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Integration- Definite Integral

Updated: Jun 22, 2021



In this note, you will learn:


· What are definite integrals?

· Properties of definite integrals

· Evaluation of definite integrals

What are definite integrals?




In the past few articles on Integration, we have learnt many different integration techniques and formulae to solve a variety of integration problems involving power functions, trigonometry, exponential and logarithm.



Those are what we call indefinite integrals, since they were all expressed without limits, and hence have infinite solutions since of the arbitrary constant, c, it contains.



Today, we will be learning about what we call definite integrals. Definite integrals are integrals expressed as the difference between the values of the integral at specified upper and lower limits of the independent variable.



As you can see from the diagram above, the evaluation of a definite integral is written as such, and what we have to the right of the integration sign are the upper (larger value) and lower (smaller value) limits, which are b and a respectively. Given that the integrand is f(x), we can read it as, “the integration of f(x) from a to b”.





Like what we have mentioned back in the “Introduction to Integration”, having specified upper and lower bounds can help use to solve questions when we are trying to values like area and volume. In the diagram above, we have a curve of y = 4x^2, and if we want to solve for the area bounded by y = 4x^2, x = 1, x = 2 and the x-axis, or the integral of y = 4x^2 from x = 1 to x = 2, we can evaluate the following definite integral as such:


2 2

∫ 1 4x^2 dx = [(4/3) x^3] 1

= [(4/3) (2)^3] – [(4/3) (1)^3]

= (32/3) – (4/3)

= 28/3

Properties of definite integrals


Since we are aware that definite integrals represent the area or difference in area bounded within a graph, hence we can conclude the following properties of definite integrals:

b

· If a = b, then ∫a f(x) dx = 0

b

· ∫ba f(x) dx = - ∫a f(x) dx



b c c

∫a f(x) dx + ∫b f(x) dx = ∫a f(x) dx


(Area of F + Area of G = Total Area)


· Definite integral of scalar multiple of function:


b b

a) ∫a k f(x) dx = k ∫a f(x) dx


b b

b) [k f(x)] a = k [f(x)] a



The Scalar Multiple Rule suggest that considering the function k f(x), where k is a constant and f(x) is a function, k will hence be called a scalar multiple. Just like for differentiation, this means that k, the scalar multiple, can be moved to the front of the term and be omitted from the integration of the term. The scalar multiple can then be multiplied back into the result once the integration of the term is done.

· Definite integral of addition/subtraction of functions:


b b b

∫a [f(x) +/- g(x)] dx = ∫a f(x) dx +/- ∫a g(x) dx


The Addition/Subtraction Rule suggests that if f(x) and g(x) are functions, then the integration when there is an addition or subtraction of the two functions can be split into the integration of two individual functions.


Evaluation of definite integrals


Now that we have learnt about definite integrals and its properties, let’s take a look at some example questions down below to help you understand the concept of definite integrals:

· Evaluating definite integrals of ax^n


Evaluate each of the following.

5

a) ∫1 x^4 dx

3

b) ∫-1 7x^2 dx



Solution:


5 5

a) ∫1 x^4 dx = [(1/5) x^5] 1

5

= (1/5) [x^5] 1 (Application of [k f(x)] ba = k [f(x)] ba)

= (1/5) [(5)^5 – (1)^5]

= (1/5) (3124)

= 3124/5


3 3

b) ∫-1 7x^2 dx = 7 ∫-1 x^2 dx (Application of ∫ba k f(x) dx = k ∫ba f(x) dx)

3

= 7 [(1/3) x^3] -1

3

= 7(1/3) [x^3] -1 (Application of [k f(x)] ba = k [f(x)] ba)

= 7(1/3) [(3)^3 – (-1)^3]

= 7(1/3) (27 + 1)

= 7(1/3) (28)

= 196/3


· Evaluation of definite integrals of (ax + b)^n


Evaluate each of the following:

a) ∫20 (5x + 4)^3 dx


4

b) ∫-2 √ (3x + 8) dx



Solution:

a) ∫20 (5x + 4)^3 dx = [(5x + 4)^4/ (4)(5)] 20

= (1/20) [(5x + 4)^4] 20 (Application of [k f(x)] ba = k [f(x)] ba)

= (1/20) {[5(2) + 4]^4 + [5(0) + 4]^4}

= (1/20) [(14)^4 + (4)^4]

= 18/20

= 38672/20

= 9668/5


4 4

b) ∫-2 √ (3x + 8) dx = ∫-2 (3x + 8)^1/2 dx

4

= [(3x + 8)^3/2 / (3/2) (3)] -2

4

= (2/9) [(3x + 8)^3/2] -2 (Application of [k f(x)] ba = k [f(x)] ba)

= (2/9) {[3(4) + 8]^3/2 – [3(-2) + 8]^3/2}

= (2/9) [(20)^3/2 – (2)^3/2]

= 19.2486 (4 d.p)

= 19.2 (3 s.f)

· Evaluation of definite integrals of trigonometric functions


Evaluate each of the following:

π

a) ∫0 (3 cos x + 2) dx


π

b) ∫0 (4 sin 2x + 5 tan^2 3x) dx


Solution:


π π

a) ∫0 (3 cos x + 2) dx = [3 sin x + 2x] 0

= {[3 sin (π) + 2(π)] – [3 sin (0) + 2(0)]}

= [(0) + 2π] – (0)

= 2π


π

b) ∫0 (4 sin 2x + 5 tan^2 3x) dx

π

= ∫0 (4 sin 2x + 5 sec^2 3x - 5) dx (Recall: tan^2 x = sec^2 x – 1)

π π π

= ∫0 (4 sin 2x) dx + ∫0 (5 sec^2 3x) dx - ∫0 5 dx

π π π

= 4 ∫0 sin 2x dx + 5 ∫0 sec^2 3x dx - 5 ∫0 1 dx

π π

= 4 [-(1/2) cos 2x] π0 + 5 [(1/3) tan 3x] 0 – 5 [x] 0

= 4 {[(-1/2) cos 2(π)] – [(-1/2) cos 2(0)]} + 5 {[(1/3) tan 3(π)] – [(1/3) tan 3(0)]} – 5 [(π) – (0)]

= 4 [(-1/2) (1) – (-1/2) (1)] + 5 [(1/3) (0) – (1/3) (0)] – 5(π)

= 4(0) + 5(0) - 5π

= -5π


· Evaluation of definite integrals of exponential functions


Evaluate each of the following, giving your answer correct to 2 decimal places.

3

a) ∫0 e^(3+x) dx


5

b) ∫2 (4e^(3x) + 7) dx

Solution:

3

a) ∫30 e^(3+x) dx = [e^(3+x)] 0

= [e^(3+(3))] – [e^(3+(0))]

= e^6 – e^3

= 383.34 (2 d.p)

5 5

b) ∫2 (4e^(3x) + 7) dx = [(4/3) e^(3x) + 7x] 2

= [(4/3) e^3(5) + 7(5)] – [(4/3) e^3(2) + 7(2)]

= 4358172.93 (2 d.p)

And that’s all for today, students! Math Lobby hopes that after this article, you have a clear understanding on definite integrals, and is equipped with the necessary skills to deal with questions that involve finding definite integrals of areas bounded on the graph!



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