![A central topic in integral calculus is the definite integral. In this lesson, you will discover some of its
applications that will enable you to solve various problems.
∫ 𝒇(𝒙) 𝒅𝒙
𝒃
𝒂
FUNDAMENTAL THEOREM OF CALCULUS
If 𝑓 is a continuous on an interval [𝑎, 𝑏] then 𝑔(𝑥) = ∫ 𝑓(𝑡) 𝑑𝑡
𝑥
𝑎
, where 𝑎 ≤ 𝑥 ≤ 𝑏 is continuous on an
interval [𝑎, 𝑏] and differentiate on (𝑎, 𝑏) and
𝑑
𝑑𝑥
∫ 𝑓(𝑡) 𝑑𝑡 = 𝑓(𝑥)
𝑥
𝑎
.
EXAMPLE 1: Find the derivative of 𝒈(𝒙) = ∫
𝟏
𝟏+𝒕𝟐 𝒅𝒕
𝒙
𝟎
SOLUTION:
𝒈(𝒙) = ∫
𝟏
𝟏 + 𝒕𝟐
𝒅𝒕
𝒙
𝟎
𝒈′(𝒙) =
𝒅
𝒅𝒙
∫
𝟏
𝟏 + 𝒙𝟐
∙ 𝟏
𝒙
𝟎
𝒈′(𝒙) =
𝟏
𝟏 + 𝒙𝟐
EXAMPLE 2: Find the derivative of 𝒈(𝒙) = ∫ 𝒄𝒐𝒔 𝒕 𝒅𝒕
𝒙𝟐
𝟎
SOLUTION:
𝒈(𝒙) = ∫ 𝒄𝒐𝒔 𝒕 𝒅𝒕
𝒙𝟐
𝟎
𝒈′(𝒙) =
𝒅
𝒅𝒙
∫ 𝒄𝒐𝒔 𝒙𝟐
∙ 𝟐𝒙
𝒙𝟐
𝟎
𝒈′(𝒙)
= 𝟐𝒙 𝒄𝒐𝒔 𝒙𝟐](https://image.slidesharecdn.com/moudule9thefundamentaltheoremofcalculus-210328102708/85/Moudule9-the-fundamental-theorem-of-calculus-1-320.jpg)
![DEFINITE INTEGRAL is defined to be exactly the limit and summation
of the net area between a function and the x-axis.
∫ 𝒇(𝒙) 𝒅𝒙
𝒃
𝒂
EXAMPLE 3: Find the area under the curve of the function 𝑓(𝑥) = √𝑥 within an interval of 0 ≤ 𝑥 ≤ 1
SOLUTION:
= ∫ √𝒙 𝒅𝒙
𝟏
𝟎
= ∫ 𝒙
𝟏
𝟐 𝒅𝒙
𝟏
𝟎
= ∫
𝒙
𝟏
𝟐
+𝟏
𝟏
𝟐 + 𝟏
𝟏
𝟎
=
𝒙
𝟑
𝟐
𝟑
𝟐
𝒐𝒓
𝟐
𝟑
𝒙
𝟑
𝟐
=
𝟐
𝟑
(𝟏)
𝟑
𝟐 −
𝟐
𝟑
(𝟎)
𝟑
𝟐
=
𝟐
𝟑
𝒖𝒏𝒊𝒕𝟐
EXAMPLE 3: Find the area under the curve of the function 𝑓(𝑥) = (𝑥 + 2)within an interval of [0,3]
SOLUTION:
= ∫ (𝒙 + 𝟐) 𝒅𝒙
𝟑
𝟎
= ∫ 𝒙 𝒅𝒙 +
𝟑
𝟎
∫ 𝟐 𝒅𝒙
𝟑
𝟎
= ∫
𝒙𝟏+𝟏
𝟏 + 𝟏
+ 𝟐𝒙 𝒅𝒙
𝟑
𝟎
= ∫
𝒙𝟐
𝟐
+ 𝟐𝒙
𝟑
𝟎](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Moudule9 the fundamental theorem of calculus
- 1. A central topic in integral calculus is the definite integral. In this lesson, you will discover some of its applications that will enable you to solve various problems. ∫ 𝒇(𝒙) 𝒅𝒙 𝒃 𝒂 FUNDAMENTAL THEOREM OF CALCULUS If 𝑓 is a continuous on an interval [𝑎, 𝑏] then 𝑔(𝑥) = ∫ 𝑓(𝑡) 𝑑𝑡 𝑥 𝑎 , where 𝑎 ≤ 𝑥 ≤ 𝑏 is continuous on an interval [𝑎, 𝑏] and differentiate on (𝑎, 𝑏) and 𝑑 𝑑𝑥 ∫ 𝑓(𝑡) 𝑑𝑡 = 𝑓(𝑥) 𝑥 𝑎 . EXAMPLE 1: Find the derivative of 𝒈(𝒙) = ∫ 𝟏 𝟏+𝒕𝟐 𝒅𝒕 𝒙 𝟎 SOLUTION: 𝒈(𝒙) = ∫ 𝟏 𝟏 + 𝒕𝟐 𝒅𝒕 𝒙 𝟎 𝒈′(𝒙) = 𝒅 𝒅𝒙 ∫ 𝟏 𝟏 + 𝒙𝟐 ∙ 𝟏 𝒙 𝟎 𝒈′(𝒙) = 𝟏 𝟏 + 𝒙𝟐 EXAMPLE 2: Find the derivative of 𝒈(𝒙) = ∫ 𝒄𝒐𝒔 𝒕 𝒅𝒕 𝒙𝟐 𝟎 SOLUTION: 𝒈(𝒙) = ∫ 𝒄𝒐𝒔 𝒕 𝒅𝒕 𝒙𝟐 𝟎 𝒈′(𝒙) = 𝒅 𝒅𝒙 ∫ 𝒄𝒐𝒔 𝒙𝟐 ∙ 𝟐𝒙 𝒙𝟐 𝟎 𝒈′(𝒙) = 𝟐𝒙 𝒄𝒐𝒔 𝒙𝟐
- 2. DEFINITE INTEGRAL is defined to be exactly the limit and summation of the net area between a function and the x-axis. ∫ 𝒇(𝒙) 𝒅𝒙 𝒃 𝒂 EXAMPLE 3: Find the area under the curve of the function 𝑓(𝑥) = √𝑥 within an interval of 0 ≤ 𝑥 ≤ 1 SOLUTION: = ∫ √𝒙 𝒅𝒙 𝟏 𝟎 = ∫ 𝒙 𝟏 𝟐 𝒅𝒙 𝟏 𝟎 = ∫ 𝒙 𝟏 𝟐 +𝟏 𝟏 𝟐 + 𝟏 𝟏 𝟎 = 𝒙 𝟑 𝟐 𝟑 𝟐 𝒐𝒓 𝟐 𝟑 𝒙 𝟑 𝟐 = 𝟐 𝟑 (𝟏) 𝟑 𝟐 − 𝟐 𝟑 (𝟎) 𝟑 𝟐 = 𝟐 𝟑 𝒖𝒏𝒊𝒕𝟐 EXAMPLE 3: Find the area under the curve of the function 𝑓(𝑥) = (𝑥 + 2)within an interval of [0,3] SOLUTION: = ∫ (𝒙 + 𝟐) 𝒅𝒙 𝟑 𝟎 = ∫ 𝒙 𝒅𝒙 + 𝟑 𝟎 ∫ 𝟐 𝒅𝒙 𝟑 𝟎 = ∫ 𝒙𝟏+𝟏 𝟏 + 𝟏 + 𝟐𝒙 𝒅𝒙 𝟑 𝟎 = ∫ 𝒙𝟐 𝟐 + 𝟐𝒙 𝟑 𝟎
- 3. = [ (𝟑)𝟐 𝟐 + 𝟐(𝟑)] − [ (𝟎)𝟐 𝟐 + 𝟐(𝟎)] = ( 𝟗 𝟐 + 𝟔) − 𝟎 = 𝟐𝟏 𝟐 𝒖𝒏𝒊𝒕 𝟐