Definit integration

1. 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. The Definite Integral
Definition
Note: The function f(x) must be continuous on the interval [a, b].

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

7. The Definite Integral
Rules of the Definite Integral
𝑎
𝑏
𝑐 𝑑𝑥 = 𝑐(𝑏 − 𝑎)
𝑎
𝑏
𝑥 𝑑𝑥 =
𝑏2
2
−
𝑎2
2
𝑎
𝑏
𝑥2
𝑑𝑥 =
𝑏3
3
−
𝑎3
3
Examples
2
6
4 𝑑𝑥 = 4 6 − 2 = 16
4
8
𝑥 𝑑𝑥 =
82
2
−
42
2
= 32 − 8 = 24
3
5
𝑥2 𝑑𝑥 =
53
3
−
33
3
=
125
3
−
27
3
=
98
3
= 32.67
3
4
𝑥2
+ 3𝑥 − 2 𝑑𝑥 =
3
4
𝑥2
𝑑𝑥 + 3
3
4
𝑥 𝑑𝑥 −
3
4
2 𝑑𝑥 =
43
3
−
33
3
+ 3
42
2
−
32
2
− 2 4 − 3 =
64
3
−
27
3
+ 3
16
2
−
9
2
− 2 1 = 20.83

8. The Fundamental Theorem of Calculus
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 =
5𝑥2
2
=
5
1
5(5)2
2
−
5 1 2
2
=
125
2
−
5
2
=
120
2
= 60
𝜋
6
2𝜋
3
𝑠𝑖𝑛𝑥 𝑑𝑥 = −𝑐𝑜𝑠𝑥 =
5
1
−𝑐𝑜𝑠
2𝜋
3
− −𝑐𝑜𝑠
𝜋
6
= − −
1
2
− −
3
2
= 0.866

9. The Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus
𝑎
𝑏
𝑓 𝑥 𝑑𝑥 = 𝐹 𝑏 − 𝐹(𝑎) .
3
4
𝑥2 + 3𝑥 − 2 𝑑𝑥 =
Examples
𝑥3
3
+
3𝑥2
2
− 2𝑥 =
4
3
43
3
+
3(4)2
2
− 2(4) −
33
3
+
3(3)2
2
− 2(3)
37.33 − 16.5 = 20.83
1
32
1
𝑥
6
5
𝑑𝑥 =
𝑥
−1
5
−
1
5
=
32
1
−5
𝑥
1
5
= −
5
2
− −
5
1
= 2.5
1
32
𝑥
−6
5 𝑑𝑥 =
32
1

10. 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
𝑡2 𝑑𝑡 = 𝑥2
𝑑
𝑑𝑥
𝑡3
3
=
𝑥
1
𝑑
𝑑𝑥
𝑥3
3
−
13
3
=
𝑑
𝑑𝑥
𝑥3
3
−
1
3
=
𝑑
𝑑𝑥
1
4𝑥
𝑡2 𝑑𝑡 = 64𝑥2
𝑑
𝑑𝑥
𝑡3
3
=
4𝑥
1
𝑑
𝑑𝑥
(4𝑥)3
3
−
13
3
=
𝑑
𝑑𝑥
64𝑥3
3
−
1
3
=
𝑑
𝑑𝑥
1
𝑥2
𝑡2 𝑑𝑡 =
6𝑥5
3
=
𝑑
𝑑𝑥
𝑡3
3
=
𝑥2
1
𝑑
𝑑𝑥
(𝑥2
)3
3
−
13
3
=
𝑑
𝑑𝑥
𝑥6
3
−
1
3
= 2𝑥5

11. 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
𝑡2 𝑑𝑡 = 𝑥 2 1 =
𝑑
𝑑𝑥
1
4𝑥
𝑡2 𝑑𝑡 = 64𝑥2
𝑑
𝑑𝑥
1
𝑥2
𝑡2 𝑑𝑡 = 𝑥2 2
2𝑥 =
𝑥2
4𝑥 2
4 =
2𝑥5
𝑑
𝑑𝑥
1
𝑥2
sin 𝑡 𝑑𝑡 = 2𝑥 𝑠𝑖𝑛 𝑥2
𝑑
𝑑𝑥
1
𝑥2+𝑥
tan 𝑡 𝑑𝑡 = 2𝑥 + 1 𝑡𝑎𝑛 𝑥2 + 2
𝑑
𝑑𝑥
𝑡𝑎𝑛𝑥
𝑠𝑖𝑛𝑥
𝑡2 𝑑𝑡 = (𝑠𝑖𝑛2𝑥) 𝑐𝑜𝑠𝑥 −(𝑡𝑎𝑛2𝑥) 𝑠𝑒𝑐2𝑥

12. The Fundamental Theorem of Calculus
Mean Value Theorem for Integrals
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13. The Definite Integral
Mean Value for Definite Integral
Find the mean (average) value of the function 𝑓 𝑥 = 𝑥2 + 2 over the interval 1, 3
𝐴𝑉 =
1
3 − 1
1
3
𝑥2 + 2 𝑑𝑥
𝐴𝑉 =
1
2
𝑥3
3
+ 2𝑥
𝐴𝑉 =
1
2
27
3
+ 6 −
1
3
+ 2
𝐴𝑉 =
1
2
15 −
7
3
𝐴𝑉 =
19
3
= 6.33
3
1

14. 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
Copyright  2010 Pearson Education, Inc.
Value of the Definite Integral
0
2𝜋
𝑠𝑖𝑛𝑥 𝑑𝑥 = −𝑐𝑜𝑠𝑥 =
2𝜋
0
−𝑐𝑜𝑠 2𝜋 − −𝑐𝑜𝑠 0 =
− 1 − −1 = 0
Total Area
0
𝜋
𝑠𝑖𝑛𝑥 𝑑𝑥 −
𝜋
2𝜋
𝑠𝑖𝑛𝑥 𝑑𝑥 =
−𝑐𝑜𝑠𝑥 −
𝜋
0
−𝑐𝑜𝑠 𝜋 − −𝑐𝑜𝑠 0 −
− −1 − −1 −
4
0
2𝜋
𝑠𝑖𝑛𝑥 𝑑𝑥 =
−𝑐𝑜𝑠 2𝜋 − −𝑐𝑜𝑠 𝜋
−𝑐𝑜𝑠𝑥
2𝜋
𝜋
− 1 − 1 =