What is Calculus?

Calculus is all about the comparison of quantities which vary in a one-liner way. It has significant applications in Science and Engineering. Many of the topics that we study like acceleration, velocity and current in a circuit do not behave in a linear fashion. If quantities are changing continually, we need calculus to study about it. Calculus is the branch of mathematics that deals with continuous change.

Table of Contents:

  • Calculus Topics
  • Definitions of Important Topics in Calculus
  • Applications of Calculus
  • Examples

Basic Calculus

List of some of the important Calculus topics :

Let us discuss definitions for some of the important calculus topics: 

Differential Calculus Basics


The degree of closeness to any value or the approaching term. A limit is normally expressed using the limit formula as-

\(\lim_{x \rightarrow c}f(x)=A\)

It is read as “the limit of f of x as x approaches c equals A”.


Instantaneous rate of change of a quantity with respect to the other. The derivative of a function is represented as:

\(\lim_{x \to h}\frac{f(x+h)-f(x)}{h}=A\)



A function is said to be continuous at a particular point if the following three conditions are satisfied-

  • f(a) is defined
  • \(lim_{x \to a}f(x)\) exists
  • \(lim_{x \to a^-}f(x) = lim_{x \to a^+}f(x) =f(a)\)

Read More: Limits And Continuity

Continuity and Differentiability

A Function is always continuous if it is differentiable at any point, whereas the vice-versa condition is not always true.

Quotient Rule

The Quotient rule is a method for determining the derivative (differentiation) of a function which is in fractional form.

Chain Rule

The rule applied for finding the derivative of the composition of a function is basically known as the chain rule.

Integral Calculus Basics

Study about integrals and their properties. It is mostly useful for the following two purposes:

  • To calculate f from f’(i.e from its derivative). If a function f is differentiable in the interval of consideration, then f’ is defined in that interval.
  • To calculate the area under a curve.


Integration is the reciprocal of differentiation. As differentiation can be understood as dividing a part into many small parts, integration can be said as a collection of small parts in order to form a whole. It is generally used for calculating area.

Definite Integral

A definite integral has a specific boundary within which function needs to be calculated. The lower limit and upper limit of the independent variable of a function is specified, its integration is described using definite integrals. A definite integral is denoted as:

\(\int_a^b\ f(x).dx = F(x)\)

Indefinite Integral

An indefinite integral does not have a specific boundary, i.e. no upper and lower limit is defined. Thus the integration value is always accompanied by a constant value (C). It is denoted as:

\(\int\) f(x).dx = F(x) + C

Calculus Applications

In real life, concepts of calculus play a major role either it is related to solving area of complicated shapes, safety of vehicles, to evaluate survey data for business planning, credit cards payment records, or to find how the changing conditions of a system affect us, etc. Calculus often used by physicians, economists, biologists, architects and statisticians. For example, Architects and engineers use concepts of calculus to determine the size and shape of the curves to design bridges, roads and tunnels etc.

Calculus Problems

Below are some examples based on calculus topics:

Problem 1: Let f(y) = ey and g(y) = 10y. Use the chain rule to calculate h′(y) where h(y) = f(g(y)).

Solution: Given,

f(y) = ey and

g(y) = 10y

First derivative above functions are

f'(y) = ey and

g'(y) = 10

To find: h′(y)

Now, h(y) = f(g(y))

h'(y) = f'(g(y))g'(y)

h'(y) = f'(10y)10

By substituting the values.

h'(y) = e10y x 10

or h'(y) = 10 e10y


Problem 2: Integrate sin 3x + 2x with respect to x.

Solution: Given instructions can be written as:

∫ sin 3x + 2x dx

Use the sum rule, which implies

∫  sin 3x dx+ ∫ int2x dx ……… Equation 1

Solve ∫  sin 3x dx first.

use substitution method,

let 3x = u => 3 dx = du (after derivation)

or dx = 1/3 du

=> ∫  sin 3x dx turned as∫ sin u  X 1/3 du

or 1/3 ∫  sin u du

which is 1/3 (-cos u) + C, where C= constant of integration

Substituting values again, we get

∫  sin 3x dx= -cos(3x)/3 + C ……… Equation 2

Solve∫ 2x dx

∫ 2x dx = 2∫  x dx = 2 * x2/2 + C = x2 + C ……. Equation 3

Equation (1) => ∫  sin 3x dx+ ∫ 2x dx

= -cos(3x)/3 + x2 + C

Practice more questions on calculus such as limits and derivatives, integration using different methods, continuity etc. To learn more about calculus in maths, Register with BYJU’S today.

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