Integrals

Integrals in Mathematics

Integrals are a way to add up small pieces to find the whole. For example, if you want to know how much water fills a pool or how much area is under a curve, integrals help you do that. It’s like taking tiny slices of a shape and adding them together to get the full picture. In math, when you know how something changes (its derivative), you can use integrals to find the original quantity or function.

Integrals are everywhere – from calculating distance travelled, to figuring out areas, and even in science and engineering problems. Once you understand how they work, you’ll see how useful they are in real-life situations!

Integral calculus is mainly divided into indefinite integrals and definite integrals.

Types of Integrals: Indefinite Integrals

Indefinite integrals are a way to find the original function when you know how it’s changing. It’s like working backwards from the derivative to get the function itself. This process is called indefinite integration, and it forms the basic definition of integration or antiderivatives.

Definition of Indefinite Integrals

If $\frac{d}{dx} F(x) = f(x)$, then we write: $\int f(x)dx = F(x) + C$

Here, $C$ is the constant of integration, and it shows that there are infinitely many solutions because you can choose any value for $C$. All these functions differ only by that constant.

Properties of Indefinite Integrals

(i) Inverse Relationship
Differentiation and integration are opposite processes:
$\frac{d}{dx}\int f(x)dx = f(x)$
$\int f'(x)dx = f(x) + C$

(ii) Family of Curves
Different integrals with the same derivative represent a group of curves that differ only by a constant. So, if two functions have the same derivative, their integrals are considered equivalent.

(iii) Sum Rule
You can break an integral of a sum into the sum of integrals:
$\int (f(x) + g(x))dx = \int f(x)dx + \int g(x)dx$

(iv) Constant Factor Rule
A constant can be taken out of the integral:
$\int a f(x)dx = a \int f(x)dx$
where $a$ is a constant.

Types of Integrals: Definite Integrals

While indefinite integrals give us a general solution, definite integrals help calculate the exact value of a function between two points. This is very useful in real-world problems where you need precise answers.

Definition of Definite Integrals

Let’s say $f(x)$ is a function defined on the interval $[a, b]$ and $F(x)$ is its antiderivative such that $\frac{d}{dx}F(x) = f(x)$. The definite integral is written as: $\int_a^b f(x)dx = F(b) - F(a)$

Here, $a$ is the lower limit and $b$ is the upper limit. This formula helps in calculating things like area under a curve or the total accumulated change in a quantity.

Applications of Inequalities in Definite Integration

Inequalities are useful to compare two functions, especially when solving problems related to application of integrals class 12.

1. Bounded Functions

If $m \leq f(x) \leq M$ for all $x$ in $[a, b]$, then: $m(b-a) \leq \int_a^b f(x)dx \leq M(b-a)$

2. Mean Value Theorem

If $f(x)$ is continuous on $[a, b]$, then there’s at least one $c \in [a, b]$ where: $f(c) = \frac{1}{b-a} \int_a^b f(x)dx$

3. Comparing Functions

If $f(x) \geq g(x)$ for all $x \in [a, b]$, then: $\int_a^b f(x)dx \geq \int_a^b g(x)dx$

4. Jensen’s Inequality

For convex functions $\phi(x)$ and a probability distribution: $\phi\left(\int_a^b f(x)dx\right) \leq \int_a^b \phi(f(x))dx$

5. Minkowski’s Inequality

For functions in $L^p$ space: $\left(\int_a^b |f(x)+g(x)|^p dx\right)^{1/p} \leq \left(\int_a^b |f(x)|^p dx\right)^{1/p} + \left(\int_a^b |g(x)|^p dx\right)^{1/p}$

Properties of Definite Integrals

The properties of definite integrals make it easier to solve problems in integrals class 12 and understand the behavior of functions across intervals.

1. $\int_a^b f(x)dx = \int_a^b f(t)dt$

2. $\int_a^b f(x)dx = -\int_b^a f(x)dx$ and $\int_a^a f(x)dx = 0$

3. $\int_a^b f(x)dx = \int_a^c f(x)dx + \int_c^b f(x)dx$

4. $\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$

5. $\int_0^a f(x)dx = \int_0^a f(a-x)dx$

6. $\int_0^{2a} f(x)dx = \int_0^a f(x)dx + \int_0^a f(2a-x)dx$

7. $\int_0^{2a} f(x)dx = \begin{cases}
   2 \int_0^a f(x)dx, & \text{if } f(2a-x) = f(x) \\
   0, & \text{if } f(2a-x) = -f(x)
   \end{cases}$
8. (i) $\int_{-a}^a f(x)dx = 2 \int_0^a f(x)dx$, if $f(x)$ is even, i.e., $f(-x) = f(x)$
   (ii) $\int_{-a}^a f(x)dx = 0$, if $f(x)$ is odd, i.e., $f(-x) = -f(x)$

These properties are not just theoretical; they’re applied in many problems when working through class 12 integrals solutions or solving real-life problems using the application of integrals class 12.

Difference between Definite and Indefinite Integrals

Integrals Formulae

This section provides all the important formulas of integrals, including basic functions and rules, to help you solve problems easily and strengthen your understanding of integrals. These formulas are essential for both theory and practice in exams like JEE Main.

Indefinite Integral Formulae

We have provided below a list of important Indefinite Integral Formulas:

Definite Integral Formulae

We have provided the formulae for definite integrals:

Rules of Integration

In addition to memorising formulas, understanding the rules of integration helps simplify problems and work with complex functions.

(a) Constant Multiple Rule

If $f(x)$ has an antiderivative, then multiplying it by a constant $k$ gives:
$\int kf(x)dx = k\int f(x)dx$

(b) Sum Rule

You can split the integral of a sum into the sum of integrals:
$\int (f(x) + g(x))dx = \int f(x)dx + \int g(x)dx$

(c) Difference Rule

Similarly, the difference between functions can be split:
$\int (f(x) - g(x))dx = \int f(x)dx - \int g(x)dx$

Integration by Parts

Integration by Parts Formula: Sometimes you’ll need to integrate products of functions. For this, we use integration by parts, a powerful technique in the application of integrals class 12.

If two functions, $u(x)$ and $v(x)$, are given, then:
$\int u v ,dx = u \times \int v ,dx - \int \left(\frac{du}{dx} \times \int v ,dx \right)dx$

This method helps when simple formulas aren’t enough, especially in problems involving products of algebraic, logarithmic, or trigonometric functions.

ILATE Rule – Choosing $u$

When selecting which function to assign as $u$, we use the ILATE rule:
Inverse > Logarithmic > Algebraic > Trigonometric > Exponential

For example, if your integral includes both a logarithmic and an algebraic function, choose the logarithmic function as $u$ because it comes first in ILATE.

Applications of Integrals Class 12

Integrals are more than just formulas — they are essential tools that help solve real problems in mathematics. Below are some of the key areas where integrals are used, along with simple explanations to help you understand their importance.

1. Finding Area Under Curves

This is one of the most common uses of integrals. With definite integrals, you can calculate the exact area between a curve and the x-axis over a given interval.

For example, to find the area under the curve defined by the function $y = x^2$ from $x = 1$ to $x = 3$, you would calculate: $\int_1^3 x^2 dx$

This method is widely applied in geometry and algebra problems.

2. Calculating Volumes of Solids of Revolution

Integrals help you find the volume of 3D objects created by rotating a curve around an axis.

For example, the volume of a solid formed by rotating $y = f(x)$ around the x-axis from $x = a$ to $x = b$ is: $V = \pi \int_a^b [f(x)]^2 dx$

This is useful in geometry and engineering applications.

3. Finding Length of Curves

Integrals are used to calculate the length of a curve between two points, which helps in areas like design, architecture, and engineering.

The formula for the length of a curve is: $L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$

4. Calculating Work Done by a Force

In mathematics, integrals are used to find the total work done when a force varies across a distance.

For example: $W = \int_a^b F(x) dx$

This application connects mathematics with physics and engineering.

5. Finding the Average Value of a Function

Integrals can be used to find the average value of a function over an interval, which is useful in statistics and analysis.

The formula is: $\text{Average} = \frac{1}{b-a} \int_a^b f(x) dx$

6. Solving Differential Equations

Integrals help in finding solutions to equations that describe how quantities change. These are known as differential equations.

For example, if $\frac{dy}{dx} = f(x)$, then the solution is: $y = \int f(x) dx + c$

This is widely applied in mathematical modelling and real-world problems.

7. Finding Moments and Centroids

Integrals are used to calculate the centre of mass or centroid of an object by integrating over its shape.

For example: $M_x = \int y f(x) dx$
$M_y = \int x f(x) dx$

These calculations are important in mechanics and geometry.

8. Probability and Statistics

Integrals are used to calculate probabilities when dealing with continuous random variables by integrating probability density functions.

For example: $P(a \leq x \leq b) = \int_a^b f(x) dx$

This forms the mathematical foundation for probability and statistical analysis.

9. Solving Problems Involving Rate of Change

Integrals are used wherever you need to calculate accumulated quantities like distance, growth, or decay over time.

By reversing the process of differentiation, integrals allow you to find the total amount of something from its rate of change.

Why These Applications Matter

• Integrals help you solve real-world problems in geometry, physics, engineering, economics, and more.

• They provide exact answers rather than approximations.

• They form the basis for advanced mathematical techniques like differential equations and statistical analysis.

• Mastering integrals allows you to understand how mathematical principles apply to everyday situations.

Integrals in Mathematics: Solved Previous Year Questions

Question 1:

$4 \int_{0}^{1}\left(\frac{1}{\sqrt{3+\mathrm{x}^{2}}+\sqrt{1+\mathrm{x}^{2}}}\right) \mathrm{dx}-3 \log _{\mathrm{e}}(\sqrt{3})$ is equal to:

Solution:

Given the expression:

$4 \int_0^1 \frac{1}{\sqrt{3 + x^2} + \sqrt{1 + x^2}} \, dx - 3 \log_e(\sqrt{3})$

First, rationalize the denominator inside the integral by multiplying numerator and denominator by

$\sqrt{3 + x^2} - \sqrt{1 + x^2}$

This gives:

$\int_0^1 \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{(\sqrt{3 + x^2})^2 - (\sqrt{1 + x^2})^2} \, dx$

Since

$(\sqrt{3 + x^2})^2 - (\sqrt{1 + x^2})^2 = (3 + x^2) - (1 + x^2) = 2$

the integral becomes

$\int_0^1 \frac{\sqrt{3 + x^2} - \sqrt{1 + x^2}}{2} \, dx = \frac{1}{2} \int_0^1 \left(\sqrt{3 + x^2} - \sqrt{1 + x^2}\right) dx$

Therefore, the original expression is

$4 \times \frac{1}{2} \int_0^1 \left(\sqrt{3 + x^2} - \sqrt{1 + x^2}\right) dx - 3 \log_e(\sqrt{3}) = 2 \int_0^1 \left(\sqrt{3 + x^2} - \sqrt{1 + x^2}\right) dx - \frac{3}{2} \log_e 3$

Split the integral:

$2 \left[ \int_0^1 \sqrt{3 + x^2} \, dx - \int_0^1 \sqrt{1 + x^2} \, dx \right] - \frac{3}{2} \log_e 3$

Recall the integral formula

$\int \sqrt{x^2 + c} \, dx = \frac{x}{2} \sqrt{x^2 + c} + \frac{c}{2} \ln \left( x + \sqrt{x^2 + c} \right) + C$

Apply for $c=3$:

$\int_0^1 \sqrt{3 + x^2} \, dx = \left[ \frac{x}{2} \sqrt{x^2 + 3} + \frac{3}{2} \ln \left( x + \sqrt{x^2 + 3} \right) \right]_0^1$

Evaluate at limits:

$= \frac{1}{2} \sqrt{4} + \frac{3}{2} \ln (1 + 2) - \left( 0 + \frac{3}{2} \ln \sqrt{3} \right) = 1 + \frac{3}{2} \ln 3 - \frac{3}{4} \ln 3 = 1 + \frac{3}{4} \ln 3$

Similarly for $c=1$:

$\int_0^1 \sqrt{1 + x^2} \, dx = \left[ \frac{x}{2} \sqrt{x^2 + 1} + \frac{1}{2} \ln \left( x + \sqrt{x^2 + 1} \right) \right]_0^1$

Evaluate at limits:

$= \frac{1}{2} \sqrt{2} + \frac{1}{2} \ln (1 + \sqrt{2}) - 0 = \frac{\sqrt{2}}{2} + \frac{1}{2} \ln (1 + \sqrt{2})$

Substitute back:

$2 \left( 1 + \frac{3}{4} \ln 3 - \frac{\sqrt{2}}{2} - \frac{1}{2} \ln (1 + \sqrt{2}) \right) - \frac{3}{2} \ln 3$

Simplify:

$= 2 - \sqrt{2} + \frac{3}{2} \ln 3 - \ln (1 + \sqrt{2}) - \frac{3}{2} \ln 3 = 2 - \sqrt{2} - \ln (1 + \sqrt{2})$

Hence, the value of the original expression is

$2 - \sqrt{2} - \ln (1 + \sqrt{2})=2-\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})$

Hence, the correct answer is $2-\sqrt{2}-\log _{\mathrm{e}}(1+\sqrt{2})$.

Question 2:

The integral $\int_0^\pi \frac{(x+3) \sin x}{1+3 \cos ^2 x} d x$ is equal to:

Solution:

$
I = \int_0^\pi \frac{(x+3) \sin x}{1 + 3 \cos^2 x} \, dx
$

Let $I = \int_0^\pi \frac{x \sin x}{1 + 3 \cos^2 x} \, dx + 3 \int_0^\pi \frac{\sin x}{1 + 3 \cos^2 x} \, dx
$

$
I = I_1 + 3 I_2
$

$
I_1 = \int_0^\pi \frac{x \sin x}{1 + 3 \cos^2 x} \, dx
$

Use substitution $t = \pi - x \implies dt = -dx
$

$
\cos x = \cos(\pi - t) = -\cos t, \quad \sin x = \sin(\pi - t) = \sin t
$

$
I_1 = \int_\pi^0 \frac{(\pi - t) \sin t}{1 + 3 \cos^2 t} (-dt) = \int_0^\pi \frac{(\pi - t) \sin t}{1 + 3 \cos^2 t} \, dt
$

$
\Rightarrow I_1 = \int_0^\pi \frac{\pi \sin t}{1 + 3 \cos^2 t} \, dt - \int_0^\pi \frac{t \sin t}{1 + 3 \cos^2 t} \, dt
$

$
I_1 = \pi I_2 - I_1
$

$
2 I_1 = \pi I_2 \implies I_1 = \frac{\pi}{2} I_2
$

Now evaluate $I_2 = \int_0^\pi \frac{\sin x}{1 + 3 \cos^2 x} \, dx
$

Substitute $t = \cos x \implies dt$

$= -\sin x \, dx \implies -dt = \sin x \, dx
$

$
I_2 = \int_{\cos 0}^{\cos \pi} \frac{-dt}{1 + 3 t^2}$

$= \int_1^{-1} \frac{-dt}{1 + 3 t^2} = \int_{-1}^1 \frac{dt}{1 + 3 t^2}
$

$
I_2 = \int_{-1}^1 \frac{dt}{1 + 3 t^2}
$

Since the integrand is even, $I_2 = 2 \int_0^1 \frac{dt}{1 + 3 t^2}
$

$
\int \frac{dt}{1 + a^2 t^2} = \frac{1}{a} \tan^{-1}(a t)
$

$
I_2 = 2 \cdot \frac{1}{\sqrt{3}} \left[ \tan^{-1}(\sqrt{3} t) \right]_0^1 = \frac{2}{\sqrt{3}} \tan^{-1}(\sqrt{3})
$

$
\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}
$

$
I_2 = \frac{2}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{2 \pi}{3 \sqrt{3}}
$

$
I_1 = \frac{\pi}{2} \times \frac{2 \pi}{3 \sqrt{3}} = \frac{\pi^2}{3 \sqrt{3}}
$

$
I = I_1 + 3 I_2 = \frac{\pi^2}{3 \sqrt{3}} + 3 \times \frac{2 \pi}{3 \sqrt{3}}$

$= \frac{\pi^2}{3 \sqrt{3}} + \frac{2 \pi}{\sqrt{3}} = \frac{\pi^2 + 6 \pi}{3 \sqrt{3}}$

$=\frac{\pi}{3 \sqrt{3}}(\pi+6)
$

Hence, the correct answer is $\frac{\pi}{3 \sqrt{3}}(\pi+6)$.

Question 3:

The integral $\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x}$ is equal to:

Solution:

$
I=\int_0^\pi \frac{8 x d x}{4 \cos ^2 x+\sin ^2 x}
$

Apply property
$
I=\int_0^\pi \frac{8(\pi-x) d x}{4 \cos ^2 x+\sin ^2 x}
$

Add
$
\begin{aligned}
& 2 I=\int_0^\pi \frac{8 \pi}{4 \cos ^2 x+\sin x} d x \\
& I=4 \pi \int_0^\pi \frac{1}{4 \cos ^2 x+\sin ^2 x}
\end{aligned}$

$\begin{aligned} & \because \frac{1}{4 \cos ^2 x+\sin ^2 x}=\frac{1}{4 \cos ^2(\pi-x)+\sin ^2(\pi-x)} \\ & \therefore I=4 \pi \times 2 \int_0^{\pi / 2} \frac{1}{4 \cos ^2 x+\sin ^2 x} d x \\ & =8 \pi \int_0^{\pi / 2} \frac{\sec ^2 x d x}{4+\tan ^2 x} \\ & \text { let } \tan x=t \\ & \sec ^2 x d x=d t \\ & =8 \pi \int_0^{\infty} \frac{d t}{4+t^2} \\ & =\frac{8 \pi}{2}\left[\tan ^{-1} t\right]_0^{\infty} \\ & \Rightarrow \frac{8 \pi}{2}(\pi / 2-0) \\ & \Rightarrow \frac{4 \pi^2}{2} \Rightarrow 2 \pi^2\end{aligned}$

Hence, the correct answer is $2 \pi^2$.

Question 4:

The integral $80 \int_0^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16 \sin 2 \theta}\right) d \theta$ is equal to:

Solution:

Simplify the denominator using the identity,

$
\begin{aligned}
& \sin 2 \theta=2 \sin \theta \cos \theta \\
& 9+16 \sin 2 \theta=9+16(2 \sin \theta \cos \theta) \\
& 9+16 \sin 2 \theta=25-16(1-2 \sin \theta \cos \theta) \\
& 9+16 \sin 2 \theta=25-16\left(\sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta\right) \\
& 9+16 \sin 2 \theta=25-16(\sin \theta-\cos \theta)^2
\end{aligned}
$

Substitute $u=\sin \theta-\cos \theta$

$
d u=(\cos \theta+\sin \theta) d \theta
$

When $\theta=0, u=\sin 0-\cos 0=-1$
When $\theta=\frac{\pi}{4}, u=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=0$
The integral becomes $80 \int_{-1}^0 \frac{1}{25-16 u^2} d u$

Simplify the integral

$
\begin{aligned}
& 80 \int_{-1}^0 \frac{1}{25-16 u^2} d u=80 \int_{-1}^0 \frac{1}{5^2-(4 u)^2} d u \\
& 80 \int_{-1}^0 \frac{1}{5^2-(4 u)^2} d u=80 \cdot \frac{1}{4} \int_{-1}^0 \frac{1}{5^2-(4 u)^2} d(4 u) \\
& 20 \int_{-1}^0 \frac{1}{5^2-(4 u)^2} d(4 u)=\left.20 \cdot \frac{1}{2 \cdot 5} \ln \left|\frac{5+4 u}{5-4 u}\right|\right|_{-1} ^0 \\
& \left.2 \ln \left|\frac{5+4 u}{5-4 u}\right|\right|_{-1} ^0=2\left(\ln \left|\frac{5}{5}\right|-\ln \left|\frac{1}{9}\right|\right) \\
& 2\left(\ln 1-\ln \frac{1}{9}\right)=2\left(0-\ln \frac{1}{9}\right)=-2 \ln \frac{1}{9}=2 \ln 9 \\
& 2 \ln 9=2 \ln 3^2=4 \ln 3
\end{aligned}
$

Hence, the correct answer is $4 \log _e 3$.

Question 5:

The value of $\int_{e^2}^{e^4} \frac{1}{x}\left(\frac{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}}{e^{\left(\left(\log _e x\right)^2+1\right)^{-1}}+e^{\left(\left(6-\log _{\mathrm{e}} x\right)^2+1\right)^{-1}}}\right) d x$ is:

Solution:

Let us make the substitution:
$
\ln x = t \implies \frac{dx}{x} = dt
$

Given the integral:
$
I = \int_2^4 \frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}} + e^{\frac{1}{1+(6 - t)^2}}} \, dt
$

Observe that the denominator is symmetric in \( t \) and \( 6 - t \). Therefore, consider the integral:
$
I = \int_2^4 \frac{e^{\frac{1}{1+t^2}}}{e^{\frac{1}{1+t^2}} + e^{\frac{1}{1+(6 - t)^2}}} \, dt
$

Now define a new integral by changing the variable inside the integral:
$
J = \int_2^4 \frac{e^{\frac{1}{1+(6 - t)^2}}}{e^{\frac{1}{1+t^2}} + e^{\frac{1}{1+(6 - t)^2}}} \, dt
$

Adding the two integrals, we get:
$
I + J = \int_2^4 \frac{e^{\frac{1}{1+t^2}} + e^{\frac{1}{1+(6 - t)^2}}}{e^{\frac{1}{1+t^2}} + e^{\frac{1}{1+(6 - t)^2}}} \, dt = \int_2^4 1 \, dt = 4 - 2 = 2
$

But by substituting \( u = 6 - t \), the limits of \( J \) also change from \( t=2 \) to \( 4 \) to \( u=4 \) to \( 2 \), so:
$
J = \int_4^2 \frac{e^{\frac{1}{1+u^2}}}{e^{\frac{1}{1+(6 - u)^2}} + e^{\frac{1}{1+u^2}}} \, (-du) = \int_2^4 \frac{e^{\frac{1}{1+u^2}}}{e^{\frac{1}{1+u^2}} + e^{\frac{1}{1+(6 - u)^2}}} \, du = I
$

Therefore, \( J = I \).

From the sum,
$
I + J = 2I = 2 \implies I = 1
$

Hence, the correct answer is 1.

Integrals in Different Exams

Integrals are an important part of Class 12 Mathematics and are asked in various school and competitive examinations. The level of questions varies from basic formula-based problems to application-oriented and analytical questions, depending on the exam.

List of Topics related to Integrals according to NCERT/JEE Mains

This section covers all the important topics on integrals that are part of the NCERT syllabus and frequently appear in the JEE Main exam.

Important Books and Resources for Class 12 Integrals

Here, you’ll find recommended books and study materials that help build a strong understanding of integrals and prepare effectively for exams.

NCERT Resources for Integrals

This section lists NCERT textbooks and reference materials that explain integrals clearly and form the foundation for solving related problems.

• NCERT Maths Notes for Class 12th Chapter 7 Integrals

• NCERT Maths Solutions for Class 12th Chapter 7 Integrals

• NCERT Maths Exemplar Solutions for Class 12th Chapter 7 Integrals

NCERT Subjectwise Resources

Explore subject-wise NCERT resources that cover different study materials such as notes, solutions and exemplar problems to support your preparation in a structured way.

Practice Questions based on Integrals

This section includes important practice questions and problems based on integrals to help you test your understanding and improve problem-solving skills.

Conclusion

The Integrals chapter builds a strong foundation for higher mathematics and helps in understanding real-life problems involving area, volume, and accumulation. By practising standard formulas, properties of definite integrals, and application-based questions, students can gain confidence and perform well in board exams as well as competitive exams like JEE Main and Advanced.