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Integrals

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.

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List of Properties of Definite Integrals

1. ab f(x) dx = ∫ab f(t) dt

2. ∫ab f(x) dx = – ∫ba f(x) dx … [Also, ∫aa f(x) dx = 0]

3. ∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx

4. ∫ab f(x) dx = ∫ab f(a + b – x) dx

5. ∫0a f(x) dx = ∫0a f(a – x) dx … [this is derived from P04]

6. ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

7. Two parts

  1. 02a f(x) dx = 2 ∫0a f(x) dx … if f(2a – x) = f(x).
  2. 02a f(x) dx = 0 … if f(2a – x) = – f(x)

8. Two parts

  1. -aa f(x) dx = 2 ∫0a f(x) dx … if f(- x) = f(x) or it is an even function
  2. -aa f(x) dx = 0 … if f(- x) = – f(x) or it is an odd function

Definite Integrals

Proofs of Definite Integrals Properties

Property 1: ∫ab f(x) dx = ∫ab f(t) dt

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

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Property 2: ∫ab f(x) dx = – ∫ba f(x) dx … [Also, ∫aa f(x) dx = 0]

Let I = ∫ab f(x) dx. If ‘F’ is the anti-derivative of ‘f’, then by using the second fundamental theorem of calculus, we have I = F(b) – F(a) = – [F(a) – F(b)] = – ∫ba f(x) dx. Also, if a = b, then I = F(b) – F(a) = F(a) – F(a) = 0. Hence, ∫aa f(x) dx = 0.

Video on Definite Integrals

Property 3: ∫ab f(x) dx = ∫ac f(x) dx + ∫cb f(x) dx

If ‘F’ is the anti-derivative of ‘f’, then by using the second fundamental theorem of calculus, we have

  • ab f(x) dx = F(b) – F(a) … (1)
  • ac f(x) dx = F(c) – F(a) … (2)
  • cb f(x) dx = F(b) – F(c) … (3)

Adding equations (2) and (3), we get
ac f(x) dx + ∫cb f(x) dx = F(c) – F(a) + F(b) – F(c)
= F(b) – F(a) = ∫ab f(x) dx

Property 4: ∫ab f(x) dx = ∫ab f(a + b – x) dx

Let, t = (a + b – x), or x = (a + b – t), so that dt = – dx … (4)

Also, observe that when x = a, t = b and when x = b, t = a. Hence, ∫ab will be replaced by ∫ba when we replace x by t. Therefore,
ab f(x) dx = – ∫ba f(a + b – t) dt … from equation (4)

From Property 2, we know that ∫ab f(x) dx = – ∫ba f(x) dx. Using this property, we get
ab f(x) dx = ∫ab f(a + b – t) dt

Next, using Property 1, we get
ab f(x) dx = ∫ab f(a + b – x) dx

Property 5: ∫0a f(x) dx = ∫0a f(a – x) dx

Let, t = (a – x) or x = (a – t), so that dt = – dx … (5)

Also, observe that when x = 0, t = a and when x = a, t = 0. Hence, ∫0a will be replaced by ∫a0 when we replace x by t. Therefore,
0a f(x) dx = – ∫a0 f(a – t) dt … from equation (5)

From Property 2, we know that ∫ab f(x) dx = – ∫ba f(x) dx. Using this property, we get
0a f(x) dx = ∫0a f(a – t) dt

Next, using Property 1, we get
0a f(x) dx = ∫0a f(a – x) dx

Property 6: ∫02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

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

Therefore, ∫02a f(x) dx = ∫0a f(x) dx + ∫a2a f(x) dx = I1 + I2 … (6)
Where, I1 = ∫0a f(x) dx and I2 = ∫a2a f(x) dx

Let, t = (2a – x) or x = (2a – t), so that dt = – dx … (7)

Also, observe that when x = a, t = a, and when x = 2a, t = 0. Hence, ∫a2a will be replaced by ∫a0 when we replace x by t. Therefore,
I2 = ∫a2a f(x) dx = – ∫a0 f(2a – t) dt … from equation (7)

From Property 2, we know that ∫ab f(x) dx = – ∫ba f(x) dx. Using this property, we get
I2 = ∫0a f(2a – t) dt

Next, using Property 1, we get
I2 = ∫0a f(2a – x) dx

Replacing the value of I2 in equation (6), we get
02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx

Property 7: ∫02a f(x) dx = 2 ∫0a f(x) dx … if f(2a – x) = f(x) and
02a f(x) dx = 0 … if f(2a – x) = – f(x)

From Property 5, we know that
02a f(x) dx = ∫0a f(x) dx + ∫0a f(2a – x) dx … (8)

Now, if f(2a – x) = f(x), then equation (8) becomes
02a f(x) dx = ∫0a f(x) dx + ∫0a f(x) dx
= 2 ∫0a f(x) dx

And, if f(2a – x) = – f(x), then equation (8) becomes
02a f(x) dx = ∫0a f(x) dx – ∫0a f(x) dx = 0

Property 8: ∫-aa f(x) dx = 2 ∫0a f(x) dx … if f(- x) = f(x) or it is an even function and ∫-aa f(x) dx = 0 … if f(- x) = – f(x) or it is an odd function

Using Property 3, we have
-aa f(x) dx = ∫-a0 f(x) dx + ∫0a f(x) dx = I1 + I2 … (9)
Where, I1 = ∫-a0 f(x) dx I2 = ∫0a f(x) dx

Consider I1

Let, t = – x or x = – t, so that dt = – dx … (10)

Also, observe that when x = – a, t = a, and when x = 0, t = 0. Hence, ∫-a0 will be replaced by ∫a0 when we replace x by t. Therefore,
I1 = ∫-a0 f(x) dx = – ∫a0 f(– t) dt … from equation (10)

From Property 2, we know that ∫ab f(x) dx = – ∫ba f(x) dx. Using this property, we get
I1 = ∫-a0 f(x) dx = ∫0a f(– t) dt

Next, using Property 1, we get
I1 = ∫-a0 f(x) dx = ∫0a f(– x) dx

Replacing the value of I2 in equation (9), we get
-aa f(x) dx = I1 + I2 = ∫0a f(– x) dx + ∫0a f(x) dx … (11)

Now, if ‘f’ is an even function, then f(– x) = f(x). Therefore, equation (11) becomes
-aa f(x) dx = ∫0a f(x) dx + ∫0a f(x) dx = 2 ∫0a f(x) dx

And, if ‘f’ is an odd function, then f(– x) = – f(x). Therefore, equation (11) becomes
-aa f(x) dx = – ∫0a f(x) dx + ∫0a f(x) dx = 0

Solved Examples on Definite Integrals

Question 1: Evaluate ∫-12 |x3 – x| dx

Answer : Observe that, (x3 – x) ≥ 0 on [– 1, 0], (x3 – x) ≤ 0 on [0, 1] and (x3 – x) ≥ 0 on [1, 2]

Hence, using Property 3, we can write
-12 |x3 – x| dx = ∫-10 (x3 – x) dx + ∫01 – (x3 – x) dx + ∫12 (x3 – x) dx
= ∫-10 (x3 – x) dx + ∫01 (x – x3) dx + ∫12 (x3 – x) dx

Solving the integrals, we get
-12 |x3 – x| dx = [(x4/4 – (x2/2)]-10 + [(x2/2 – (x4/4)]01 + [(x4/4 – (x2/2)]12
= – [1/4 – ½] + [1/2 – ¼] + [4 – 2] – [1/4 – ½] = 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 function’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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