Integration can be used to find areas, volumes, and many other useful things. Integration is a method of adding slices to find the whole. It refers to summation. It is also known as the antiderivative. It is the reverse process of differentiation. As far as the JEE exam is concerned, integration is a very important topic. Students can easily score marks from this chapter if they are thorough with the concept.

The process of finding an indefinite integral of a given function is called the integration of the function. If we want to find the integral of a function f(x), it means we have to find a function Φ(x) such that (d/dx)Φ(x) = f(x). The symbol for integration is ∫. In this article, we will discuss **definite and indefinite integration****. **

## Notation

We use the symbol ∫ for integration. After the Integral symbol, we write the function which has to be integrated. It is called the integrand. Then we put dx. The following example shows how to write the function which is to be integrated.

∫ x^{2} dx = x^{3}/3 + C. Here C represents the constant of integration. We use C because of all the functions whose derivative is x^{3}/3:

## Different methods of Integration

- Substitution
- Integration by partial fractions
- By parts
- Euler substitution
- Reduction method

### Some Properties of Indefinite integral

- The method of differentiation and integration are inverses of each other.

(d/dx) ∫ f(x) dx = f(x)

∫ f’(x) dx = f(x) + C

- Two indefinite integrals having the same derivative lead to the same family of curves. Hence they are equivalent.
- The integral of the sum of two functions is equal to the sum of the integral of each function.

∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx

Let us have a look at some problems with indefinite integration.

Example 1:

Find ∫ 2x sin x^{2} dx.

Solution:

Here we will substitute x^{2} = t

So 2x dx = dt

Hence the function becomes ∫ sin t dt.

∫ sin t dt = cos t + C

= cos x^{2} + C

Example 2.

Find ∫ (x^{3} -1)/x^{2}

Solution:

We split the given function.

x^{3}/x^{2} = x

1/x^{2} = x^{-2}

∫ (x^{3} -1)/x^{2} = ∫ x dx – ∫ x^{-2} dx

= (x^{2}/2) – (x^{-1}/-1) + C

= (x^{2}/2) + (1/x) + C

### Definite Integral

A definite integral has upper and lower limits. It has a unique value. It is represented as ∫_{a}^{b} f(x) dx. This denotes the area bounded by curve y = f(x) the ordinates are x = a, x = b and the x-axis. Here a is the lower limit of the integral and b is the upper limit of the integral. It is introduced as the limit of a sum in interval [a,b].

∫_{a}^{b} f(x) dx = F(b) – F(a).

There is a connection between indefinite integral and definite integral which makes the definite integral as a practical tool for different science and engineering. It is also used to solve many interesting problems from various disciplines like economics, finance, and probability. JEE aspirants can expect one question from definite integrals. Students are recommended to practise **Definite Integrals Previous Year Questions With Solutions**. This will help students to have an idea of the question pattern. One example is given below.

Example 1.

Let [ ] denote greatest integer function, then find the value of ∫_{0}^{1.5} x[x^{2}]dx.

(1) 5/4

(2) 0

(3) 3/2

(4) ¾

Solution:

∫_{0}^{1.5} x[x^{2}]dx = ∫_{0}^{1} 0dx+ ∫_{1}^{√2}x.1dx+ ∫_{√2}^{1.5}2x dx

= (x^{2}/2)_{1}^{√2}+ (x^{2})_{√2}^{1.5}

= (2/2) -(½)+2.25 -2

= (½) + (¼)

= ¾

Hence option (4) is the answer.