Integrals Class 12th Notes - Free NCERT Class 12 Maths Chapter 7 Notes

Integrals 12th Notes - Free NCERT Class 12 Maths Chapter 7 Notes - Download PDF

Edited By Ramraj Saini | Updated on Sep 13, 2022 10:59 AM IST

Integrals belong to the 7 chapter of NCERT. The NCERT Class 12 Maths Chapter 7 notes entirely cover up the main portions of the chapter integrals. Class 12 Maths chapter 7 notes are predominantly focused on the important formulas and their needed derivations. A Class 12 Maths chapter 7 notes enable you to find the entire chapter in an easy way. Notes for Class 12 Maths chapter 7 is made by following the sequence of the chapter starting from the indefinite integrals to definite integrals. NCERT Notes for Class 12 Maths chapter 7 not only covers the NCERT notes but covers CBSE Class 12 Maths chapter 7 notes also.

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This Story also Contains

Indefinite Integral
Geometrical Interpretation of Indefinite Integral
Methods of Integration
Integration by Substitution
Integrals of Some Particular Function
Integration by Partial Fraction
Integration by Parts
Definite Integral
Definite Integral as a Limit of the Sum
Some Properties of Definite Integral
Significance of NCERT Class 12 Maths Chapter 7 Notes

After going through Class 12 integrals notes students can also refer to,

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Adding or summing up the small parts to find the whole sum of the function, is the basic meaning of integral calculus.

Indefinite integral and definite integral are the two types of integrals that are found in integral calculus.

Indefinite Integral

Thus the formula that gives this anti-derivative is called indefinite integral.

Integration as an Inverse Process of Differentiation

Integration is defined as the inverse of differentiation. Here we are given a function and asked to find it is primitive (the original function). Such a process is called anti-derivative or indefinite integral.

Let us see an example:-

We know that

$\frac{d}{dx}(sinx+c)=cosx$

$\int cosx \ dx =sinx+c$

Here the actual function is “sinx +c” using indefinite integral we found the actual equation from the derived one.

Thus we came to know that the integration of some functions is written as

$\int f(x) \ dx =F(x)+c$

Here f(x) is “integrand”, f(x)dx is “element of integration” , x is “variable of integration” and “c” is the integration constant.

Some of the formulas are listed below which will help us to solve the problems.

$\\ \int x^ndx=\frac{x^{n+1}}{n+1}+c, (n\neq -1)\\ \int dx=x+c \\ \int cos x dx=sin x+c \\ \int sin x dx=-cos x+c \\ \int sec^2x dx=tan x+c \\ \int cosec^2x dx= -cot x+c\\ \int sec x tan x=sec x+c\\ \int cosec x cot x= -cosec x+c\\ \int \frac{dx}{\sqrt{1-x^2}}=sin^{-1}x+c\\ \int \frac{dx}{\sqrt{1-x^2}}=-cos^{-1}x+c\\ \int \frac{dx}{1+x^2}= tan^{-1}x+c\\ \int \frac{dx}{1+x^2}= -cot^{-1}x+c\\ \int \frac{dx}{x\sqrt{x^2-1}}= sec^{-1}x+c\\ \int \frac{dx}{x\sqrt{x^2-1}}= -cosec^{-1}+c\\ \int e^xdx=e^x+c\\ \int \frac{1}{x}dx=log |x|+c\\ \int a^xdx= \frac{a^x}{log a} +c\\$

Geometrical Interpretation of Indefinite Integral

Let us assume that y=f(x) be a curve such that $f'(x)= 2x \Rightarrow f(x)= x^2+c$

1645082112385

Now if we give values like 4, 3, 2, 1, 0, -1, -2, -3 to arbitrary constant c then the equation will give y= x^2+4, x^2+3, x^2+2, x^2+1, x^2, x^2-1, x^2-2, x^2-3

having loci as a parabola.

Thus the equation $\int f(x)dx=F(x)+c=y$ represents a family of curves.

Some properties of indefinite integral

Here we shall derive some properties of indefinite integrals.

(I) The process of integration and differentiation are inverses of each other in the sense of the following results:

$\frac{d}{dx} \int f(x)dx=f(x)$

And

$\int f'(x)dx=f(x)+c$

where c is the arbitrary constant.

(II) Two indefinite integrals with the equal derivative lead to the same family of curves thus they are equivalent.

$\\ \ \frac{d}{dx} \int f(x)dx= \frac{d}{dx} \int g(x)dx\\ \\ (III) \ \int [f(x)+g(x)]dx= \int f(x)dx+\int g(x)dx \\ \\ (IV) \ \text{For any real number k},\ \ \int kf(x)dx= k \int f(x)dx$

(V) Using property (III) and (IV) can be generalised to a finite number of functions f1, f2, f3, …, fn and the real numbers k1, k2, k3, …, kn gives

$\int [k_1f_1(x) + k_2f_2(x)+\ . \ . \ . k_nf_n(x)] dx \\ = \int k_1f_1(x) dx + \int k_2f_2(x) dx+\ . \ . \ .\ \int k_nf_n(x) dx$

Example

Find the anti-derivative (integral) of $\frac{1}{x} , \ \ x\neq 0$

Solution:

$\\ \text{We know that}\ \frac{d}{dx}(log x)=\frac{1}{x} , x > 0 \\ and \ \frac{d}{dx}(log (-x))=\frac{1}{-x}(-1)=\frac{1}{x}, x<0 \\ \text{Thus combining the above }\ \frac{d}{dx}log x = \frac{1}{x} , x \neq 0 \\ \text{Therefore}\ \frac{1}{x}dx=log |x| +c\$

Methods of Integration

We have many ways of solving integration. Mostly we solve by using

Integration by substitution
Integration by Partial Fraction
Integration by Parts

Integration by Substitution

The given integral $\int f(x)dx$ can be transferred into another form by changing the independent variable x to t by substituting x = g (t).

Let us assume that

$I=\int f(x)dx$

Substitute

x=g(t) , so that dx/dt=g'(t)

Then we write dx=g’(t) dt

$Thus \ \ I =\int f(x)dx= \int f(g(t))g'(t) dt$

Example: Integrate w.r.t x, sin mx

Solution:

We know that by differentiating mx we get m. Thus we can substitute mx=t so that mdx =dt

$Thus \ \int sin(mx) dx= \int \frac{1}{m}sin(t) dt= -\frac{1}{m}cos t+c= -\frac{1}{m}cos mx+c$

Integration by Trigonometric Identities

The identities are used to find the integral when integration involves some trigonometric functions.

Example: $Find \ \int cos^2x dx$

Solution: We know that $cos^2x= \frac{1+cos 2x}{ 2}$

Using this we get that $cos^2 x dx=\frac{1}{2}(1+cos 2x) dx= \frac{1}{2}[dx+cos 2x dx]= \frac{x}{2} +\frac{1}{4}sin 2x+c$

Integrals of Some Particular Function

Some of the formulas are listed below

$\\ \int \frac{dx}{x^2-a^2}dx = \frac{1}{2a}log \frac{|x-a|}{|x+a|}+c \\ \\ \int \frac{dx}{a^2-x^2} = \frac{1}{2a}log \frac{|a+x|}{|a-x|}+c \\ \\ \int \frac{dx}{x^2+a^2}= \frac{1}{a}tan^{-1}\frac{x}{a}+c \\ \\ \int \frac{dx}{\sqrt{x^2-a^2}}=log | x+\sqrt{x^2-a^2} |+c \\ \int \frac{dx}{\sqrt{x^2+a^2}}=log | x+\sqrt{x^2+a^2}|+c \\ \int \frac{dx}{\sqrt{a^2-x^2}}=sin^{-1}\frac{x}{a}+c \\$

$\\ \text{Let us see an example}\\ \int \frac{dx}{(4+9x^2)}dx\\ =\int \frac{ dx}{9(\frac{4}{9}+x2)}\\ \text{Now using special formula}\\ \int \frac{dx}{x^2+a^2}= \frac{1}{a}tan^{-1}\frac{x}{a}+c\\ =\frac{1}{9} \int \frac{dx}{x^2+(\frac{2}{3})^2}\\ \\ =\frac{1}{9}\frac{1}{2/3} tan^{-1}\frac{3x}{2}+c \\ \\ =\frac{1}{6}tan^{-1} \frac{3x}{2}+c$

Integration by Partial Fraction

The integration of rational function can be solved by using partial fraction.

S. No	Form of a rational function	formation of partial function
1	$\frac{px+q}{(x-a)(x-b)}, \ a\neq b$	$\frac{A}{x-a}+\frac{B}{x-b}$
2	$\frac{px+q}{(x-a)^2}$	$\frac{A}{x-a}+\frac{B}{(x-a)^2}$
3	$\frac{px^2+qx+r}{(x-a)(x-b)(x-c)}$	$\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}$
4	$\frac{px^2+qx+r}{(x-a)^2(x-b)}$	$\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}$
5	$\frac{px^2+qx+r}{(x-a)(x^2+bx+c)}$	$\frac{A}{x-a}+\frac{Bx+c}{x^2+bx+c}$
6		$Where \ \ x^2+bx+c \ \ \ \text{cannot be factorised further}$

$\text{Let us see the following example}\\ \int \frac{x^2}{x^2+x+6}dx\\ =\int (1-\frac{x-6}{x^2+x+6}dx\\ =\int dx-\int \frac{x-6}{x2+x+6}dx\\ =x-\frac{9}{5}\int \frac{dx}{x+3}+\frac{4}{5}\int \frac{dx}{x-2}\\ =x-\frac{9}{5}ln|x+3| +\frac{4}{5}ln |x-2| +c$

Integration by Parts

ILATE rule needs to be followed while doing integration by parts. According to ILATE rule, we can determine which function will be the 1st function and which will be the 2nd function.

ILATE

I → Inverse function

L → Logarithmic function

A → Algebraic function

T → Trigonometric function

E → Exponential function

From the above rule, as an example, we can say that there are two functions like trigonometric function and logarithmic function. The logarithmic function will be treated as the first function and the trigonometric function will be treated as the second function.

The general expression follows as

$\int f(x)g(x)dx=f(x)\int g(x)dx-\int \left [ { \left \{ \int g(x)dx \right \}} \frac{d}{dx}{f(x)} \right ] dx$

“The integral of the product of two functions = (1st function) × (integral of the 2nd function) – Integral of [(differential coefficient of the 1st function) × (integral of the 2nd function)]”

Example:

$\\ Solve \ \int ln x dx \\ Solution:\\ \int ln x .1 dx \\ \Rightarrow ln x \int 1 dx- \int [(\int 1dx)\frac{d}{dx}(lnx)]dx\\ \Rightarrow xln x- \int x.\frac{1}{x}dx\\ \Rightarrow xln x-x+c$

Integration of special types function

The formulas are listed below

$\\ \int \sqrt{x^2+a^2} \ dx= \frac{1}{2}\ x \ \sqrt{x^2+a^2} +\frac{a^2}{2}log |x+\sqrt{x^2+a^2}| +c \\ \\ \int \sqrt{x^2-a^2} \ dx= \frac{1}{2}\ x \ \sqrt{x^2-a^2} +\frac{a^2}{2}sin^{-1} \ \frac{x}{a} +c$

Definite Integral

In indefinite, we saw that the result of the integration was not a unique value. But here indefinite integral we will find a unique value and it is well defined in a limited boundary. Generally, definite integral is denoted by

$\int_{a}^{b}f(x)dx$

Her “a” is the lower limit and “b” is the upper limit of the integral. The value of the definite integral is computed by F(b) - F(a).

Basically in Definite integral consist of two cases as discussed below:

Definite Integral as a Limit of the Sum

If the definite integral $\int_{a}^{b}f(x)dx$

is the area bounded by the curve y = f(x), the ordinates x = a, x = b and the x-axis. To evaluate this area, consider the region PRSQP between this curve, x-axis, and the ordinates x = a and x =b

1645082112127

From the figure, we find the general equation of definite integral as a limit of sum

$\int _{a}^{b}f(x)dx=h[f(a)+ f(a+h)+\ . \ . \ . + f(a+(n-1)h]\\ \Rightarrow \int_{a}^{b}f(x)dx=\frac{(b-a)}{n}[f(a)+ f(a+h)+.\ . \ .+ f(a+(n-1)h]\\ Where\ h=\frac{b-a}{n}\rightarrow 0 \ as \ n \rightarrow \infty$

Some Properties of Definite Integral

$\\ P_0: \int_{a}^{b}f(x)dx= \int _{a}^{b}f(t)dt\\ \\ P_1 : \int _{a}^{b}f(x)dx= -\int _{b}^{a}f(x)dx \ ; in\ particular, \int_{a}^{a}f(x)dx =0\\ \\ P_2: \int_{a}^{b}f(x)dx= \int_{a}^{c}f(x)dx + \int_{c}^{b}f(x)dx\\ \\ P_3: \int_{a}^{c}f(x)dx= \int_{a}^{c}f(a+b-x)dx \\ \\ P_4 : \int_{0}^{b}f(x)dx= \int_{0}^{b}f(b-x)dx\\ \\$

$\\ P_5 : \int_{0}^{2b}f(x)dx= \int_{0}^{b}f(x)dx+ \int _{0}^{b}f(2b-x)dx\\ \\ P_6 : \int _{0}^{2b}f(x)dx= 2\int_{0}^{b}f(x)dx \ \ if \ \ f(2b-x)=f(x) \\ and\ \ \int _{0}^{2b}f(x)dx =0 \ \ \ if \ \ \ \ f(2b-x)=-f(x)\\ \\$

$\\ P_7 : (i) \int_{-a}^{a} f(x)dx=2 \int _{0}^{a}f(x)dx ,\ \text{if f is even function, i.e., if f(-x)=f(x)} \\ (ii) \int _{-a}^{a}f(x)dx=0 ,\text{ if f is odd function, i.e., f(-x)=-f(x)}$

Let us see an example

$\\ \int _{0}^{\frac{\pi}{2}}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx\\ \\ Let \ \int _{0}^{\frac{\pi}{2}}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx=I--------1\\ \\$

$\\ I=\int _{0}^{\frac{\pi}{2}}\frac{\sqrt{sin (\frac{\pi}{2}-x)}}{\sqrt{sin(\frac{\pi}{2}-x)}+\sqrt{cos(\frac{\pi}{2}-x)}}dx \\ \\ I=\int _{0}^{\frac{\pi}{2}}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx----------2\\ \\$

$\\ \text{Now 1+2}\\ \int _{0}^{\frac{\pi}{2}}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx+\int _{0}^{\frac{\pi}{2}}\frac{\sqrt{cosx}}{\sqrt{sinx}+\sqrt{cosx}}dx =2I \\ \\ 2I=\int_{0}^{\frac{\pi}{2}} dx \\ \\ I=\frac{\pi}{4}$

This brings us to the end of the chapter.

Significance of NCERT Class 12 Maths Chapter 7 Notes

Class 12 integrals notes will be really helpful to revise the chapter and get a brief overview of the important topics. Also, Class 12 Math Chapter 7 Notes is useful for covering Class 12 CBSE syllabuses and also for competitive exams like BITSAT, JEE MAINS. Class 12 Math chapter 7 notes pdf download can be used for preparing in offline mode.

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Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is