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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

Updated on Apr 09, 2025 11:39 PM IST

Imagine a farmer who wants to build a fence around his land that is not in a regular geometric shape. He will need the help of integrals to calculate the area of curved surfaces to estimate his spending on the fencing materials. Integral is an important mathematics topic we studied in the previous chapter. The Class 12 Maths Chapter 8 question answers cover the application of integrals, focusing on the area under simple curves, the area between lines and arcs of circles, parabolas, and ellipses in standard forms only. All these topics are essential not only for the class 12 board exam but also for higher competitive exams like JEE Main, JEE Advanced, etc.

This Story also Contains
  1. NCERT Class 12 Chapter 8 Notes
  2. Importance of NCERT Class 12 Maths Chapter 8 Notes:
  3. Subject Wise NCERT Exemplar Solutions
  4. Subject Wise NCERT Solutions
  5. NCERT Books and Syllabus

The application of integral class 12 notes is a crucial part of a student's learning phase. After completing the class 12 maths chapter 8 NCERT solutions, students need a study material from which they can frequently revise important concepts and formulas. These notes are prepared by Careers360 experts closely following the latest CBSE syllabus. Students should also practice the NCERT Exemplar Class 12 Maths Chapter 8 Application of Integral for more exposure to this chapter.

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:

y=f(x)

Area under simple curves:
The area under the curve is y=f(x) with coordinates x=a and x=b.

Graph for Area Under Curves:

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

1645012679810

Now, these are steps to find the integral of the function:

Step 1: Write down the equation

A = y = f(x)

Here, A denotes area

Step 2: Take integration on both sides.

abA=abydx=abf(x)dx

Step 3: Find the integration

Step 4: Apply the limits and solve the equation.

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

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

Area of the above equation:

abA=abxdy=abg(y)dy

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

But if we get a area negative, then as the area doesn’t exist in negative value so we take its absolute value, which is nothing but,
|abf(x)dx|

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

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

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

Graph for Curve Bounded by Line:

1645012723407

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].

Formulae for Area Under Two Curves:

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

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

A=ab(f(x)g(x))dx

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

A=ac(f(x)g(x))dx+cb(g(x)f(x))dx

Graph Between Two Curves:

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

1645012754737

Here is the link to the NCERT textbook PDF.

URL: ncert.nic.in/ncerts/l/lemh202.pdf

Importance of NCERT Class 12 Maths Chapter 8 Notes:

NCERT Class 12 Maths chapter 8 notes are very valuable for the aspirants to score good marks in the board exams as well as in competitive exams. These are the important factors of why students need these notes.

  • These notes cover all the necessary concepts and explain them clearly and concisely.
  • Experts who have years of experience in this field have made these notes, keeping in mind the difficulty level of this chapter. That is why step-by-step problem-solving methods have been provided in these notes.
  • The latest Class 12 CBSE Mathematics Syllabus has been followed to help students learn the topics that are required to get good grades in the exam.
  • These notes will give conceptual clarity to students and will prove to be a handy tool during revision.
  • After going through these notes thoroughly, students will feel confident about their preparation, and these positive vibes will only help them to prepare well in other subjects.
  • These notes also contain the previous year’s questions and the NCERT TextBook PDF.
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Subject Wise NCERT Exemplar Solutions

After completing the textbook exercises, students can practice NCERT exemplars for further reference.

Subject Wise NCERT Solutions

These are the links to the solutions of NCERT textbooks subject-wise.

NCERT Books and Syllabus

It will always be recommended to students to check the latest syllabus before making the study routine. Here are three links to the 2025-26 CBSE syllabus and a reference book.

Articles

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)

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

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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