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 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:

  • Definite Integral
  • 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

pq f(a) da = pq f(t) dt
Property 2 pq f(a) d(a) = – qp f(a) d(a), Also pp f(a) d(a) = 0
Property 3 pf(a) d(a) = pr f(a) d(a) + rq f(a) d(a)
Property 4 pq f(a) d(a) = pq f( p + q – a) d(a)
Property 5 op f(a) d(a) = op f(p – a) d(a)
Property 6 02p f(a)da = ∫0f(a)da +∫0f(2p-a)da…if f(2p-a) = f(a)
Property 7 2 parts

  • 0f(a)da = 2∫0a f(a) da … if f(2p-a) = f(a)
  • 0p f(a)da = 0 … if f(2p-a) = -f(a)
Property 8 2 parts

  • -pf(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

Proofs of Properties of Definite Integrals

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.

Solved Example

Question: 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] -[\(\frac{1}{4}\) -\(\frac{1}{2}\) = \(\frac{11}{4}\)

 

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