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Calculus is an important subject for NEET aspirants. In 2021, a total of 12 questions asked in Physics and Chemistry were related to Calculus. Calculus is a branch of Mathematics that deals with mainly two operations, i.e, Differentiation and Integration. Calculus is widely used as a tool to determine quantities like the rate of change, gradient of curve, and area under curves. In Physics, most of the chapters in Class 11 and 12 have some topics relating to Calculus. Similarly, in Chemistry, a few chapters like Thermodynamics and Chemical Kinetics have some topics that demand a basic understanding of Calculus. But NEET aspirants don't need to master Calculus because it is very vast and the exam demands only a very basic knowledge of that. This article explains the physical significance of Calculus, the difference between differential and integral Calculus, and also the relation between them. This article also includes an analysis of the previous five year's papers to help you understand the concepts of Calculus easily.
Whenever we differentiate any mathematical equation, we are basically finding the slope of the equation. Integral Calculus refers to the summation of uncountable or infinitely many elemental areas. Consider an example given in the graph below. Derivative of f(x) at point ‘a’ gives slope or rate of change of f(x) at point ‘a’. Another graph illustrates integration as integrating f(x) in limits b to c and gives the area under that curve.

Integration is of two types, indefinite and definite. If lower and upper limits are given then it is called a definite integral.
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Let there is a variable y that is changing with reference variable x. If there is a change in x that is ∆x and a corresponding change in y that is ∆y. The average rate of change of y with respect to x is given by ∆y/∆x. When the change in x is infinitely small then the rate can be represented as dy/dx or y’, known as differentiation of y with respect to x.
Integration is the opposite process of differentiation represented as the symbol ”∫ “.
∫ (differentiation) = Initial function ⇒ ∫(dy/dx)dx = ∫dy = y.
Let consider an example y = sinx
Differentiation of y = dy/dx = cosx
Integration = ∫cosx dx = sinx = Initial function.
There are some fundamental formulas to compute differentiation and integration, in which a few of them are listed below.
Differential Calculus | Indefinite Integral Calculus |
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The NEET paper includes a total of 200 questions and out of that 50 questions each are from Physics and Chemistry and 100 questions from Biology. Students need to attempt only 180 questions which include 45 from Physics, 45 from Chemistry, and 90 from Biology. Careers360 has analyzed the last 5 years' papers and extracted some important information, which is tabulated below.
Year | Physics | Chemistry |
2021 | 9 | 3 |
2020 | 7 | 2 |
2019 | 10 | 2 |
2018 | 9 | 0 |
2017 | 7 | 3 |
It is observed that in Physics out of 45 questions around 7-10 questions demand a basic understanding of Calculus whereas, in Chemistry out of 45 questions, 2-3 questions demand basic knowledge of Calculus. This means around 15-20 % of Physics sections required a basic understanding of Calculus. As Calculus accounts for around 20% of Physics questions, it becomes very important for NEET aspirants.
Here are some previous year's questions with their solution in which concepts of Calculus were used. After going through these questions and analyzing the previous year's papers you will realize that the NEET doesn't go deep to Calculus. Most of the questions are formulas-based or very normal calculations with basic Differentiation and Integration.
Q.1 (NEET - 2021)
A capacitor of capacitance C is connected across an ac source of voltage V.given by
. The displacement current between the plates of the capacitor would then be given by:
Solution:
We know that charge on the capacitor’s plate is given by Q = CV. Current is the rate of flow of charge which means the differentiation of charge function with respect to time gives current.

Concept Used From Differentiation-
We know the chain rule is

![]()
That is chain rule and differentiation of sin function
Q.2 (NEET - 2020)
The phase difference between displacement and acceleration of a particle in a simple harmonic motion is:
Solution:
We know that rate of change of displacement is velocity and rate of change of velocity is acceleration. This means the differentiation of the displacement function with time gives velocity and also that acceleration can be calculated by differentiating the velocity function with time.

(Since -sin⍵t=sin(⍵t+π))
Concept Used From Calculus-

That is chain rule and differentiation of sin and cos function.
Q.3 (NEET - 2017)
The x and y coordinates of the particle at any time are x = 5t - 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2s is.
Solution:
We know that velocity is the rate of change of displacement and acceleration is the rate of change of velocity. When we differentiate the displacement function we get the velocity function and when we differentiate the velocity function we get the acceleration function.
x = 5t-2t2 | y = 10t |
dx/dt=5-4t | dy/dt=10 |
vx=5 - 4t | vy = 10 |
ax= -4 | ay= 0 |
Acceleration of particle at (t=2)= -4m/s2
Concept Used From Differentiation-
![]()
and the concept that differentiation of a constant function is zero is also used.
Q.1 (NEET - 2019)
A force F = 20 + 10 y acts on a particle in y-direction where F is in newton and y in meter Work done by this force to move the particle from y = 0 to y = 1 is:
Solution:
Definition of work done by variable force -
![]()
wherein
is variable force and
is small displacement
![\\ F = 20+ 10 y \\\\ W = F\cdot ds \\\\ here \ W = \int_{0}^{1}F\cdot dy =\int_{0}^{1} ( 20 + 10 y ) dy \\\\ W = [ 20 y ]^1 _0 + \left [ \frac{10 y^2}{2} \right ]^1_0 \\\\ W = 20+5 = 25 J](https://cache.careers360.mobi/media/articles/uploads/froala_editor/images/2022/5/19/1652953324606.png)
Concept Used From Integral Calculus-
Here we used the integration formula

Putting upper and lower limits and then subtracting them, we can calculate the value of the definite integral.
Q.2 (NEET - 2017)
A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy
of the gas in joules will be:
Solution:
Work Done for Irreversible Isothermal Expansion of an ideal gas -

wherein
may be equal to
or may not be equal to
, but work done is always calculated by
and in Adiabatic Process, Heat exchange between system and surrounding is zero i.e. ![]()
wherein
![]()
![]()
![]()
Work done in this process will be

= -2.5[4.5 - 2.5] = -5 L atm
= - 5
101 J = -505J
since the system is well-insulated
q = 0
![]()
or ΔU = +W = -505 J
Two words “Why” and “How” are widely mentioned. If the reason or “why” is clear it means a student is aware of why s/he must have command on Calculus then s/he can find a way or “How” to get it. First of all stop fearing the Calculus part in calculations. Just face it, set a goal, work smart and practise well. Then Calculus becomes easy.
As you have already gone through the previous years’ questions you will understand that only tricky questions are asked in NEET. Having a good command of Physics, Chemistry, and Biology concepts as well as a basic understanding of Calculus is essential for premiere exams like NEET. A basic understanding of Calculus cannot be ignored because without it would be difficult to obtain good marks in Physics. Students can easily get hold of Calculus concepts through disciplined study and practise.
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