Application of Integrals
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Application of Integrals

Team Careers360Updated on 17 Sep 2025, 07:21 PM IST

Imagine you are tracking the distance a car travels when its speed changes every second. To find the total distance covered, you need to add up all the small distances over time, this is where integrals become very useful. The application of integrals helps in calculating total quantities from varying rates, such as distance, area, volume, and more. In application of integrals class 12, students learn to solve problems using class 12 application of integrals formulas, important questions, and solutions, making it easier to understand how these concepts apply in real-world situations. Application of integrals class 12 notes provide a clear summary of key methods and formulas for quick revision. In this article, we will discuss the main applications of integrals, their formulas, methods, and solved examples to help you master class 12 maths effectively.

This Story also Contains

  1. Integrals
  2. Formulae for Integration
  3. Key Properties of Definite Integrals
  4. Application of Integrals Class 12
  5. Application of Integrals Formula
  6. List of Topics related to applications of Integrals according to NCERT/JEE MAIN
  7. Important Books and Resources for Class 12 Integrals
  8. Practice Questions based on Applications of Integrals
Application of Integrals
Applications of Integrals

Integrals

Integration is the reverse process of differentiation. In integration, we determine the function whose derivative is given. Class 12 application of integrals formulas rely on this principle to solve real-life and mathematical problems.

Integrals are based on a limiting process that approximates the area of a curvilinear region by dividing it into thin vertical slabs and summing them up.

For example: $\frac{d}{dx} (\sin x) = \cos x$
$\frac{d}{dx} (x^2) = 2x$
$\frac{d}{dx} (e^x) = e^x$

In these examples, $\cos(x)$ is the derivative of $\sin(x)$. So, $\sin(x)$ is an antiderivative (or integral) of $\cos(x)$. Similarly, $x^2$ and $e^x$ are antiderivatives (or integrals) of $2x$ and $e^x$, respectively.

Role of the Constant of Integration

The derivative of a constant ($C$) is zero. Therefore, we can also write:

$\frac{d}{dx} (\sin x + C) = \cos x$
$\frac{d}{dx} (x^2 + C) = 2x$
$\frac{d}{dx} (e^x + C) = e^x$

This shows that the antiderivatives (or integrals) of a function are not unique. There are infinitely many integrals for each function, obtained by choosing $C$ arbitrarily from real numbers. This principle forms the foundation of many application of integrals class 12 solutions and exercises.

Formulae for Integration

This section provides all essential formulas for indefinite and definite integrals, covering polynomials, trigonometric, exponential, logarithmic, and inverse functions. Mastering these formulas is key for solving application of integrals class 12 problems efficiently.

Formulas for Indefinite Integrals

This section lists all key formulas for indefinite integrals, including polynomials, trigonometric, exponential, and logarithmic functions, essential for solving class 12 integrals problems.

Function / RuleIntegral Formula
Constant function$\int 0 ,dx = C$
Constant multiplied$\int k ,dx = kx + C$
Power function$\int x^n ,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$
Reciprocal function$\int \frac{1}{x} ,dx = \log x + C$
Sine function$\int \sin x ,dx = -\cos x + C$
Cosine function$\int \cos x ,dx = \sin x + C$
Secant squared$\int \sec^2 x ,dx = \tan x + C$
Cosecant squared$\int \csc^2 x ,dx = -\cot x + C$
Secant × tangent$\int \sec x \tan x ,dx = \sec x + C$
Cosecant × cotangent$\int \csc x \cot x ,dx = -\csc x + C$
Exponential $e^x$$\int e^x ,dx = e^x + C$
Exponential $a^x$$\int a^x ,dx = \frac{a^x}{\log a} + C$
Inverse sine$\int \frac{1}{\sqrt{1-x^2}} ,dx = \sin^{-1} x + C$
Inverse tangent$\int \frac{1}{1+x^2} ,dx = \tan^{-1} x + C$
Logarithmic Function$\int \frac{1}{x} ,dx = \log x + C$

Formulas for Definite Integrals

Here, you’ll find important formulas for definite integrals to calculate areas, volumes, and accumulated quantities in class 12 application of integrals problems.

Function / RuleDefinite Integral Formula
Constant function$\int_a^b k ,dx = k(b-a)$
Power function$\int_a^b x^n ,dx = \frac{b^{n+1} - a^{n+1}}{n+1}, \quad n \neq -1$
Reciprocal function$\int_a^b \frac{1}{x} ,dx = \log b - \log a$
Sine function$\int_a^b \sin x ,dx = -\cos b + \cos a$
Cosine function$\int_a^b \cos x ,dx = \sin b - \sin a$
Secant squared$\int_a^b \sec^2 x ,dx = \tan b - \tan a$
Cosecant squared$\int_a^b \csc^2 x ,dx = -\cot b + \cot a$
Secant × tangent$\int_a^b \sec x \tan x ,dx = \sec b - \sec a$
Cosecant × cotangent$\int_a^b \csc x \cot x ,dx = -\csc b + \csc a$
Exponential $e^x$$\int_a^b e^x ,dx = e^b - e^a$
Exponential $a^x$$\int_a^b a^x ,dx = \frac{a^b - a^a}{\log a}$
Inverse sine$\int_a^b \frac{1}{\sqrt{1-x^2}} ,dx = \sin^{-1} b - \sin^{-1} a$
Inverse tangent$\int_a^b \frac{1}{1+x^2} ,dx = \tan^{-1} b - \tan^{-1} a$

Key Properties of Definite Integrals

  1. Change of variable: $\int_a^b f(x) ,dx = \int_a^b f(t) ,dt$

  2. Reversing limits: $\int_a^b f(x) ,dx = -\int_b^a f(x) ,dx$

  3. Integral over same limits: $\int_a^a f(x) ,dx = 0$

  4. Breaking into parts: $\int_a^b f(x) ,dx = \int_a^c f(x) ,dx + \int_c^b f(x) ,dx$

  5. Even function: $\int_{-a}^a f(x) ,dx = 2 \int_0^a f(x) ,dx, \quad f(-x) = f(x)$

  6. Odd function: $\int_{-a}^a f(x) ,dx = 0, \quad f(-x) = -f(x)$

Application of Integrals Class 12

The application of Integrals class 12 notes includes - the area along the X-axis and Y-axis, the area of piecewise function and the area bounded by two curves.

Area along X-axis

If the function $f(x) ≥ 0 ∀ x ∈ [a, b]$ then $\int_a^{\infty} f(x) d x$ represents the area bounded by $y = f(x), x-$axis and lines $x = a$ and $x = b$.

If the function $f(x) ≤ 0 ∀ x ∈ [a, b]$, then the area by bounded $4y = f(x)$, x-axis and lines $4x = a$ and $x = b$ is $\int_a^b f(x) d x$.

Area along Y-axis

Area along y axis

The area by bounded $x = g(y)$ [with $g(y)>0$], $y$-axis and the lines $y = a$ and $y = b$ is $\int_a^b x d y=\int_a^b g(y) d y$

Area of Piecewise Function

If the graph of the function $f(x)$ is of the following form, then

image-20240912172242-5image-20240912172242-6

then $\int_a^b f(x) d x$ will equal $A_1-A_2+A_3-A_4$ and not $A_1+A_2+A_3+A_4$.

If we need to evaluate $A_1+A_2+A_3+A_1$ (the magnitude of the bounded area) we will have to calculate

$ \underbrace{\int_a^x f(x) d x}_{\mathrm{A}_1}+\underbrace{\left|\int_x^y f(x) d x\right|}_{\mathrm{A}_2}+\underbrace{\int_5^z f(x) d x}_{\mathrm{A}_3}+\underbrace{\left|\int_z^b f(x) d x\right|}_{\mathrm{A}_4} $

The area bounded by the curve when the curve intersects $X$-axis

The graph $y=f(x) \forall x \in[a, b]$ intersects $x-a x i s$ at $x=c$.

If the function $f(x) \geq 0 \forall x \in[a, c]$ and $f(x) \leq 0 \forall x \in[c, b]$ then area bounded by curve and $x$-axis, between lines $x=a$ and $x=b$ is

$ \int_a^b|f(x)| d x=\int_a^c f(x) d x-\int_c^b f(x) d x$

image-20240912172242-7image-20240912172242-8

Area Bounded by Two Curves

Area bounded by the curves $y=f(x), y=g(x) $ and the lines $ x = a$ and $x = b$, and it is given that $f(x) ≤ g(x). $

image-20240912172242-9image-20240912172242-10

From the figure, it is clear that,

Area of the shaded region = Area of the region $ABEF$ - Area of the region $ABCD$

$\int_a^b g(x) d x-\int_a^b f(x) d x=\int_a^b(\underbrace{g(x)}_{\begin{array}{c}\text { upper } \\ \text { curve }\end{array}}-\underbrace{f(x)}_{\begin{array}{c}\text { lower } \\ \text { curve }\end{array}}) d x$

Area Bounded by Curves When Intersects at More Than One Point

Area bounded by the curves $y = f(x), y = g(x)$ which intersect each other in the interval $[a, b]$

First find the point of intersection of these curves $y = f(x)$ and $y = g(x)$ by solving the equation $f(x) = g(x)$, let the point of intersection be $x = c $

image-20240912172242-11image-20240912172242-12

$=\int_a^c\{f(x)-g(x)\} d x+\int_c^b\{g(x)-f(x)\} d x$

When two curves intersects more than one point

Area bounded by the curves $y=f(x), y=g(x)$ which intersect each other at three points at $x = a, x = b$ and $x = c. $

To find the point of intersection, solve $f(x) = g(x). $

For $x ∈ (a, c), f(x) > g(x)$ and for $x ∈ (c, b),g(x) > f(x).$

Area bounded by curves,

$\begin{aligned} A & =\int_a^b|f(x)-g(x)| d x \\ & =\int_a^c(f(x)-g(x)) d x+\int^b(g(x)-f(x)) d x\end{aligned}$

Now, let us summarize and recall the application of integrals class 12 formulas.

Application of Integrals Formula

Application of integrals class 12 formulas include the formulas for the topics area along the X-axis and Y-axis, area of piecewise function, and are bounded by two curves.

Area along X-axis

If the function $f(x) ≥ 0 ∀ x ∈ [a, b]$ then $\int_a^{\infty} f(x) d x$ represents the area bounded by $y = f(x), x-$axis and lines $x = a$ and $x = b$.

If the function $f(x) ≤ 0 ∀ x ∈ [a, b]$, then the area by bounded $4y = f(x)$, x-axis and lines $4x = a$ and $x = b$ is $\int_a^b f(x) d x$.

Area along Y-axis

The area by bounded $x = g(y)$ [with $g(y)>0$], $y$-axis and the lines $y = a$ and $y = b$ is $\int_a^b x d y=\int_a^b g(y) d y$

Area of Piecewise Function

If the function $f(x) \geq 0 \forall x \in[a, c]$ and $f(x) \leq 0 \forall x \in[c, b]$ then area bounded by curve and $x$-axis, between lines $x=a$ and $x=b$ is

$ \int_a^b|f(x)| d x=\int_a^c f(x) d x-\int_c^b f(x) d x$

Area Bounded by Two Curves

Area bounded by the curves $y=f(x), y=g(x) $ and the lines $ x = a$ and $x = b$, and it is given that $f(x) ≤ g(x). $

Area bounded by the curves = $\int_a^b g(x) d x-\int_a^b f(x) d x=\int_a^b(\underbrace{g(x)}_{\begin{array}{c}\text { upper } \\ \text { curve }\end{array}}-\underbrace{f(x)}_{\begin{array}{c}\text { lower } \\ \text { curve }\end{array}}) d x$

Important Books and Resources for Class 12 Integrals

Here, you’ll find the best books and reference materials to strengthen your understanding of integrals and practice effectively for exams.

Book Title

Author / Publisher

Description

NCERT Class 12 Mathematics

NCERT

Official textbook with fundamental theory and exercises on integrals.

Mathematics for Class 12

R.D. Sharma

Detailed explanations, solved examples, and practice on integration techniques.

Objective Mathematics

R.S. Aggarwal

Provides practice problems and objective questions on integrals.

Arihant All-In-One Mathematics

Arihant

Comprehensive theory and unsolved/specially solved integration problems for JEE and board exams.

Integral Calculus

M.L. Khanna

Advanced textbook covering integral calculus concepts and applications.

NCERT Resources

This section highlights NCERT textbooks and reference material that clearly explain integrals, forming the foundation for solving problems.

NCERT Subjectwise Resources

Explore subject-wise NCERT resources that cover notes, solutions and exemplar problems, supporting structured and focused learning.

Frequently Asked Questions (FAQs)

Q: How is the area between two curves calculated using integrals?
A:

The area is calculated as $\int_a^b |f(x) - g(x)| , dx$, where $f(x)$ and $g(x)$ are the functions forming the boundaries

Q: How are integrals applied in economics?
A:

They are used to calculate total revenue, total cost, consumer and producer surplus, and accumulated profit over time.

Q: What are the main applications of integrals in mathematics?
A:

Integrals are used to find areas under curves, volumes of solids, total distance from velocity, work done by a force, and accumulated quantities.

Q: Is it feasible to have a negative area between two curves?
A:

No, there will be no negative area between the two curves. This is due to the fact that the area between the two curves differs from the region beneath the curve. As a result, the area between the two curves must be positive at all times.

Q: What does the area under the curve mean?
A:

A definite integral between two locations can be used to compute the area under a curve that exists between them. To get the area under the curve y = f(x) between x = a and x = b, integrate y = f(x) between a and b's limits.

Q: What does it mean to have a polar curve?
A:

A polar curve is a form that is created with the use of the polar coordinate system. They are distinguished by the presence of points at varying distances from the pole or origin.