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 the different properties of definite integrals, which will help to solve integration problems based on them.

Table of Contents:

• Definite Integral Definition
• Properties
• Proofs
• Example
• FAQs

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;

∫_a^b f(x) dx

Definite Integrals Properties

Following is the list of important properties of definite integrals which is easy to read and understand.

Properties of Definite Integrals Proofs

Property 1: _p∫^q f(a) da = _p∫^q 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: _p∫^q f(a) d(a) = – _q∫^p f(a) d(a), Also _p∫^p f(a) d(a) = 0

Suppose I = _p∫^q 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)] = – _q∫^p(a)da. Also, if p = q, then I= f’(q)-f’(p) = f’(p) -f’(p) = 0. Hence, _a∫^af(a)da = 0.

 

Property 3: _p∫^q f(a) d(a) = _p∫^r f(a) d(a) + _r∫^q f(a) d(a)

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

_p∫^q f(a)da = f’(q)-f’(p)… (1)

_p∫^rf(a)da = f’(r) – f’(p)… (2)

_r∫^qf(a)da = f’(q) – f’(r) … (3)

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

_p∫^r f(a)daf(a)da + _r∫^q f(a)daf(a)da = f’(r) – f’(p) + f’(q)

= f’(q) – f’(p) = _p∫^q f(a)da

 

Property 4: _p∫^q f(a) d(a) = _p∫^q 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, _p∫^q wil be replaced by _q∫^p when we replace a by t. Therefore,

_p∫^q f(a)da = –_q∫^p f(p+q-t)dt … from equation (4)

From property 2, we know that _p∫^q f(a)da = – _q∫^p f(a)da. Use this property, to get

_p∫^q f(a)da =_p∫^q f(p+q-t)da

Now use property 1 to get

_p∫^q f(a)da = _p∫^q f(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

I_{2 =} \(\int_{0}^{a}\)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, I_{1 =}\(\int_{-a}^{0}\)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,

I_{1 =}\(\int_{-p}^{0}\)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 on Properties of Definite Integrals

Example 1:

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

Solution:

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

Proof:

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

By using the property of definite integral

₀∫^a f(x) dx = ₀∫^a f(a-x) dx 

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

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

The above expression can be written as

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

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

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

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

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

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

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

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

Now, apply the logarithm property, we get

I = ₀∫^π/2log[(sinx. cos x)/sin2x]dx

We know that sin2x= 2 sinx cos x)

Now, the integral expression can be written as

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

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

I = ₀∫^π/2 log(1/2)dx

It can be written as

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

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

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

I = -log 2 ₀∫^π/2dx

Now, integrate the function

I = -log 2 [x]₀^π/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. ₀∫^π/2 (2log sinx – log sin 2x)dx  = – (π/2) log 2 is proved.

 

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