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Application of Integrals 12th Notes - Free NCERT Class 12 Maths Chapter 8 Notes - Download PDF

Application of Integrals 12th Notes - Free NCERT Class 12 Maths Chapter 8 Notes - Download PDF

Edited By Ramraj Saini | Updated on Apr 23, 2022 12:48 PM IST

Introduction: Class 12 Math chapter 8 notes are regarding Applications of integrals. In chapter 7 integrals we already learned about integrations, now in Applications of integrals Class 12 notes contain about finding the area of the functions and curves that is nothing but finding the total area of the curve. This Class 12 Maths chapter 8 notes contains the following topics: area under a single curve, area under two curves, area of the region bounded by a curve, and a line. NCERT Class 12 Math chapter 8 notes also contain standard formulas that are to be remembered for the implementation in problems.

NCERT Class 12 Math chapter 8 contains a detailed explanation of topics, examples, exercises. By going through the document students can cover all the topics that are in NCERT Notes for Class 12 Math chapter 8 textbook. It also contains examples, exercises, a few interesting points and most importantly contains FAQs that are frequently asked questions by students which can clarify many other students with the same doubt. Every concept that is in CBSE Class 12 Maths chapter 8 notes is explained here in a simple and understanding way that can reach students easily. All these concepts can be downloaded from Class 12 Maths chapter 8 notes pdf download, Class 12 notes Applications of integrals, Applications of integrals Class 12 notes pdf download.

Also, students can refer,

NCERT Class 12 Chapter 8 Notes

Application of integrals is about integrating and finding areas under curves. Integrating is adding or summation of all the points.

General equation of area under curves:


Area under simple curves:

Area under the curve is y=f(x) with coordinates x=a, x=b.

Graph for Area Under Curves:

According to class 12 Applications of integrals notes, graph for area under curves is:


Now finding the integral of the function:

Step1: Write down the equation


Here A denotes area

Step2: Take integration on both sides.

\int_{a}^{b} A= \int_{a}^{b}y dx= \int_{a}^{b} f(x) dx

Step3: Find the integration

Step4: Apply the limits and solve the equation.

Similarly, if it is for x terms then equation is:

x=g(y) with coordinates y=a, y=b.

Area of the above equation:

\int_{a}^{b} A= \int_{a}^{b} xdy= \int_{a}^{b} g(y) dy

Meanwhile, when we get an area as positive after solving the equation then we have no issues.

But if we get area negative then as the area doesn’t exist in negative value so we take their absolute value that is nothing but,

\left | \int_{a}^{b} f(x) dx\right |

Finding the area of the region bounded by a curve and a line:

It is the same as finding an area under the curves.

But the only difference is we find the area of the region that is only enclosed by a line and a circle.

Graph for Curve Bounded by Line:


Area between two curves:

Here we have two curves:

a) y=f(x)

b) y=g(x), where f(x)>g(x) in the interval [a, b].


if y=f(x) and y=g(x)

a) when f(x)>g(x) in interval [a, b]

A= \int_{a}^{b} \left (f(x)-g(x) \right ) dx

b) when f(x) > g(x) in [a, c] and f(x) < g(x) in [c, b]

A= \int_{a}^{c} \left (f(x)-g(x) \right ) dx +\int_{c}^{b} \left (g(x)-f(x) \right ) dx

Graph Between Two Curves:

In NCERT notes for Class 12 Maths chapter 8, the graph between two curves is given as:


With this topic we conclude NCERT Class 12 chapter 8 notes.

The link for the NCERT textbook pdf is given below:


Significance of NCERT Class 12 Maths Chapter 8 Notes:

NCERT Class 12 Maths chapter 8 notes will be very much helpful for students to score maximum marks in their 12 board exams. In Applications of integrals Class 12 chapter 8 notes, we have discussed many topics: area under single curve, area under two curves, area of the region bounded by a curve, and a line like with their mathematical representations, along with the graphs. NCERT Class 12 MMathematics chapter 8 is also very useful to cover major topics of the Class 12 CBSE Mathematics Syllabus.

The CBSE Class 12 Maths chapter 8 will help to understand the formulas, statements, rules with their conditions in detail. This pdf also contains previous year’s questions and NCERT TextBook pdf. The next part contains FAQ’s most frequently asked questions along with topic-wise explanations. By referring to the document you can get a complete idea of all the topics of Class12 chapter 8 Applications of integrals pdf download.

NCERT Class 12 Notes Chapter Wise.

Go through the above link to learn more examples topic-wise.

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Frequently Asked Question (FAQs)

1. What is the role of Applications of integrals Class 12 notes?

They are used to calculate the area of the curves, combine or integrate functions, and many more.

2. Difference between integration and differentiation in Class 12 Math chapter 8 notes?

Integration is just the reverse of differentiation.

Differentiation calculates the slope whereas integration calculates the area under the curve.

3. What is the use of Class 12 notes Applications of integrals in our daily life?

To calculate area, cost, maximum and minimum values, amount of material needed, profit, and loss.

4. What is the meaning of calculus?

It is generally a study on shapes, algebra using arithmetic operations. It is classified into two branches: Differential and integral calculus. Differentiations, integrations, limits all come under this.

5. How many types of techniques do we have in Class 12 Maths chapter 8 notes?

Integration by dividing into parts, substitution, reverse chain rule, solving trigonometric equations, etc…..

These solutions can be obtained from Class 12 Maths chapter 8 notes pdf download.


Get answers from students and experts

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 3.8×107 J 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)


Option 2)

\; K\;

Option 3)


Option 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)


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)


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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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