Properties of Definite Integrals: Definition, Properties & Examples 

1.0Introduction to Definite Integrals

Definite integrals are one of the most crucial subjects in JEE Mathematics. They can be found in both JEE Main and JEE Advanced, frequently in conjunction with ideas from symmetry, differentiation, limits, and continuity.

Understanding the characteristics of definite integrals can greatly simplify computations, even though solving a definite integral directly might occasionally be challenging. These characteristics save time, streamline integration processes, and are especially helpful for objective-type problems.

2.0Definition of Definite Integrals

A definite integral represents the signed area under a curve between two given limits.

If f(x) is a continuous function on [a,b], then the definite integral is defined as:

${\int }_a^{b}f(x)dx$

• Here, a = lower limit, b = upper limit.
• The result is a number (not a function).

3.0Standard Properties of Definite Integrals

Let’s go through the main properties of definite integrals that are frequently used in JEE.

Property 1: Interchanging Limits

${\int }_a^{b}f(x),dx=-{\int }_b^{a}f(x),dx$

Explanation:
Reversing the order of limits changes the sign of the integral.

Property 2: Additivity Over Intervals

${\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx={\int }_a^{c}f(x)dx$

Explanation:
The total integral from ( a ) to ( c ) can be broken into intervals.

Property 3: Linearity

${\int }_a^b[k_{1}f(x)+k_{2}g(x)],dx=k_1{\int }_a^{b}f(x),dx+k_2{\int }_a^{b}g(x),dx$

Explanation:
The integral of a sum (or difference) is the sum (or difference) of the integrals. Constants can be taken outside the integral sign.

Property 4: Integration of Even and Odd Functions

When integrating over symmetric limits ( [-a, a] ):

• Even Function: ( f(-x) = f(x) )

${\int }_{-a}^{a}f(x),dx=2{\int }_0^{a}f(x),dx$

• Odd Function: ( f(-x) = -f(x) )
  ${\int }_{-a}^{a}f(x),dx=0$

Explanation:
Even functions double up over symmetric limits, odd functions cancel out.

Property 5: Periodicity

If ( f(x) ) is periodic with period ( T ), then:

${\int }_a^{a+T}f(x),dx={\int }_0^{T}f(x),dx$

Explanation:
The value of the integral over any interval of one period is the same.

Property 6: Symmetry

${\int }_0^{a}f(x),dx={\int }_0^{a}f(a-x),dx$

Explanation:
This property is very useful for simplifying integrals by substituting ( x ) with ( a-x ).

Property 7: Substitution

If $(x=\varphi (t))$, then:

${\int }_a^{b}f(x),dx={\int }_{\alpha }^{\beta }f(\varphi (t))$ I am running a few minutes late; my previous meeting is running over.\phi'(t), dt
where $(\alpha ={\varphi }^{-1}(a)),(\beta ={\varphi }^{-1}(b))$.

Explanation:
Change of variables can often simplify the integral.

4.0Important Results and Shortcuts

• ${\int }_a^{b}f(x),dx={\int }_a^{b}f(a+b-x),dx$
• If ( f(x) + f(a + b - x) = k ), then ${\int }_a^{b}f(x),dx=k(b-a)/2$
• If ( f(x) ) is odd about ( (a + b)/2 ), then ${\int }_a^{b}f(x),dx=0$

Apply these shortcuts wherever possible to save time in JEE exams.

5.0Solved Examples

Example 1: Using Symmetry

Evaluate $({\int }_0^2(x^2+(2-x{)}^2),dx)$.

Using property:

${\int }_0^a[f(x)+f(a-x)],dx=2{\int }_0^{a}f(x),dx$

Let $(f(x)=x^2),(a=2)$:

${\int }_0^2(x^2+(2-x{)}^2),dx=2{\int }_0^2{x}^2,dx=2{\left[\frac{x^3}{3}\right]}_0^2=2\times \frac{8}{3}=\frac{16}{3}$

Example 2: Integration of Even Function

Evaluate $({\int }_{-3}^3(5x^2+2),dx)$.

Split into even and odd parts:

• $(5x^2)$ is even, ( 2 ) is constant (even).

${\int }_{-3}^{3}5x^2,dx=2{\int }_0^{3}5x^2,dx=2\times 5{\left[\frac{x^3}{3}\right]}_0^3=2\times 5\times \frac{27}{3}=90$

${\int }_{-3}^{3}2,dx=2\times 3-(-3)=2\times 6=12$

Total: ( 90 + 12 = 102 )

Example 3: Changing Order of Limits

Evaluate $({\int }_4^1{e}^{2x},dx)$.

Interchanging limits:

${\int }_4^1{e}^{2x},dx=-{\int }_1^4{e}^{2x},dx$

Now integrate:

$-{\left[\frac{e^{2x}}{2}\right]}_1^4=-\left(\frac{e^8-e^2}{2}\right)=\frac{e^2-e^8}{2}$

Example 4: Substitution Method

Evaluate $({\int }_0^1\frac{dx}{\sqrt{1-x^2}})$.

Substitute $(x=\sin \theta ),(dx=\cos \theta ,d\theta )$:

When $(x=0,\theta =0)$; when $(x=1,\theta =\frac{\pi }{2})$:

${\int }_0^{\pi /2}\frac{\cos \theta }{\sqrt{1-{\sin }^2\theta }},d\theta ={\int }_0^{\pi /2}\frac{\cos \theta }{\cos \theta },d\theta ={\int }_0^{\pi /2}d\theta =\frac{\pi }{2}$

6.0Tips and Tricks for JEE

• Always check for symmetry when the limits are symmetric about the origin or a point.
• For polynomials or trigonometric expressions, check if the integrand is even, odd, or periodic.
• Use substitution to simplify complex integrals.
• Remember the properties and shortcuts—they can turn a lengthy problem into a one-step solution.