Objectives: By the end of the lesson, the learner should be able to
▪ illustrate an antiderivative of a function,
▪ compute the general antiderivative of polynomial, radical, exponential, and trigonometric functions,
▪ compute the antiderivative of the function using substitution rule.
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Antiderivative of a Function
Given 𝐹(𝑥) = 5𝑥 3 and 𝑓(𝑥) = 15𝑥 2. Notice that 𝑓(𝑥) is the derivative of 𝐹(𝑥). We call 𝐹(𝑥) as an antiderivative of
𝑓(𝑥).
Definition
A function 𝐹 is an antiderivative of 𝑓 on an interval 𝐼 if 𝐹 ′ (𝑥) = 𝑓(𝑥) for all x in 𝐼.
Example
Given 𝐹(𝑥) = 3𝑥 2 + 2𝑥 + 1 and 𝑓(𝑥) = 6𝑥 + 2 . Then, we say that 𝐹′(𝑥) = 6𝑥 + 2 and therefore 𝐹(𝑥) is an
antiderivative of 𝑓(𝑥).
The function 𝐺(𝑥) = 3𝑥 2 + 2𝑥 − 5 and 𝐻(𝑥) = 3𝑥 2 + 2𝑥 are both antiderivatives of f (x ) as well.
Example
Consider the function 𝐹(𝑥) = 𝑠𝑖𝑛 𝑥 and 𝑓(𝑥) = 𝑐𝑜𝑠 𝑥. The derivative of the function 𝐹(𝑥) is 𝐹′(𝑥) = 𝑐𝑜𝑠 𝑥. Hence, 𝐹(𝑥)
is an antiderivative of 𝑓(𝑥).
Also, the function 𝐺(𝑥) = + 𝑠𝑖𝑛 𝑥 and 𝐻(𝑥) = 𝑠𝑖𝑛 𝑥 − 3 are both antiderivatives of f (x ) .
1
2
Both examples show how derivatives and antiderivatives are related. They also give a clear indication that the
antiderivative of a function is not unique, and that they vary with some constant 𝐶. Therefore, every antiderivative of
a function may be expressed in the following form:
𝑭(𝒙) + 𝑪
where 𝐶 is a constant.
For the function 𝑓(𝑥) = 6𝑥 + 2 , its antiderivative takes the form 𝐹(𝑥) = 3𝑥 2 + 2𝑥 + 𝐶 and for 𝑓(𝑥) = 𝑐𝑜𝑠 𝑥 , the
antiderivative takes the form 𝐹(𝑥) = 𝑠𝑖𝑛 𝑥 + 𝐶.
Definition
The process of finding an antiderivative of 𝑓(𝑥) is called antidifferentiation or indefinite integration. This process
is denoted by the operator
and the notation ∫ 𝒇(𝒙)𝒅𝒙 indicates the process of integration. In symbols,
∫ 𝒇(𝒙)𝒅𝒙 = 𝑭(𝒙) + 𝑪
where
is the integral sign
where 𝒇(𝒙) is the expression to be integrated called the integrand
where dx indicates the variable of integration
where C is now called the constant of integration
Basic Integration Rules
1. ∫ 𝑑𝑥 = 𝑥 + 𝐶
2. ∫ 𝑘𝑑𝑥 = 𝑘 ∫ 𝑑𝑥 = 𝑘𝑥 + 𝐶 where k is any constant
3. ∫ 𝑘𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥
, 4. ∫(𝑓1 (𝑥) ± 𝑓2 (𝑥)±. . . ±𝑓𝑛 (𝑥))𝑑𝑥 = ∫ 𝑓1 (𝑥)𝑑𝑥 ± ∫ 𝑓2 (𝑥)𝑑𝑥 ±. . . ± ∫ 𝑓𝑛 (𝑥)𝑑𝑥
𝑥 𝑛+1
5. If n is any rational number not equal to -1, then ∫ 𝑥 𝑛 𝑑𝑥 = + 𝐶. (Power Rule)
𝑛+1
6. ∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
𝑎𝑥
7. ∫ 𝑎 𝑥 𝑑𝑥 = 𝑙𝑛 𝑎 + 𝐶
𝑑𝑥
8. ∫ 𝑥 −1 𝑑𝑥 = ∫ = 𝑙𝑛|𝑥| + 𝐶
𝑥
9. ∫ 𝑐𝑜𝑠 𝑥 𝑑𝑥 = 𝑠𝑖𝑛 𝑥 + 𝐶
10. ∫ 𝑠𝑖𝑛 𝑥 𝑑𝑥 = − 𝑐𝑜𝑠 𝑥 + 𝐶
11. ∫ 𝑠𝑒𝑐 2 𝑥 𝑑𝑥 = 𝑡𝑎𝑛 𝑥 + 𝐶
12. ∫ 𝑐𝑠𝑐 2 𝑥 𝑑𝑥 = − 𝑐𝑜𝑡 𝑥 + 𝐶
13. ∫ 𝑠𝑒𝑐 𝑥 𝑡𝑎𝑛 𝑥 𝑑𝑥 = 𝑠𝑒𝑐 𝑥 + 𝐶
14. ∫ 𝑐𝑠𝑐 𝑥 𝑐𝑜𝑡 𝑥 𝑑𝑥 = − 𝑐𝑠𝑐 𝑥 + 𝐶
Note
There may be a need to rewrite the integrand to a form like the ones given so that the integration process
may be carried out. (Rewrite, Integrate, Simplify)
Example
Find the integral of the following:
1. ∫(𝟑𝒙𝟐 + 𝟒𝒙 − 𝟓)𝒅𝒙
Solution:
∫(3𝑥 2 + 4𝑥 − 5)𝑑𝑥 = 3 ∫ 𝑥 2 𝑑𝑥 + 4 ∫ 𝑥𝑑𝑥 − 5 ∫ 𝑑𝑥
𝑥 2+1 𝑥 1+1
∫(3𝑥 2 + 4𝑥 − 5)𝑑𝑥 = 3 ( 2+1 ) + 4 ( 1+1 ) − 5𝑥 + 𝐶
𝑥3 𝑥2
∫(3𝑥 2 + 4𝑥 − 5)𝑑𝑥 = 3 ( 3 ) + 4 ( 2 ) − 5𝑥 + 𝐶
∫(𝟑𝒙𝟐 + 𝟒𝒙 − 𝟓)𝒅𝒙 = 𝒙𝟑 + 𝟐𝒙𝟐 − 𝟓𝒙 + 𝑪
2. ∫(𝒙 + 𝟏)(𝟐𝒙 − 𝟑)𝒅𝒙
Solution:
∫(𝑥 + 1)(2𝑥 − 3)𝑑𝑥 = ∫(2𝑥 2 − 𝑥 − 3)𝑑𝑥
∫(𝑥 + 1)(2𝑥 − 3)𝑑𝑥 = 2 ∫ 𝑥 2 𝑑𝑥 − ∫ 𝑥𝑑𝑥 − 3 ∫ 𝑑𝑥
𝑥 2+1 𝑥 1+1
∫(𝑥 + 1)(2𝑥 − 3)𝑑𝑥 = 2 ( 2+1 ) − ( 1+1 ) − 3𝑥 + 𝐶
𝑥3 𝑥2
∫(𝑥 + 1)(2𝑥 − 3)𝑑𝑥 = 2 ( 3 ) − ( 2 ) − 3𝑥 + 𝐶
𝟐 𝟏
∫(𝒙 + 𝟏)(𝟐𝒙 − 𝟑)𝒅𝒙 = 𝟑 𝒙𝟑 − 𝟐 𝒙𝟐 − 𝟑𝒙 + 𝑪