Properties of Definite Integrals

As we know Integration is the opposite of differentiation and there are two types of Integrals:

  • Definite Integral
  • Indefinite Integral

We will be exploring some of the important properties of definite integrals and their proofs as we move forward in this article to get a better understanding.

Note: \(\int_{a}^{b}\)f(x)dx, In this article Variable x is replaced by a and limits a to b is replaced by p to q.

\(\int_{p}^{q}\)f(a)da

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. 

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 \(\int_{0}^{2p}\) f(a)da = \(\int_{0}^{p}\)f(a)da +\(\int_{0}^{p}\)f(2p-a)da…if f(2p-a) = f(a).
Property 7 2 parts

  • \(\int_{0}^{2}\)f(a)da = 2\(\int_{0}^{a}\) f(a) da … if f(2p-a) = f(a)
  • \(\int_{0}^{2}\)p f(a)da = 0 … if f(2p-a) = -f(a)
Property 8 2 parts

  • \(\int_{-p}^{p}\)f(a)da = 2\(\int_{0}^{p}\) f(a) da … if f(-a) = f(a) or it’s an even function
  • \(\int_{-p}^{p}\)f(a)da = 0 … if f(2p-a) = -f(a) or it’s an odd function

Properties of Definite Integrals Proofs

Property 1: \(\int_{p}^{q}\)f(a)da = \(\int_{p}^{q}\)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: \(\int_{p}^{q}\)f(a)da = – \(\int_{q}^{p}\)f(a)da… Also, \(\int_{a}^{a}\)f(a)da=0]

Suppose I = \(\int_{p}^{q}\)f(a)da.

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)] = – \(\int_{q}^{p}\)f(a)da. Also, if p = q, then I= f’(q)-f’(p) = f’(p) -f’(p) = 0. Hence, \(\int_{a}^{a}\)f(a)da = 0.

 

Property 3: \(\int_{p}^{q}\)f(a)da = \(\int_{p}^{r}\)f(a)da + \(\int_{r}^{q}\)f(a)da.

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

\(\int_{p}^{q}\)f(a)da = f’(q)-f’(p)… (1)

\(\int_{p}^{r}\)f(a)da = f’(r) – f’(p)… (2)

\(\int_{r}^{q}\)f(a)da = f’(q) – f’(r) … (3)

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

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

= f’(q) – f’(p) = \(\int_{p}^{q}\)f(a)da

 

Property 4: \(\int_{p}^{q}\)f(a)da = \(\int_{p}^{q}\)f(p+q- a)da

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, \(\int_{p}^{q}\) wil be replaced by \(\int_{q}^{p}\) when we replace a by t. Therefore,

\(\int_{p}^{q}\)f(a)da = -\(\int_{q}^{p}\)f(p+q-t)dt … from equation (4)

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

\(\int_{p}^{q}\)f(a)da =\(\int_{p}^{q}\)f(p+q-t)da

Now use property 1 to get

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

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 Examples on Properties of Definite Integrals.

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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