# Properties of Definite Integrals

In this article, we will be looking at some important properties of definite integrals which will be useful in evaluating such integrals effectively. We will also look at the proofs of each of these properties to gain a better understanding of them.

## List of Properties of Definite Integrals

1.Â âˆ«ab f(x) dx = âˆ«ab f(t) dt

2.Â âˆ«ab f(x) dx = â€“ âˆ«ba f(x) dx â€¦ [Also, âˆ«aa f(x) dx = 0]

3.Â âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx

4.Â âˆ«ab f(x) dx = âˆ«ab f(a + b â€“ x) dx

5.Â âˆ«0a f(x) dx = âˆ«0a f(a â€“ x) dx â€¦ [this is derived from P04]

6.Â âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a â€“ x) dx

7.Â Two parts

1. âˆ«02a f(x) dx = 2 âˆ«0a f(x) dx â€¦ if f(2a â€“ x) = f(x).
2. âˆ«02a f(x) dx = 0 â€¦ if f(2a â€“ x) = â€“ f(x)

8.Â Two parts

1. âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx â€¦ if f(- x) = f(x) or it is an even function
2. âˆ«-aa f(x) dx = 0 â€¦ if f(- x) = – f(x) or it is an odd function

## Proofs of Definite Integrals Properties

### Property 1: âˆ«ab f(x) dx = âˆ«ab f(t) dt

The proof for this property is not needed since simply by substituting x = t, the desired output is achieved.

### Property 2: âˆ«ab f(x) dx = â€“ âˆ«ba f(x) dx â€¦ [Also, âˆ«aa f(x) dx = 0]

Let I = âˆ«ab f(x) dx. If â€˜Fâ€™ is the anti-derivative of â€˜fâ€™, then by using the second fundamental theorem of calculus, we have I = F(b) â€“ F(a) = â€“ [F(a) â€“ F(b)] = â€“ âˆ«ba f(x) dx. Also, if a = b, then I = F(b) â€“ F(a) = F(a) â€“ F(a) = 0. Hence, âˆ«aa f(x) dx = 0.

### Property 3: âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx

If â€˜Fâ€™ is the anti-derivative of â€˜fâ€™, then by using the second fundamental theorem of calculus, we have

• âˆ«ab f(x) dx = F(b) â€“ F(a) â€¦ (1)
• âˆ«ac f(x) dx = F(c) â€“ F(a) â€¦ (2)
• âˆ«cb f(x) dx = F(b) â€“ F(c) â€¦ (3)

Adding equations (2) and (3), we get
âˆ«ac f(x) dx + âˆ«cb f(x) dx = F(c) â€“ F(a) + F(b) â€“ F(c)
= F(b) â€“ F(a) = âˆ«ab f(x) dx

### Property 4: âˆ«ab f(x) dx = âˆ«ab f(a + b â€“ x) dx

Let, t = (a + b â€“ x), or x = (a + b â€“ t), so that dt = â€“ dx â€¦ (4)

Also, observe that when x = a, t = b and when x = b, t = a. Hence, âˆ«ab will be replaced by âˆ«ba when we replace x by t. Therefore,
âˆ«ab f(x) dx = â€“ âˆ«ba f(a + b â€“ t) dt â€¦ from equation (4)

From Property 2, we know that âˆ«ab f(x) dx = â€“ âˆ«ba f(x) dx. Using this property, we get
âˆ«ab f(x) dx = âˆ«ab f(a + b â€“ t) dt

Next, using Property 1, we get
âˆ«ab f(x) dx = âˆ«ab f(a + b â€“ x) dx

### Property 5: âˆ«0a f(x) dx = âˆ«0a f(a â€“ x) dx

Let, t = (a â€“ x) or x = (a â€“ t), so that dt = â€“ dx â€¦ (5)

Also, observe that when x = 0, t = a and when x = a, t = 0. Hence, âˆ«0a will be replaced by âˆ«a0 when we replace x by t. Therefore,
âˆ«0a f(x) dx = â€“ âˆ«a0 f(a â€“ t) dt â€¦ from equation (5)

From Property 2, we know that âˆ«ab f(x) dx = â€“ âˆ«ba f(x) dx. Using this property, we get
âˆ«0a f(x) dx = âˆ«0a f(a â€“ t) dt

Next, using Property 1, we get
âˆ«0a f(x) dx = âˆ«0a f(a â€“ x) dx

### Property 6: âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a â€“ x) dx

From Property 3, we know that
âˆ«ab f(x) dx = âˆ«ac f(x) dx + âˆ«cb f(x) dx

Therefore, âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«a2a f(x) dx = I1 + I2 â€¦ (6)
Where, I1 = âˆ«0a f(x) dx and I2 = âˆ«a2a f(x) dx

Let, t = (2a â€“ x) or x = (2a â€“ t), so that dt = â€“ dx â€¦ (7)

Also, observe that when x = a, t = a, and when x = 2a, t = 0. Hence, âˆ«a2a will be replaced by âˆ«a0 when we replace x by t. Therefore,
I2 = âˆ«a2a f(x) dx = â€“ âˆ«a0 f(2a â€“ t) dt â€¦ from equation (7)

From Property 2, we know that âˆ«ab f(x) dx = â€“ âˆ«ba f(x) dx. Using this property, we get
I2 = âˆ«0a f(2a â€“ t) dt

Next, using Property 1, we get
I2 = âˆ«0a f(2a â€“ x) dx

Replacing the value of I2 in equation (6), we get
âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a â€“ x) dx

### Property 7: âˆ«02a f(x) dx = 2 âˆ«0a f(x) dx â€¦ if f(2a â€“ x) = f(x) and âˆ«02a f(x) dx = 0 â€¦ if f(2a â€“ x) = â€“ f(x)

From Property 5, we know that
âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(2a â€“ x) dx â€¦ (8)

Now, if f(2a â€“ x) = f(x), then equation (8) becomes
âˆ«02a f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx
= 2 âˆ«0a f(x) dx

And, if f(2a â€“ x) = â€“ f(x), then equation (8) becomes
âˆ«02a f(x) dx = âˆ«0a f(x) dx â€“ âˆ«0a f(x) dx = 0

### Property 8: âˆ«-aa f(x) dx = 2 âˆ«0a f(x) dx â€¦ if f(- x) = f(x) or it is an even function and âˆ«-aa f(x) dx = 0 â€¦ if f(- x) = – f(x) or it is an odd function

Using Property 3, we have
âˆ«-aa f(x) dx = âˆ«-a0 f(x) dx + âˆ«0a f(x) dx = I1 + I2 â€¦ (9)
Where, I1 = âˆ«-a0 f(x) dx I2 = âˆ«0a f(x) dx

Consider I1

Let, t = â€“ x or x = â€“ t, so that dt = â€“ dx â€¦ (10)

Also, observe that when x = â€“ a, t = a, and when x = 0, t = 0. Hence, âˆ«-a0 will be replaced by âˆ«a0 when we replace x by t. Therefore,
I1 = âˆ«-a0 f(x) dx = â€“ âˆ«a0 f(â€“ t) dt â€¦ from equation (10)

From Property 2, we know that âˆ«ab f(x) dx = â€“ âˆ«ba f(x) dx. Using this property, we get
I1 = âˆ«-a0 f(x) dx = âˆ«0a f(â€“ t) dt

Next, using Property 1, we get
I1 = âˆ«-a0 f(x) dx = âˆ«0a f(â€“ x) dx

Replacing the value of I2 in equation (9), we get
âˆ«-aa f(x) dx = I1 + I2 = âˆ«0a f(â€“ x) dx + âˆ«0a f(x) dx â€¦ (11)

Now, if â€˜fâ€™ is an even function, then f(â€“ x) = f(x). Therefore, equation (11) becomes
âˆ«-aa f(x) dx = âˆ«0a f(x) dx + âˆ«0a f(x) dx = 2 âˆ«0a f(x) dx

And, if â€˜fâ€™ is an odd function, then f(â€“ x) = â€“ f(x). Therefore, equation (11) becomes
âˆ«-aa f(x) dx = â€“ âˆ«0a f(x) dx + âˆ«0a f(x) dx = 0

## Solved Examples on Definite Integrals

Question 1: Evaluate âˆ«-12 |x3 â€“ x| dx

Answer : Observe that, (x3 â€“ x) â‰¥ 0 on [â€“ 1, 0],Â (x3 â€“ x) â‰¤ 0 on [0, 1] andÂ (x3 â€“ x) â‰¥ 0 on [1, 2]

Hence, using Property 3, we can write
âˆ«-12 |x3 â€“ x| dx = âˆ«-10 (x3 â€“ x) dx + âˆ«01 â€“ (x3 â€“ x) dx + âˆ«12 (x3 â€“ x) dx
= âˆ«-10 (x3 â€“ x) dx + âˆ«01 (x â€“ x3) dx + âˆ«12 (x3 â€“ x) dx

Solving the integrals, we get
âˆ«-12 |x3 â€“ x| dx = [(x4/4 â€“ (x2/2)]-10 + [(x2/2 â€“ (x4/4)]01 + [(x4/4 â€“ (x2/2)]12
= â€“ [1/4 â€“ Â½] + [1/2 â€“ Â¼] + [4 â€“ 2] â€“ [1/4 â€“ Â½] = 11/4.

Question 2: What is meant by definite integral?

Answer: A definite integral refers to an integral with upper and lower limits. If it is restricted to exist on the real line, the definite integral is called by the name of Riemann integral.

Question 3: Differentiate between indefinite and definite integral?

Answer: A definite integral is characterized by upper and lower limits. Moreover, the reason why it is called definite is because it provides a definite answer at the end of the problem. Indefinite integral, in contrast, refers to a form of integration that is more general in nature. Furthermore, the interpretation of the indefinite integral is as the considered functionâ€™s anti-derivative.

Question 4: Is it possible for definite integrals to be positive?

Answer: Yes, it is possible for a definite integral to be positive. Integrals measure the area between the curve in question and the x-axis over a specified interval. Furthermore, if all of the area that is within the interval exists above the curve and below the x-axis then the result shall certainly be negative.

Question 5: Do definite integrals require constant of integration?

Answer: No, definite integrals have no requirement of a constant of integration.

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