# Properties of Definite Integrals

We will be exploring some of the important properties of definite integrals and their proofs in this article to get a better understanding. Integration is the estimation of an integral. It is just the opposite process of differentiation.  Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc.  There are two types of Integrals namely, definite integral and indefinite integral. Here, we will learn about definite integrals and its properties, which will help to solve integration problems based on them.

## Definite Integral Definition

If an integral has upper and lower limits, it is called a Definite Integral. There are many definite integral formulas and properties. Definite Integral is the difference between the values of the integral at the specified upper and lower limit of the independent variable. It is represented as;

ab f(x) dx

## Definite Integral Properties

Following is the list of definite integrals in the tabular form which is easy to read and understand.

 Properties Description Property 1 p∫q f(a) da = p∫q f(t) dt Property 2 p∫q f(a) d(a) = – q∫p f(a) d(a), Also p∫p f(a) d(a) = 0 Property 3 p∫q f(a) d(a) = p∫r f(a) d(a) + r∫q f(a) d(a) Property 4 p∫q f(a) d(a) = p∫q f( p + q – a) d(a) Property 5 o∫p f(a) d(a) = o∫p f(p – a) d(a) Property 6 ∫02p f(a)da = ∫0p f(a)da +∫0p f(2p-a)da…if f(2p-a) = f(a) Property 7 2 parts ∫02 f(a)da = 2∫0a f(a) da … if f(2p-a) = f(a) ∫02 p f(a)da = 0 … if f(2p-a) = -f(a) Property 8 2 parts ∫-pp f(a)da = 2∫0p f(a) da … if f(-a) = f(a) or it’s an even function ∫-ppf(a)da = 0 … if f(2p-a) = -f(a) or it’s an odd function

## Properties of Definite Integrals Proofs

Property 1: pq f(a) da = pq f(t) dt

This is the simplest property as only a is to be substituted by t, and the desired result is obtained.

Property 2: pq f(a) d(a) = – qp f(a) d(a), Also pp f(a) d(a) = 0

Suppose I = pq f(a) d(a)

If f’ is the anti-derivative of f, then use the second fundamental theorem of calculus, to get I = f’(q)-f’(p) = – [f’(p) – f’(q)] = – qp(a)da. Also, if p = q, then I= f’(q)-f’(p) = f’(p) -f’(p) = 0. Hence, aaf(a)da = 0.

Property 3: pf(a) d(a) = pr f(a) d(a) + rq f(a) d(a)

If f’ is the anti-derivative of f, then use the second fundamental theorem of calculus, to get;

pf(a)da = f’(q)-f’(p)… (1)

prf(a)da = f’(r) – f’(p)… (2)

rqf(a)da = f’(q) – f’(r) … (3)

Let’s add equations (2) and (3), to get

pf(a)daf(a)da + rf(a)daf(a)da = f’(r) – f’(p) + f’(q)

= f’(q) – f’(p) = pf(a)da

Property 4: pq f(a) d(a) = pq f( p + q – a) d(a)

Let, t = (p+q-a), or a = (p+q – t), so that dt = – da … (4)

Also, note that when a = p, t = q and when a = q, t = p. So, pq wil be replaced by qp when we replace a by t. Therefore,

pf(a)da = –qf(p+q-t)dt … from equation (4)

From property 2, we know that pf(a)da = – qf(a)da. Use this property, to get

pf(a)da =pf(p+q-t)da

Now use property 1 to get

pf(a)da = pf(p + q – a )da

Property 5: $\int_{0}^{p}$f(a)da = $\int_{0}^{p}$f(p-a)da

Let, t = (p-a) or a = (p – t), so that dt = – da …(5)

Also, observe that when a = 0, t =p and when a = p, t = 0. So, $\int_{0}^{p}$ will be replaced by $\int_{o}^{p}$ when we replace a by t. Therefore,

$\int_{0}^{p}$f(a)da = – $\int_{p}^{0}$f(p – t)da … from equation (5)

From Property 2, we know that $\int_{p}^{q}$f(a)da = -$\int_{q}^{p}$f(a)da. Using this property , we get

$\int_{0}^{p}$f(a)da = $\int_{0}^{p }$f(p-t)dt

Next, using Property 1, we get

$\int_{0}^{a}$f(a)da = $\int_{0}^{p}$f(p – a)da

Property 6: $\int_{0}^{2p}$f(a)da = $\int_{0}^{p}$f(a)da + $\int_{0}^{p}$f(2p – a))da

From property 3, we know that

$\int_{p}^{q}$f(a)da = $\int_{p}^{r}$f(a)da + $\int_{r}^{q}$f(a)da

Therefore, $\int_{0}^{2p}$f(a)da = $\int_{0}^{p}$f(a)da + $\int_{p}^{2p}$f(a)da = I1 + I2 … (6)

Where, I1 = $\int_{0}^{p}$f(a)da and I2 =$\int_{p}^{2p}$f(a)da

Let, t = (2p – a) or a = (2p – t), so that dt = -da …(7)

Also, note that when a = p, t = p, and when a =2p, t= 0. Hence, $\int_{a}^{0}$ when we replace a by t. Therefore,

I2 = $\int_{p}^{2p}$f(a)da = – $\int_{p}^{0}$f(2p-0)da… from equation (7)

From Property 2, we know that $\int_{p}^{q}$f(a)da =- $\int_{q}^{p}$f(a)da. Using this property, we get I2 = $\int_{0}^{p}$f(2p-t)dt

Next, using Property 1, we get

I2 = $\int_{0}^{a}$f(a)da + $\int_{0}^{a}$f(2p-a)da

Replacing the value of I2 in equation (6), we get

$\int_{0}^{2p}$f(a)da = $\int_{0}^{p}$f(a)da + $\int_{o}^{p}$f(2p – a)da

Property 7: $\int_{0}^{2a}$f(a)da = 2 $\int_{0}^{a}$f(a)da … if f(2p – a) = f(a) and

$\int_{0}^{2a}$f(a)da = 0 … if f(2p- a) = -f(a)

we know that

$\int_{0}^{2p}$f(a)da =$\int_{0}^{p}$f(a)da + $\int_{0}^{p}$f(2p – a)da … (8)

Now, if f(2p – a) = f(a), then equation (8) becomes

$\int_{0}^{2p}$f(a)da = $\int_{0}^{p}$f(a)da + $\int_{0}^{p}$f(a)da

=2$\int_{0}^{p}$f(a)da

And, if f(2p – a) = – f(a), then equation (8) becomes

$\int_{0}^{2p}$f(a)da = $\int_{0}^{p}$f(a)da -$\int_{0}^{p}$f(a)da = 0

Property 8: $\int_{-p}^{p}$f(a)da = 2$\int_{0}^{p}$f(a)da … if f(-a) =f(a) or it is an even function and $\int_{-a}^{a}$f(a)da = 0, … if f(-a) = -f(a) or it is an odd function.

Using Property 3, we have

$\int_{-p}^{p}$f(a)da = $\int_{-a}^{0}$f(a)da + $\int_{0}^{p}$f(a)da = I1 + I,2 …(9)

Where, I1 =$\int_{-a}^{0}$f(a)da, I2 =$\int_{0}^{p}$f(a)da

Consider I1

Let, t = -a or a = -t, so that dt = -dx … (10)

Also, observe that when a = -p, t = p, when a = 0, t =0. Hence, $\int_{-a}^{0}$ will be replaced by $\int_{a}^{0}$ when we replace a by t. Therefore,

I1 = $\int_{-a}^{0}$f(a)da = – $\int_{a}^{0}$f(-a)da … from equation (10)

From Property 2, we know that$\int_{p}^{q}$f(a)da = – $\int_{q}^{p}$f(a)da, use this property to get,

I1 =$\int_{-p}^{0}$f(a)da = $\int_{0}^{p}$f(-a)da

Next, using Property 1, we get

I1 = $\int_{-p}^{0}$f(a)da = $\int_{0}^{p}$f(-a)da

Replacing the value of I2 in equation (9), we get

$\int_{-p}^{p}$f(a)da = I1 + I2 = $\int_{0}^{p}$f(-a)da + $\int_{0}^{p}$f(a)da =2 $\int_{0}^{p}$f(a)da … (11)

Now, if ‘f’ is an even function, then f(– a) = f(a). Therefore, equation (11) becomes

$\int_{-p}^{p}$f(a)da = $\int_{0}^{p}$f(a)da +$\int_{0}^{p}$f(a)da =2$\int_{0}^{p}$f(a)da

And, if ‘f’ is an odd function, then f(–a) = – f(a). Therefore, equation (11) becomes

$\int_{-p}^{p}$f(a)da = – $\int_{0}^{a}$f(a)da + $\int_{0}^{p}$f(a)da = 0

Now, let us evaluate Definite Integral through a problem sum.

### Example

Question 1: Evaluate $\int_{-1}^{2}$f(a3 – a)da

Solution: Observe that, (a3 – a) ≥ 0 on [– 1, 0], (a3 – a) ≤ 0 on [0, 1] and (a3 – a) ≥ 0 on [1, 2]

Hence, using Property 3, we can write

$\int_{-1}^{2}$f(a3 – a)da =$\int_{-1}^{0}$f(a3 – a)da + $\int_{0}^{1}$f-(a3 – a)da + $\int_{1}^{2}$f(a3 – a)da = $\int_{-1}^{0}$f(a3 – a)da +$\int_{0}^{1}$f(a – a3 )da + $\int_{1}^{2}$f(a3 – a)da

$\int_{-1}^{0}$f(a3 – a)da +$\int_{0}^{1}$f(a – a3 )da +$\int_{1}^{2}$f(a3 – a)da

Solving the integrals, we get

$\int_{-1}^{2}f(a3 – a)da = \frac{x4}{4} – (\frac{x2}{2}) ]-10 + [(\frac{x2}{2} – (\frac{x4}{4}))01 + [ \frac{x4}{4} -(\frac{x2}{2}) ]12$

= – [$\frac{1}{4}$ – $\frac{1}{2}$] + [$\frac{}{}$ – $\frac{1}{4}$] + [ 4 – 2] -[$\frac{1}{4}$ -$\frac{1}{2}$ = $\frac{11}{4}$

Question 2:

Prove that 0π/2 (2log sinx – log sin 2x)dx  = – (π/2) log 2 using the properties of definite integral

Solution:

To prove: 0π/2 (2log sinx – log sin 2x)dx  = – (π/2) log 2

Proof:

Let take I = 0π/2 (2log sinx – log sin 2x)dx …(1)

By using the property of definite integral

0a f(x) dx = 0a f(a-x) dx

Now, apply the property in (1), we get

I = 0π/2 2log sin[(π/2)-x] – log sin 2[(π/2)-x])dx

The above expression can be written as

I = 0π/2 [2log cosx- log sin(π-2x)]dx (Since, sin (90-θ = cos θ)

I = 0π/2 [2log cosx- log sin2x]dx ..(2)

Now, add the equation (1) and (2), we get

I+ I = 0π/2 [(2log sinx – log sin 2x) +(2log cosx- log sin2x)]dx

2I =  0π/2 [2log sinx -2log 2sinx + 2log cos x]dx

2I = 2 0π/2 [log sinx -log 2sinx + log cos x]dx

Now, cancel out 2 on both the sides, we get

I = 0π/2 [log sinx + log cos x- log 2sinx]dx

Now, apply the logarithm property, we get

I = 0π/2log[(sinx. cos x)/sin2x]dx

We know that sin2x= 2 sinx cos x)

Now, the integral expression can be written as

I = 0π/2log[(sinx. cos x)/(2 sinx cos x)]dx

Cancel the terms which are common in both numerator and denominator, then we get

I = 0π/2 log(1/2)dx

It can be written as

I = 0π/2 (log1-log 2)dx [Since, log (a/b) = log a- log b]

I = 0π/2 -log 2 dx (value of log 1 = 0)

Now, take the constant – log 2 outside the integral,

I = -log 2 0π/2dx

Now, integrate the function

I = -log 2 [x]0π/2

Now, substitute the limits

I = -log 2 [(π/2)-0]

I = – log 2 (π/2)

I = – (π/2) log 2 = R.H.S

Therefore, L.H. S = R.H.S

Hence. 0π/2 (2log sinx – log sin 2x)dx  = – (π/2) log 2 is proved.

## Frequently Asked Questions on Properties of Definite Integral

### Define definite integral

The definite integral is defined as an integral with two specified limits called the upper and the lower limit. The definite integral of a function generally represents the area under the curve from the lower bound value to the higher bound value.

### What are the different properties of definite integral?

Some of the important properties of definite integrals are:
Interval of zero-length property
Reversing the interval property
Area above – area below property

### What is the difference between the definite integral and indefinite integral?

The definite integral f(x) is a number which defines the area under the curves within the specified limits. It has an upper limit and lower limit and it gives a definite answer. Whereas the indefinite integral f(x) is a function and it has no upper and lower limits. It gives a solution to the question “what function produces f(x) when it is differentiated?”.

### Mention the four important concepts covered in calculus?

The most important basic concepts in calculus are:
Function
Limits
Integral
Derivatives

### Define the definite integral of a function?

The definite integral of a function on the interval [a, b] is defined as the difference of antiderivative of the given function, which is calculated for the upper bound of integration minus lower bound of integration.

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