Maths Application of Integrals: Definition, Formula, Application & Class 12 Math Notes

An integral is a basic concept in mathematics used to find the area under a curve or the accumulation of quantities. It is the reverse of differentiation. 

Types of Integrals 

There are two types of integrals: definite and indefinite integrals. 

1. Indefinite Integral

•  Represents antiderivatives 

•  Limits are not used in indefinite integrals. 

• Contains a constant of integration C 
• Example: $\int x^{2}dx=\frac{x^3}{3}+C$

2. Definite Integral

•  Express the numerical value of the area under a curve between two limits 

•  Has a fixed limit of integration 
• Example: ${\int }_a^{b}f\left(x\right)dx=\text{Area under the curve}y=f\left(x\right)\text{from}x=a\text{to}x=b$

The application of integrals is an important topic in calculus. It is used to find the antiderivatives of a function. Also, it helps to solve real-life problems related to area, volume, length, and physical quantities.

Also Read: NCERT Solution for class 11 and 12

Concepts Involved

1. Definite Integral as Area: The area under the curve of f (x) between the vertical line x=a and x=b is represented by the definite integral, expressed as ${\int }_a^{b}f\left(x\right)dx$
2. Area between two curves: To calculate the area between the curves f (x) and g (x), where f(x) is above g (x), we must integrate the difference [f(x) - g(x)].
3. Area Between a Curve and a Line: It is like the area between a curve. We need to integrate the difference between the curve and the line, considering which is above, to find the area between the curve and the line.
4. Area Bounded by Multiple Curves: For the complex region, first of all, divide the region into smaller sub-regions and apply the above concepts to each region.

• Finding Areas under a curve, between two curves, and bounded by lines and curves. Example: Area under curve y=f(x) between x=a and x=b

$A={\int }_a^{b}f\left(x\right)dx$

• Finding Volumes of solids of revolution and irregular solids using integration.

$V=\pi {\int }_a^b{\left(f\left(x\right)\right)}^{2}dx$

• Finding Lengths: Finding the arc length of a curve. Example: for the curve y=f(x)

$L={\int }_a^b\sqrt{1+{\genfrac{}{}{0ex}{}{dy}{dx}}^2}dx$

• Surface Area of a solid of revolution. Example: for the curve y=f(x) rotated about x-axis,

$S=2\pi {\int }_a^{b}y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

The area under a simple curve is the region enclosed between the curve and the coordinate axis(x-axis or y-axis) over a particular interval (limits). The area is calculated using the definite integrals by adding infinitely small strips of area to find the total region.

Important Topics:

NCERT Class 11 notes	CBSE NCERT Class 11 Physics
Class 11 Chemistry notes	Class 11 Maths notes

Formula for the area under a simple curve

• Area under a curve y =f(x) between x=a and x=b:

$A={\int }_a^{b}f\left(x\right)dx$

• Area under curve x=g(y) between y=c and y=d:

$A={\int }_c^{d}g\left(y\right)dy$

• If the curve lies below the axis (negative values):

$A=\left|{\int }_a^{b}f\left(x\right)dx\right|$

• Area between two curves y = f(x) and y=g(x), where $f\left(x\right)\ge g\left(x\right)$

$A={\int }_a^b\left(f\left(x\right)-g\left(x\right)\right)dx$

We shall discuss the area problems in which we are given a single curve whose equation contains both x and y. We have the following working rule:

1. Express the equation of the given curve in the form y = f(x).
2. Sketch the region whose area is to be determined. It can be done as follows:
   1. Draw a table containing the values of x and the corresponding values of y. The values in the table can be specifically chosen as x = 0, y = 0 and given values of x and y (if any). To know the shape of the curve more accurately, we can also take some more values of x and y.
   2. Plot the points (x, y) using a table and join them smoothly to get the required curve.
   3. Draw the given values of x (if any) as the lines parallel to the y-axis.
   4. Draw the given values of y (if any) as the lines parallel to the x-axis.
   5. Shade the region satisfying all the conditions of the given problem.
3. To find the required area, use the formula

$\mathrm{Area}={\int }_a^b\left(\mathrm{Upper}\mathrm{curve}\mathrm{function}-\mathrm{Lower}\mathrm{curve}\mathrm{function}\right)dx$

Also Check: NCERT Notes for Class 11 & 12

Students can get the link here for all class 12 maths chapters.

Chapter No.	Chapter Notes
1	Relations and Functions
2	Inverse Trigonometric Functions
3	Matrices
4	Determinants
5	Continuity and Differentiability
6	Application of Derivatives
7	Integrals
8	Application of Integrals
9	Differential Equations
10	Vector Algebra
11	Three-Dimensional Geometry
12	Linear Programming
13	Probability

Here are the links for Class 12 Maths NCERT solutions.

S.No.	Name of Chapter
1	Chapter 1 Relations and Functions
2	Chapter 2 Inverse Trigonometric Functions
3	Chapter 3 Matrices
4	Chapter 4 Determinants
5	Chapter 5 Continuity and Differentiability
6	Chapter 6 Applications of Derivatives
7	Chapter 7 Integrals
8	Chapter 8 Application of Integrals
9	Chapter 9 Differential Equations
10	Chapter 10 Vector Algebra
11	Chapter 11 Three-Dimensional Geometry
12	Chapter 12 Linear Programming
13	Chapter 13 Probability