1. Section 5.3 – The Definite Integral
As the number of rectangles increased, the approximation of the area under the curve approaches a value.
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2. Section 5.3 – The Definite Integral
Definition
Note: The function f(x) must be continuous on the interval [a, b].

3. Section 5.3 – The Definite Integral
Parts of the Definite Integral
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4. Section 5.3 – The Definite Integral
Properties of the Definite Integral
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5. Section 5.3 – The Definite Integral
Geometric Interpretations of the Properties of the Definite Integral
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6. Section 5.3 – The Definite Integral
Using the Properties of the Definite Integral
1 3 𝑓 𝑥 𝑑𝑥 = 𝑓 𝑥 𝑑𝑥 = 𝑔 𝑥 𝑑𝑥 =−4
Given:
1 3 3𝑓 𝑥 𝑑𝑥 =
3 1 3 𝑓 𝑥 𝑑𝑥 =
3 6 =18
1 3 2𝑓 𝑥 −4𝑔 𝑥 𝑑𝑥 =
2 1 3 𝑓 𝑥 𝑑𝑥 −4 1 3 𝑔 𝑥 𝑑𝑥 =
2 6 −4 −4 =28
1 7 𝑓 𝑥 𝑑𝑥 =
1 3 𝑓 𝑥 𝑑𝑥 𝑓 𝑥 𝑑𝑥 =
6+9=15
3 1 𝑓 𝑥 𝑑𝑥 =
− 1 3 𝑓 𝑥 𝑑𝑥 = −6

7. Section 5.3 – The Definite Integral
Rules of the Definite Integral
𝑎 𝑏 𝑐 𝑑𝑥=𝑐(𝑏−𝑎)
𝑎 𝑏 𝑥 𝑑𝑥= 𝑏 2 2 − 𝑎 2 2
𝑎 𝑏 𝑥 2 𝑑𝑥= 𝑏 3 3 − 𝑎 3 3
Examples
2 6 4 𝑑𝑥=
4 8 𝑥 𝑑𝑥=
− =
4 6−2 = 16
32−8= 24
3 5 𝑥 2 𝑑𝑥 =
− =
125 3 − 27 3 =
98 3 =32.67
3 4 𝑥 2 +3𝑥−2 𝑑𝑥 =
3 4 𝑥 2 𝑑𝑥 𝑥 𝑑𝑥 − 𝑑𝑥 = −
− −
2 4−3 =
− 9 2 −
64 3 −
2 1 =
20.83

8. Section 5.4 – The Fundamental Theorem of Calculus
𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑎𝑡 𝑒𝑣𝑒𝑟𝑦 𝑝𝑜𝑖𝑛𝑡 𝑖𝑛 𝑎, 𝑏 𝑎𝑛𝑑 𝐹 𝑖𝑠 𝑎𝑛𝑦
𝑎𝑛𝑡𝑖𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑓 𝑜𝑛 𝑎, 𝑏 , 𝑡ℎ𝑒𝑛
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
Examples
1 5 5𝑥 𝑑𝑥 =
5 𝑥 = 5 1
5(5) 2 2
− =
125 2 − 5 2 =
120 2 =60
𝜋 6 2𝜋 3 𝑠𝑖𝑛𝑥 𝑑𝑥 = 5 1
−𝑐𝑜𝑠 2𝜋 3
− −𝑐𝑜𝑠 𝜋 6 =
− − 1 2 − − =
−𝑐𝑜𝑠𝑥 =
0.866

9. Section 5.4 – The Fundamental Theorem of Calculus
𝑎 𝑏 𝑓 𝑥 𝑑𝑥=𝐹 𝑏 −𝐹(𝑎) .
Examples
3 4 𝑥 2 +3𝑥−2 𝑑𝑥 =
𝑥 𝑥 2 2 −2𝑥 = 4 3
(4) 2 2 −2(4)
− (3) 2 2 −2(3)
37.33−16.5=
20.83
𝑥 −1 5 − = 32 1
𝑥 𝑑𝑥 = 32
1 32 𝑥 −6 5 𝑑𝑥 =
−5 𝑥 =
− 5 2 − − 5 1 =
2.5 1

10. Section 5.4 – The Fundamental Theorem of Calculus
Differentiating a Definite Integral
𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑜𝑛 𝑎, 𝑏 𝑎𝑛𝑑 𝑥 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑎, 𝑏 , 𝑡ℎ𝑒𝑛
𝐹 ′ 𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓 𝑡 𝑑𝑡=𝑓(𝑥)
𝑑 𝑑𝑥 1 𝑥 𝑡 2 𝑑𝑡=
𝑑 𝑑𝑥 𝑡 = 𝑥 1
𝑑 𝑑𝑥 𝑥 3 3 − =
𝑑 𝑑𝑥 𝑥 3 3 − 1 3 =
𝑥 2
𝑑 𝑑𝑥 1 4𝑥 𝑡 2 𝑑𝑡=
𝑑 𝑑𝑥 𝑡 = 4𝑥 1
𝑑 𝑑𝑥 (4𝑥) 3 3 − =
𝑑 𝑑𝑥 64𝑥 3 3 − 1 3 =
64𝑥 2
𝑑 𝑑𝑥 1 𝑥 2 𝑡 2 𝑑𝑡=
𝑑 𝑑𝑥 𝑡 =
𝑥 2 1
𝑑 𝑑𝑥 ( 𝑥 2 ) 3 3 − =
𝑑 𝑑𝑥 𝑥 6 3 − 1 3 =
6 𝑥 5 3 =
2𝑥 5

11. Section 5.4 – The Fundamental Theorem of Calculus
Differentiating a Definite Integral
𝐼𝑓 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑜𝑛 𝑎, 𝑏 𝑎𝑛𝑑 𝑥 𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑎, 𝑏 , 𝑡ℎ𝑒𝑛
𝐹 ′ 𝑥 = 𝑑 𝑑𝑥 𝑎 𝑥 𝑓 𝑡 𝑑𝑡=𝑓(𝑥)
𝑑 𝑑𝑥 1 𝑥 𝑡 2 𝑑𝑡=
𝑑 𝑑𝑥 1 𝑥 2 sin 𝑡 𝑑𝑡=
𝑥 =
𝑥 2
2𝑥 𝑠𝑖𝑛 𝑥 2
𝑑 𝑑𝑥 1 4𝑥 𝑡 2 𝑑𝑡=
𝑑 𝑑𝑥 1 𝑥 2 +𝑥 tan 𝑡 𝑑𝑡=
4𝑥 =
64𝑥 2
2𝑥+1 𝑡𝑎𝑛 𝑥 2 +2
𝑑 𝑑𝑥 1 𝑥 2 𝑡 2 𝑑𝑡=
𝑑 𝑑𝑥 𝑡𝑎𝑛𝑥 𝑠𝑖𝑛𝑥 𝑡 2 𝑑𝑡=
𝑥 𝑥 =
2𝑥 5
(𝑠𝑖𝑛 2 𝑥) 𝑐𝑜𝑠𝑥
−(𝑡𝑎𝑛 2 𝑥) 𝑠𝑒𝑐 2 𝑥

12. Section 5.4 – The Fundamental Theorem of Calculus
Mean Value Theorem for Integrals
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13. Section 5.3 – The Definite Integral
Mean Value for Definite Integral
Find the mean (average) value of the function 𝑓 𝑥 = 𝑥 2 +2 over the interval 1, 3
𝐴𝑉= 1 3− 𝑥 𝑑𝑥 3
𝐴𝑉= 𝑥 𝑥 1
𝐴𝑉= −
𝐴𝑉= − 7 3
𝐴𝑉= 19 3 =6.33

14. Section 5.3 – The Definite Integral
Difference between the Value of a Definite Integral and Total Area
Find the mean (average) value of the function 𝑓 𝑥 =𝑠𝑖𝑛𝑥 over the interval 0, 2𝜋
Total Area
0 2𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 =
0 𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 − 𝜋 2𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 = 𝜋 2𝜋 𝜋
Value of the Definite Integral
−𝑐𝑜𝑠𝑥 −
−𝑐𝑜𝑠𝑥
0 2𝜋 𝑠𝑖𝑛𝑥 𝑑𝑥 = 2𝜋
−𝑐𝑜𝑠𝑥 =
−𝑐𝑜𝑠 𝜋
− −𝑐𝑜𝑠 0 −
−𝑐𝑜𝑠 2𝜋 − −𝑐𝑜𝑠 𝜋
−𝑐𝑜𝑠 2𝜋
− −𝑐𝑜𝑠 0 =
− −1 − −1 −
− 1 − 1 = 4
− 1 − −1 =
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