Sales Toll Free No: 1-855-666-7446



Calculus is divided into two main parts - differential calculus and integral calculus. Differential calculus deals with differentiation of functions, while integral calculus is concerned with Integration. Integration can be understood as the reverse process of differentiation. It is also called anti-differentiation. Basically, integration is the summation of a infinitesimally small quantities to be integrated as a body. Integration is denoted by the symbol $\int $. The quantity obtained after integrating a function is known as its integral or anti-derivative.

More specifically, to find the area between the curve of a function f(x) and a fixed line say (X axis), we integrate the function f(x) within the given limits. The concept behind this is that, we are dividing the required portion of curve (for which we need to calculate area) into infinitesimally small parts as shown in the given figure.
Now, consider that the width of each part is dx. Then, find the sum of those parts using integration technique.

Integration Definition

Back to Top
There are two types of integrations: Indefinite and Definite. Indefinite integration is the type of integration, which is not defined over a particular interval. Indefinite integral of a function f(x) with respect to the independent variable x is written as $\int f(x)\ dx$. Integration is reverse of differentiation. But, one difference is that, when we find an indefinite integral of a quantity, we add a constant term.
Integration Definition
Where, g(x) is another function which is obtained after integration of f(x).

A definite integral is defined over an interval. It is the summation of a function between lower and upper limit of that interval. Let us consider a continuous function f(x) defined over a closed interval [a, b]. Definite integration of f(x) can be expressed as follows:
Integration Formula
There is no need to add constant c, while calculating definite integrals. Because, there are already specified lower and upper bounds.

Integration Rules

Back to Top
Few important integration rules are given below:
  • $\int 1\ dx = x + c$
  • $\int x^{n}dx$ = $\frac{x^{n + 1}}{n + 1}$$ + c$
  • $\int e^{x}dx$ = $e^{x} + c$
  • $\int a^{x}dx$ = $\frac{a^{x}}{\ln x}$$ +c$, for $a > 0$
  • $\int $$\frac{1}{x}$$ dx = \ln x + c$
  • $\int \sin x dx = -\cos x + c$
  • $\int \cos x dx = \sin x + c$
  • $\int \sec^{2}x dx = \tan x + c$
  • $\int \csc^{2}x dx = -\cot x + c$
  • $\int \sec x \tan x dx = \sec x + c$
  • $\int \csc x \cot x dx = -\csc x + c$
  • $\int \tan x dx = \ln \sec x + c = -\ln \cos x + c$
  • $\int \cot x dx = \ln \sin x + c$
  • $\int \sec x dx = \ln (\sec x + \tan x) + c$
  • $\int \csc x dx = \ln (\csc x - \cot x) + c$
  • $\int $$\frac{1}{\sqrt{1 - x^{2}}}$$ dx = \sin^{-1}x + c = -\cos^{-1}x + c$
  • $\int $$\frac{1}{1 + x^{2}}$$ dx = \tan^{-1}x + c = -\cot^{-1}x + c$
  • $\int $$\frac{1}{x\sqrt{x^{2} - 1}}$$ dx = \sec^{-1}x + c = -\csc^{-1}x + c$
  • $\int \sinh x dx = \cosh x + c$
  • $\int \cosh x dx = \sinh x + c$

Integration Techniques

Back to Top
There are few important techniques for integration as follows:

Multiple of Constant Term:
$\int a f(x)dx = a \int f(x) + c$

Integration by Parts:
When we have to find the integration of product of two functions, we use integration by parts formula according to which, if we have two functions u and v, then according to the formula for integration by parts, we have:
Integration Techniques
We need to choose u and v wisely. The quantity whose integration is not known should not be chosen as v.

Integration by Substitution: At times, we come across unfamiliar integrands. In order to convert them to the standard familiar forms, we use substitution technique.

Integration by Partial Fractions: Sometimes, the integrand is in the form of algebraic fraction. In this case, it has to be decomposed into partial fractions which makes it similar to standard form.

Partial Integration

Back to Top
Partial integration may be referred as integration by parts, which is discussed earlier. Let us understand this with the help of an example:

Solved Example

Question: Integrate $x\ e^{x}$.
Let u = x and v = ex

Then, by integration by parts formula:

$\int u v = u \int v - \int \left [ {u}'\ \int v \right ]$

Now, $\int x\ e^{x}\ dx$ = $x\int e^{x}-\int 1 \times e^{x}dx + c$, where c is a integration constant.

= $xe^{x} - e^{x} + c$

Applications of Integration

Back to Top
Integration has a vast area of application. It is widely applicable in physics, electronics, engineering, commerce, finance etc. Few useful applications of integration are as follows:
  • Area between Two Curves: It is useful in finding the area between two curves or area under a curve. We integrate the function for which, we need to find the area applying the appropriate limits.
  • Arc Length: While finding arc length, we take a length element and integrate it along with its length.
  • Volumes of Revolution: Integration is most commonly used in finding the volume of the shape formed by revolving a curve about a particular axis. In this case, we need to determine elementary volume and then, integrate it within suitable limits.
  • Moment of Inertia: Integration is also used to compute moment of inertia of a body.
  • Average Value: Integration is widely used in finding average value of a function, such as average temperature, average score etc.
In medical science, finance, commerce and many other fields also, integration plays an important role.

Integration Examples

Back to Top
Let us consider few examples based on integration.

Solved Examples

Question 1: $\int \sin (3x-5)dx$
Let u = 3x - 5
By integrating with respect to x, we get
du = 3 dx

$\int \sin (3x - 5)dx$ = $\int \sin u $$\frac{du}{3}$

= $\frac{1}{3}$$(-\cos u)+c$

Substituting back the value of u, we get

= $-\frac{1}{3}$$\cos (3x - 5) + c$

Question 2: Integrate $x \ln x$ with respect to x.
Let u = ln x and v = x
Then, by integration by parts formula:
$\int u v = u \int v - \int \left [ {u}'\ \int v \right ]$
Now, $\int x \ln x dx$ = $\ln x\ \times $$\frac{x^{2}}{2}$$ - \int $$\left [ \frac{1}{x}\ \times \frac{x^{2}}{2}\right ]$$dx + c$, where, c is a constant
= $\frac{x^{2}}{2}$$\ln x\ - $$\frac{x^{2}}{4}$$ + c$