Module 7 the antiderivative

1. 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

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
𝒍𝒏 𝒂, 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