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. **∫_{a}^{b} f(x) dx = ∫_{a}^{b} f(t) dt

**2.** ∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(x) dx … [Also, ∫_{a}^{a} f(x) dx = 0]

**3.** ∫_{a}^{b} f(x) dx = ∫_{a}^{c} f(x) dx + ∫_{c}^{b} f(x) dx

**4.** ∫_{a}^{b} f(x) dx = ∫_{a}^{b} f(a + b – x) dx

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

**6.** ∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(2a – x) dx

**7.** Two parts

- ∫
_{0}^{2a}f(x) dx = 2 ∫_{0}^{a}f(x) dx … if f(2a – x) = f(x). - ∫
_{0}^{2a}f(x) dx = 0 … if f(2a – x) = – f(x)

**8.** Two parts

- ∫
_{-a}^{a}f(x) dx = 2 ∫_{0}^{a}f(x) dx … if f(- x) = f(x) or it is an even function - ∫
_{-a}^{a}f(x) dx = 0 … if f(- x) = – f(x) or it is an odd function

## Proofs of Definite Integrals Properties

### Property 1: ∫_{a}^{b} f(x) dx = ∫_{a}^{b} 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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**Browse more Topics under Integrals**

- Fundamental Theorem of Calculus
- Introduction to Integration
- Properties of Indefinite Integrals
- Definite Integral as a Limit of a Sum
- Integration by Partial Fractions
- Integration by Parts
- Integration by Substitutions
- Integral of Some Particular Functions
- Integral of the Type e^x[f(x) + f'(x)]dx

### Property 2: ∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(x) dx … [Also, ∫_{a}^{a} f(x) dx = 0]

Let I = ∫_{a}^{b} 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)] = – ∫_{b}^{a} f(x) dx. Also, if a = b, then I = F(b) – F(a) = F(a) – F(a) = 0. Hence, ∫_{a}^{a} f(x) dx = 0.

### Video on Definite Integrals

### Property 3: ∫_{a}^{b} f(x) dx = ∫_{a}^{c} f(x) dx + ∫_{c}^{b} f(x) dx

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

- ∫
_{a}^{b}f(x) dx = F(b) – F(a) … (1) - ∫
_{a}^{c}f(x) dx = F(c) – F(a) … (2) - ∫
_{c}^{b}f(x) dx = F(b) – F(c) … (3)

Adding equations (2) and (3), we get

∫_{a}^{c} f(x) dx + ∫_{c}^{b} f(x) dx = F(c) – F(a) + F(b) – F(c)

= F(b) – F(a) = ∫_{a}^{b} f(x) dx

### Property 4: ∫_{a}^{b} f(x) dx = ∫_{a}^{b} 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, ∫_{a}^{b} will be replaced by ∫_{b}^{a} when we replace x by t. Therefore,

∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(a + b – t) dt … from equation (4)

From Property 2, we know that ∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(x) dx. Using this property, we get

∫_{a}^{b} f(x) dx = ∫_{a}^{b} f(a + b – t) dt

Next, using Property 1, we get

∫_{a}^{b} f(x) dx = ∫_{a}^{b} f(a + b – x) dx

### Property 5: ∫_{0}^{a} f(x) dx = ∫_{0}^{a} 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, ∫_{0}^{a} will be replaced by ∫_{a}^{0} when we replace x by t. Therefore,

∫_{0}^{a} f(x) dx = – ∫_{a}^{0} f(a – t) dt … from equation (5)

From Property 2, we know that ∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(x) dx. Using this property, we get

∫_{0}^{a} f(x) dx = ∫_{0}^{a} f(a – t) dt

Next, using Property 1, we get

∫_{0}^{a} f(x) dx = ∫_{0}^{a} f(a – x) dx

### Property 6: ∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(2a – x) dx

From Property 3, we know that

∫_{a}^{b} f(x) dx = ∫_{a}^{c} f(x) dx + ∫_{c}^{b} f(x) dx

Therefore, ∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{a}^{2a} f(x) dx = I_{1} + I_{2} … (6)

Where, I_{1} = ∫_{0}^{a} f(x) dx and I_{2} = ∫_{a}^{2a} 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, ∫_{a}^{2a} will be replaced by ∫_{a}^{0} when we replace x by t. Therefore,

I_{2} = ∫_{a}^{2a} f(x) dx = – ∫_{a}^{0} f(2a – t) dt … from equation (7)

From Property 2, we know that ∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(x) dx. Using this property, we get

I_{2} = ∫_{0}^{a} f(2a – t) dt

Next, using Property 1, we get

I_{2} = ∫_{0}^{a} f(2a – x) dx

Replacing the value of I_{2} in equation (6), we get

∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(2a – x) dx

### Property 7: ∫_{0}^{2a} f(x) dx = 2 ∫_{0}^{a} f(x) dx … if f(2a – x) = f(x) and

∫_{0}^{2a} f(x) dx = 0 … if f(2a – x) = – f(x)

From Property 5, we know that

∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(2a – x) dx … (8)

Now, if f(2a – x) = f(x), then equation (8) becomes

∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(x) dx

= 2 ∫_{0}^{a} f(x) dx

And, if f(2a – x) = – f(x), then equation (8) becomes

∫_{0}^{2a} f(x) dx = ∫_{0}^{a} f(x) dx – ∫_{0}^{a} f(x) dx = 0

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

Using Property 3, we have

∫_{-a}^{a} f(x) dx = ∫_{-a}^{0} f(x) dx + ∫_{0}^{a} f(x) dx = I_{1} + I_{2} … (9)

Where, I_{1} = ∫_{-a}^{0} f(x) dx I_{2} = ∫_{0}^{a} f(x) dx

**Consider I _{1}**

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, ∫_{-a}^{0} will be replaced by ∫_{a}^{0} when we replace x by t. Therefore,

I_{1} = ∫_{-a}^{0} f(x) dx = – ∫_{a}^{0} f(– t) dt … from equation (10)

From Property 2, we know that ∫_{a}^{b} f(x) dx = – ∫_{b}^{a} f(x) dx. Using this property, we get

I_{1} = ∫_{-a}^{0} f(x) dx = ∫_{0}^{a} f(– t) dt

Next, using Property 1, we get

I_{1} = ∫_{-a}^{0} f(x) dx = ∫_{0}^{a} f(– x) dx

Replacing the value of I_{2} in equation (9), we get

∫_{-a}^{a} f(x) dx = I_{1} + I_{2} = ∫_{0}^{a} f(– x) dx + ∫_{0}^{a} f(x) dx … (11)

Now, if ‘f’ is an even function, then f(– x) = f(x). Therefore, equation (11) becomes

∫_{-a}^{a} f(x) dx = ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(x) dx = 2 ∫_{0}^{a} f(x) dx

And, if ‘f’ is an odd function, then f(– x) = – f(x). Therefore, equation (11) becomes

∫_{-a}^{a} f(x) dx = – ∫_{0}^{a} f(x) dx + ∫_{0}^{a} f(x) dx = 0

## Solved Examples on Definite Integrals

**Question 1: Evaluate ∫ _{-1}^{2} |x^{3} – x| dx**

**Answer :** Observe that, (x^{3} – x) ≥ 0 on [– 1, 0], (x^{3} – x) ≤ 0 on [0, 1] and (x^{3} – x) ≥ 0 on [1, 2]

Hence, using Property 3, we can write

∫_{-1}^{2} |x^{3} – x| dx = ∫_{-1}^{0} (x^{3} – x) dx + ∫_{0}^{1} – (x^{3} – x) dx + ∫_{1}^{2} (x^{3} – x) dx

= ∫_{-1}^{0} (x^{3} – x) dx + ∫_{0}^{1} (x – x^{3}) dx + ∫_{1}^{2} (x^{3} – x) dx

Solving the integrals, we get

∫_{-1}^{2} |x^{3} – x| dx = [(x^{4}/4 – (x^{2}/2)]_{-1}^{0} + [(x^{2}/2 – (x^{4}/4)]_{0}^{1} + [(x^{4}/4 – (x^{2}/2)]_{1}^{2
}= – [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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