use horizontal strips (dy) when curves are more easily expressed as functions of y.
6. What are common misconceptions students face when applying integrals to find areas?
Students often make errors such as:
- Mixing up definite and indefinite integrals
- Reversing upper and lower limits
- Not identifying the correct upper and lower functions
- Missing absolute values, resulting in negative areas
7. Can you explain, with an example, how to calculate the area enclosed by a circle using integration?
For the circle x² + y² = a², the area can be found by evaluating ∫0a √(a² − x²) dx (for the first quadrant) and multiplying by 4:
- Total area = 4 × ∫0a √(a² − x²) dx = πa², as per CBSE Class 12 methods.
8. How are problems involving ellipses and parabolas different from those with circles in the Application of Integrals?
For ellipses and parabolas, limits and integrand expressions change based on their specific equations. The principle remains the same—set up definite integrals using the equations provided, but the solving process may involve transforming the integrand or changing variable limits, as expected in board exam questions.
9. Why is mastering the method for setting up definite integrals important for board preparation?
Setting up definite integrals precisely is crucial because:
- It guarantees correct evaluation of areas or required quantities
- Misplaced limits or improper setup can result in loss of marks
- Board examiners look for logical setup and clear use of boundaries
10. What real-life situations require the application of integrals as studied in Class 12?
Integrals are used in various fields, such as:
- Calculating work done in physics
- Finding center of gravity of irregular objects
- Computing the volume of domes and curved structures in engineering
- Estimating population or bacterial growth in biology
- Analyzing cost and revenue in economics
11. What should students focus on during revision to score high in Application of Integrals?
Students should:
- Practice deriving and solving definite integrals for standard curves
- Pay attention to interpretation of geometric regions
- Work through previous years’ CBSE questions
- Ensure clear and labeled sketching of regions
12. How can recognizing the difference between definite and indefinite integrals help avoid errors in the exam?
Definite integrals give a specific value between two limits, often for area or accumulated quantity, while indefinite integrals provide a general expression (antiderivative) plus a constant. Knowing when to use each prevents mistakes in board answers, especially for questions requiring a final area or numeric value.
13. What if the limits of integration are reversed accidentally in the examination?
If the upper and lower limits are reversed, the result of the definite integral becomes negative. According to CBSE evaluation, area should always be reported as a positive value, so always check the orientation of your limits before finalizing your answer.
14. How does integration connect different topics in the CBSE Class 12 Maths syllabus?
Integration is closely linked with differentiation and is used to connect calculus concepts such as the Fundamental Theorem of Calculus. It is also applied in topics like differential equations and probability, allowing students to see the subject as an integrated whole rather than isolated chapters.
