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![Just like any other mathematical operation, the process of differentiation can be reversed. For example,
when we perform the differentiation of 𝑓(𝑥) = 𝑥3
.
𝒇(𝒙) = 𝒙𝟑
𝒇′(𝒙) = 𝟑𝒙𝟑−𝟏
𝒇′(𝒙) = 𝟑𝒙𝟐
Now if you begin with the function 𝒇(𝒙) = 𝟑𝒙𝟐
, reversing the process should yield the possible functions
below:
𝒇(𝒙) = 𝒙𝟑
𝒇(𝒙) = 𝒙𝟑
+ 𝟏
𝒇(𝒙) = 𝒙𝟑
− 𝟏
The reversing of the operation of differentiation is known as ANTIDIFFERENETIATION or INDEFINITE
INTEGRATION. If the derivative of 𝑓(𝑥) = 𝑥3
is 𝑓′(𝑥) = 3𝑥2
, then we say that an antiderivative of 𝑓(𝑥) =
3𝑥2
is 𝑓(𝑥) = 𝑥3
.
𝒇(𝒙) = 𝒙𝟑
𝒇′(𝒙) = 𝟑𝒙𝟐
∫ 𝒇′(𝒙) 𝒅𝒙 = 𝒇(𝒙) + 𝑪
FINDING THE ANTIDERIVATIVE OF A FUNCTION
BASIC INTEGRATION FORMULAS
Differentiation Formulas Integration Formulas
𝒅
𝒅𝒙
(𝑪) = 𝟎 ∫ 𝟎 𝒅𝒙 = 𝑪
𝒅
𝒅𝒙
(𝒌𝒙) = 𝒌 ∫ 𝒌 𝒅𝒙 = 𝒌𝒙 + 𝑪
𝒅
𝒅𝒙
(𝒌𝑭(𝒙)) = 𝒌𝑭′(𝒙) ∫ 𝒌 𝒇(𝒙) 𝒅𝒙 = 𝒌 ∫ 𝒇( 𝒙) 𝒅𝒙
𝒅
𝒅𝒙
(𝑭(𝒙) + 𝑮(𝒙)) = 𝑭′(𝒙) + 𝑮′(𝒙) ∫[𝒇(𝒙) + 𝒈(𝒙)]𝒅𝒙 = ∫ 𝒇(𝒙) 𝒅𝒙 + ∫ 𝒈(𝒙) 𝒅𝒙
DERIVATIVE
INTEGRAL](https://image.slidesharecdn.com/module7theantiderivative-210225034500/75/Module-7-the-antiderivative-1-2048.jpg)
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Similar to Module 7 the antiderivative
Module 7 the antiderivative
- 1. Just like anyother mathematical operation, the process of differentiation can be reversed. For example, when we perform the differentiation of 𝑓(𝑥) = 𝑥3 . 𝒇(𝒙) = 𝒙𝟑 𝒇′(𝒙) = 𝟑𝒙𝟑−𝟏 𝒇′(𝒙) = 𝟑𝒙𝟐 Now if you begin with the function 𝒇(𝒙) = 𝟑𝒙𝟐 , reversing the process should yield the possible functions below: 𝒇(𝒙) = 𝒙𝟑 𝒇(𝒙) = 𝒙𝟑 + 𝟏 𝒇(𝒙) = 𝒙𝟑 − 𝟏 The reversing of the operation of differentiation is known as ANTIDIFFERENETIATION or INDEFINITE INTEGRATION. If the derivative of 𝑓(𝑥) = 𝑥3 is 𝑓′(𝑥) = 3𝑥2 , then we say that an antiderivative of 𝑓(𝑥) = 3𝑥2 is 𝑓(𝑥) = 𝑥3 . 𝒇(𝒙) = 𝒙𝟑 𝒇′(𝒙) = 𝟑𝒙𝟐 ∫ 𝒇′(𝒙) 𝒅𝒙 = 𝒇(𝒙) + 𝑪 FINDING THE ANTIDERIVATIVE OF A FUNCTION BASIC INTEGRATION FORMULAS Differentiation Formulas Integration Formulas 𝒅 𝒅𝒙 (𝑪) = 𝟎 ∫ 𝟎 𝒅𝒙 = 𝑪 𝒅 𝒅𝒙 (𝒌𝒙) = 𝒌 ∫ 𝒌 𝒅𝒙 = 𝒌𝒙 + 𝑪 𝒅 𝒅𝒙 (𝒌𝑭(𝒙)) = 𝒌𝑭′(𝒙) ∫ 𝒌 𝒇(𝒙) 𝒅𝒙 = 𝒌 ∫ 𝒇( 𝒙) 𝒅𝒙 𝒅 𝒅𝒙 (𝑭(𝒙) + 𝑮(𝒙)) = 𝑭′(𝒙) + 𝑮′(𝒙) ∫[𝒇(𝒙) + 𝒈(𝒙)]𝒅𝒙 = ∫ 𝒇(𝒙) 𝒅𝒙 + ∫ 𝒈(𝒙) 𝒅𝒙 DERIVATIVE INTEGRAL
- 2. 𝒅 𝒅𝒙 (𝒙𝒏) = 𝒏𝒙𝒏−𝟏 ∫𝒙𝒏 𝒅𝒙 = 𝒙𝒏+𝟏 𝒏 + 𝟏 + 𝑪; 𝒏 ≠ −𝟏 EXAMPLE 1: Find the ∫ 2 SOLUTION: ∫ 𝟐 = 𝟐𝒙 + 𝒄 EXAMPLE 2: Find the ∫(𝑥 − 3)(𝑥 + 4) SOLUTION: ∫(𝒙 − 𝟑)(𝒙 + 𝟒) = (𝒙 − 𝟑)(𝒙 + 𝟒) = (𝒙𝟐 + 𝒙 − 𝟏𝟐) = 𝒙𝟐+𝟏 𝟐+𝟏 + 𝒙𝟏+𝟏 𝟏+𝟏 − 𝟏𝟐𝒙 + 𝑪 = 𝒙𝟑 𝟑 + 𝒙𝟐 𝟐 − 𝟏𝟐𝒙 + 𝑪 EXAMPLE 3: Find the ∫ √𝑥(2𝑥2 − 3𝑥 + 1) SOLUTION: ∫ √𝑥(2𝑥2 − 3𝑥 + 1) = (𝒙 𝟏 𝟐)(2𝑥2 − 3𝑥 + 1) = 𝟐𝒙 𝟏 𝟐 +𝟐 − 𝟑𝒙 𝟏 𝟐 +𝟏 + 𝒙 𝟏 𝟐 = 𝟐𝒙 𝟓 𝟐 − 𝟑𝒙 𝟑 𝟐 + 𝒙 𝟏 𝟐 = 𝟐𝒙 𝟓 𝟐 +𝟏 𝟓 𝟐 +𝟏 − 𝟑𝒙 𝟑 𝟐 +𝟏 𝟑 𝟐 +𝟏 + 𝒙 𝟏 𝟐 +𝟏 𝟏 𝟐 +𝟏 + 𝑪 = 𝟐𝒙 𝟕 𝟐 𝟕 𝟐 − 𝟑𝒙 𝟓 𝟐 𝟓 𝟐 + 𝒙 𝟑 𝟐 𝟑 𝟐 + 𝑪 = 𝟒𝒙 𝟕 𝟐 𝟕 − 𝟔𝒙 𝟓 𝟐 𝟓 + 𝟐𝒙 𝟑 𝟐 𝟑 + 𝑪
- 3. = 𝟒 𝟕 √𝒙𝟕 − 𝟔 𝟓 √𝒙𝟓 + 𝟐 𝟑 √𝒙𝟑+ 𝑪 TRIGONOMETRIC FUNCTIONS INTEGRATION FORMULAS Differentiation Formulas Integration Formulas 𝒅 𝒅𝒙 (𝒔𝒊𝒏 𝒙) = 𝒄𝒐𝒔 𝒙 ∫ 𝒄𝒐𝒔 𝒙 𝒅𝒙 = 𝒔𝒊𝒏 𝒙 + 𝑪 𝒅 𝒅𝒙 (𝒄𝒐𝒔 𝒙) = −𝒔𝒊𝒏 𝒙 ∫ 𝒔𝒊𝒏 𝒙 𝒅𝒙 = −𝒄𝒐𝒔 𝒙 + 𝑪 𝒅 𝒅𝒙 (𝒕𝒂𝒏 𝒙) = 𝒔𝒆𝒄𝟐 𝒙 ∫ 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 = 𝒕𝒂𝒏 𝒙 + 𝑪 𝒅 𝒅𝒙 (𝒄𝒐𝒕 𝒙) = −𝒄𝒔𝒄𝟐 𝒙 ∫ 𝒄𝒔𝒄𝟐 𝒙 𝒅𝒙 = −𝒄𝒐𝒕 𝒙 + 𝑪 𝒅 𝒅𝒙 (𝒔𝒆𝒄 𝒙) = 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 ∫ 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝒔𝒆𝒄 𝒙 + 𝑪 𝒅 𝒅𝒙 (𝒄𝒔𝒄 𝒙) = −𝒄𝒔𝒄 𝒙 𝒄𝒐𝒕 𝒙 ∫ 𝒄𝒔𝒄 𝒙 𝒄𝒐𝒕 𝒙 𝒅𝒙 = −𝒄𝒔𝒄 𝒙 + 𝑪 EXAMPLE 4: Find the ∫(4 cos 𝑥 − 3 sin𝑥) 𝑑𝑥 SOLUTION: ∫(𝟒 𝐜𝐨𝐬 𝒙 − 𝟑 𝐬𝐢𝐧 𝒙) 𝒅𝒙 = 𝟒 ∫ 𝐜𝐨𝐬 𝒙 𝒅𝒙 − 𝟑 ∫ 𝐬𝐢𝐧 𝒙 𝒅𝒙 = 𝟒𝒔𝒊𝒏 𝒙 − 𝟑(−𝒄𝒐𝒔 𝒙) + 𝑪 = 𝟒𝒔𝒊𝒏 𝒙 + 𝟑𝒄𝒐𝒔 𝒙 + 𝑪 EXAMPLE 5: Find the ∫(𝑠𝑒𝑐2 𝑥 + 𝑐𝑠𝑐2 𝑥) 𝑑𝑥 SOLUTION: ∫(𝒔𝒆𝒄𝟐 𝒙 + 𝒄𝒔𝒄𝟐 𝒙) 𝒅𝒙 = 𝒕𝒂𝒏 𝒙 − 𝒄𝒐𝒕 𝒙 + 𝑪 EXPONENTIAL & LOGARITHMIC FUNCTIONS INTEGRATION FORMULAS Differentiation Formulas Integration Formulas 𝒅 𝒅𝒙 (𝒆𝒙 ) = 𝒆𝒙 ∫ 𝒆𝒙 𝒅𝒙 = 𝒆𝒙 + 𝑪
- 4. 𝒅 𝒅𝒙 (𝒂𝒙) = 𝒂𝒙 𝒍𝒏𝒂, 𝒂 > 𝟎 ∫ 𝒂𝒙 𝒅𝒙 = 𝒂𝒙 𝒍𝒏 𝒂 + 𝑪, 𝒂 > 𝟎 𝒅 𝒅𝒙 (𝒍𝒏 𝒙) = 𝟏 𝒙 ∫ 𝒅𝒙 𝒙 = 𝒍𝒏 |𝒙| + 𝑪 EXAMPLE 6: Find the ∫(2𝑥 − 3𝑥 ) 𝑑𝑥 SOLUTION: ∫(𝟐𝒙 − 𝟑𝒙) = 𝟐𝒙 𝒍𝒏 𝟐 − 𝟑𝒙 𝒍𝒏 𝟑 + 𝑪 EXAMPLE 7: Find the ∫( 2 𝑥 − 3𝑒3 ) 𝑑𝑥 SOLUTION: ∫ ( 𝟐 𝒙 − 𝟑𝒆𝒙 ) 𝒅𝒙 = 𝟐 ∫ 𝒅𝒙 𝒙 − 𝟑 ∫ 𝒆𝟑 = 𝟐 𝒍𝒏 |𝒙| − 𝟑𝒆𝟑