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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Section 5.3 – The Definite IntegralDefinition

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

Section 5.3 – The Definite IntegralParts of the Definite Integral

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Section 5.3 – The Definite IntegralProperties of the Definite Integral

Copyright 2010 Pearson Education, Inc.

Section 5.3 – The Definite IntegralGeometric Interpretations of the Properties of the Definite Integral

Copyright 2010 Pearson Education, Inc.

Section 5.3 – The Definite IntegralUsing 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

3 𝑓 (𝑥 )𝑑𝑥=¿3∫1

3

𝑓 (𝑥 ) 𝑑𝑥=¿3 (6 )=18

∫1

3

(2 𝑓 (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:

Section 5.3 – The Definite IntegralRules of the Definite Integral

∫𝑎

𝑏

𝑐𝑑𝑥=𝑐 (𝑏−𝑎) ∫𝑎

𝑏

𝑥 𝑑𝑥=𝑏2

2−𝑎2

2∫𝑎

𝑏

𝑥2𝑑𝑥=𝑏3

3−𝑎3

3

Examples

∫2

6

4𝑑𝑥=¿¿ 4 (6−2 )=¿16 ∫4

8

𝑥 𝑑𝑥=¿¿ 82

2−

42

2=¿32−8=¿24

∫3

5

𝑥2𝑑𝑥=¿53

3−

33

3=¿

1253−

273

=¿9 83

=32.67

∫3

4

𝑥2+3 𝑥−2𝑑𝑥=¿∫3

4

𝑥2𝑑𝑥+3∫3

4

𝑥 𝑑𝑥−∫3

4

2𝑑𝑥=¿43

3−

33

3+¿3 ( 42

2−

32

2 )−2 (4−3 )=¿

643−

273

+¿3 (162−

92 )−2 (1 )=¿20.83

Section 5.4 – The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus

∫𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥=𝐹 (𝑏)−𝐹 (𝑎) .

Examples

∫1

5

5 𝑥 𝑑𝑥=¿5𝑥2

2=¿

5

1

5(5)2

2−

5 (1 )2

2=¿

1252−

52=¿120

2=60

∫𝜋6

2𝜋3

𝑠𝑖𝑛𝑥 𝑑𝑥=¿−𝑐𝑜𝑠𝑥=¿5

1

−𝑐𝑜𝑠 ( 2𝜋3 )−(−𝑐𝑜𝑠( 𝜋6 ))=¿−(− 1

2 )−(− √32 )=¿0 .866

Section 5.4 – The Fundamental Theorem of CalculusThe Fundamental Theorem of Calculus

∫𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥=𝐹 (𝑏)−𝐹 (𝑎) .

∫3

4

𝑥2+3 𝑥−2𝑑𝑥=¿

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

321

𝑥65

𝑑𝑥=¿ 𝑥−15

−15

=¿

32

1

−5

𝑥15

=¿−52−(−5

1 )=¿2.5∫1

32

𝑥−65 𝑑𝑥=¿

32

1

Section 5.4 – The Fundamental Theorem of CalculusDifferentiating a Definite Integral

𝐼𝑓 𝑓 𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑜𝑛 [𝑎 ,𝑏 ]𝑎𝑛𝑑 𝑥𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑜𝑛 [𝑎 ,𝑏 ] , h𝑡 𝑒𝑛

𝐹 ′ (𝑥 )= 𝑑𝑑𝑥∫𝑎

𝑥

𝑓 (𝑡 ) 𝑑𝑡= 𝑓 (𝑥)

𝑑𝑑𝑥∫1

𝑥

𝑡 2𝑑𝑡=¿¿ 𝑥2𝑑𝑑𝑥 ( 𝑡

3

3 )=¿𝑥

1

𝑑𝑑𝑥 (𝑥

3

3−

13

3 )=¿𝑑𝑑𝑥 (𝑥

3

3−

13 )=¿

𝑑𝑑𝑥∫1

4𝑥

𝑡 2𝑑𝑡=¿¿ 64 𝑥2𝑑𝑑𝑥 ( 𝑡

3

3 )=¿4 𝑥

1

𝑑𝑑𝑥 ((4 𝑥)3

3−

13

3 )=¿𝑑𝑑𝑥 ( 64 𝑥3

3−

13 )=¿

𝑑𝑑𝑥∫1

𝑥2

𝑡 2𝑑𝑡=¿¿ 6 𝑥5

3=¿𝑑

𝑑𝑥 ( 𝑡3

3 )=¿𝑥2

1

𝑑𝑑𝑥 ((𝑥2)3

3−

13

3 )=¿𝑑𝑑𝑥 (𝑥

6

3−

13 )=¿ 2 𝑥5

Section 5.4 – The Fundamental Theorem of CalculusDifferentiating a Definite Integral

𝐼𝑓 𝑓 𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠𝑜𝑛 [𝑎 ,𝑏 ]𝑎𝑛𝑑 𝑥𝑖𝑠 𝑎𝑛𝑦 𝑝𝑜𝑖𝑛𝑡𝑜𝑛 [𝑎 ,𝑏 ] , h𝑡 𝑒𝑛

𝐹 ′ (𝑥 )= 𝑑𝑑𝑥∫𝑎

𝑥

𝑓 (𝑡 ) 𝑑𝑡= 𝑓 (𝑥)

𝑑𝑑𝑥∫1

𝑥

𝑡 2𝑑𝑡=¿¿(𝑥 )2 (1 )=¿

𝑑𝑑𝑥∫1

4𝑥

𝑡 2𝑑𝑡=¿¿ 64 𝑥2

𝑑𝑑𝑥∫1

𝑥2

𝑡 2𝑑𝑡=¿¿(𝑥2 )2 (2𝑥 )=¿

𝑥2

( 4 𝑥 )2 ( 4 )=¿

2 𝑥5

𝑑𝑑𝑥∫1

𝑥2

sin 𝑡 𝑑𝑡=¿¿ (2 𝑥 )𝑠𝑖𝑛 (𝑥2 )

𝑑𝑑𝑥 ∫

1

𝑥2+𝑥

tan 𝑡 𝑑𝑡=¿¿ (2 𝑥+1 ) 𝑡𝑎𝑛 (𝑥2+2 )

𝑑𝑑𝑥 ∫

𝑡𝑎𝑛𝑥

𝑠𝑖𝑛𝑥

𝑡 2𝑑𝑡=¿¿(𝑠𝑖𝑛¿¿2𝑥) (𝑐𝑜𝑠𝑥 )¿−(𝑡𝑎𝑛¿¿ 2𝑥) (𝑠𝑒𝑐2𝑥 )¿

Section 5.4 – The Fundamental Theorem of CalculusMean Value Theorem for Integrals

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Section 5.3 – The Definite IntegralMean Value for Definite Integral

Find the mean (average) value of the function over the interval

𝐴𝑉= 13−1

∫1

3

(𝑥2+2   )𝑑𝑥

𝐴𝑉=12 ( 𝑥

3

3+2𝑥)

𝐴𝑉=12 [( 27

3+6)−( 1

3+2)]

𝐴𝑉=12 [15−

73 ]

𝐴𝑉=193

=6.33

3

1

Section 5.3 – The Definite IntegralDifference between the Value of a Definite Integral and Total Area

Find the mean (average) value of the function over the interval

Copyright 2010 Pearson Education, Inc.

Value of the Definite Integral

∫0

2 𝜋

𝑠𝑖𝑛𝑥 𝑑𝑥=¿−𝑐𝑜𝑠𝑥=¿2𝜋

0

−𝑐𝑜𝑠 (2𝜋 )− (−𝑐𝑜𝑠 (0 ) )=¿

− (1 )− (−1 )=¿0

Total Area

∫0

𝜋

𝑠𝑖𝑛𝑥𝑑𝑥−∫𝜋

2 𝜋

𝑠𝑖𝑛𝑥𝑑𝑥=¿

−𝑐𝑜𝑠𝑥−𝜋

0

−𝑐𝑜𝑠 (𝜋 )− (−𝑐𝑜𝑠 (0 ) )−− (−1 )− (−1 )−

4

∫0

2 𝜋

𝑠𝑖𝑛𝑥 𝑑𝑥=¿

[(−𝑐𝑜𝑠 (2𝜋 )− (−𝑐𝑜𝑠 (𝜋 ) ) ) ]

−𝑐𝑜𝑠𝑥2𝜋

𝜋

(− (1 )− (1 ) )=¿