If a < b < c, then for any number b between a and c, the integral from a to c is the integral from a to b plus the integral from b to c.

Theorem:

∫𝑎

𝑐

𝑓 (𝑥 )𝑑𝑥=¿¿∫𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥+¿¿∫𝑏

𝑐

𝑓 (𝑥 )𝑑𝑥

Section 4.4 – Properties of Definite Integrals

Section 4.4 – Properties of Definite Integrals

𝑓 (𝑥 )={− 𝑥+6 𝑓𝑜𝑟 𝑥 ≤ 2𝑥2 𝑓𝑜𝑟 𝑥>2

Example:

Calculate the area under the given curve between and

∫−3

3

𝑓 (𝑥 ) 𝑑𝑥=¿¿∫−3

2

(−𝑥+6 )𝑑𝑥+¿¿∫2

3

𝑥2𝑑𝑥

−𝑥2

2+6 𝑥¿

𝑥3

3 |32∫−3

3

𝑓 (𝑥 ) 𝑑𝑥=¿¿

¿32.5+¿6.3333

∫−3

3

𝑓 (𝑥 ) 𝑑𝑥=38.8333

Section 4.4 – Properties of Definite Integrals

𝑓 (𝑥 )={− 2𝑥+4 𝑓𝑜𝑟 𝑥≤ 22 𝑥− 4 𝑓𝑜𝑟 𝑥>2

Example:

Calculate the area under the given curve between and

∫−1

6

𝑓 (𝑥 )𝑑𝑥=¿¿

∫−1

2

(−2 𝑥+4 )𝑑𝑥+¿¿∫2

6

(2𝑥− 4 )𝑑𝑥

−2𝑥2

2+4 𝑥 ¿ 2𝑥2

2− 4 𝑥|62

¿9+¿16 ∫−1

6

𝑓 (𝑥 )𝑑𝑥=25

𝑓 (𝑥 )=|2 𝑥− 4| 2 𝑥− 4=0𝑥=2

−𝑥2+4 𝑥 ¿𝑥2− 4 𝑥|62 →

Section 4.4 – Properties of Definite Integrals

Copyright 2010 Pearson Education, Inc.

As the number of rectangles increased, the approximation of the area under the curve approaches a value.

If a continuous function, f(x), has an antiderivative, F(x), on the interval [a, b], then

𝑨𝒓𝒆𝒂 𝑩𝒆𝒕𝒘𝒆𝒆𝒏𝑪𝒖𝒓𝒗𝒆𝒔

Copyright 2010 Pearson Education, Inc.

𝑨𝒓𝒆𝒂 𝑩𝒆𝒕𝒘𝒆𝒆𝒏𝑪𝒖𝒓𝒗𝒆𝒔

h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒= 𝑓 (𝑥 )−𝑔 (𝑥)

h𝑤𝑖𝑑𝑡 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒=𝑑𝑥h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒=𝑢𝑝𝑝𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛− 𝑙𝑜𝑤𝑒𝑟 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

If a continuous function, f(x), has an antiderivative, F(x), on the interval [a, b], then

Section 4.4 – Properties of Definite Integrals

Example:Calculate the area bounded by the graphs of and

𝐴𝑟𝑒𝑎=∫𝑎

𝑏

[ 𝑓 (𝑥 ) −𝑔 (𝑥)]𝑑𝑥

0.8333 −(−1.8333)

Section 4.4 – Properties of Definite Integrals

𝐴𝑟𝑒𝑎=∫−1

1

[ (𝑥2+1 ) − (𝑥 ) ] 𝑑𝑥

𝑥3

3+𝑥−

𝑥2

2 | 1−1

2.6667

Example:Calculate the area bounded by the graphs of

𝐴𝑟𝑒𝑎=∫𝑎

𝑏

[ 𝑓 (𝑥 ) −𝑔 (𝑥)]𝑑𝑥

10.6667− 0

Section 4.4 – Properties of Definite Integrals

4 𝑥2

2−𝑥3

3 |40

10.6667

Find the points of intersection

𝑓 (𝑥 )=𝑔 (𝑥 )𝑥2=4 𝑥

𝑥2− 4 𝑥=0𝑥 (𝑥− 4)=0𝑥=0 , 4

𝐴𝑟𝑒𝑎=∫0

4

[ ( 4 𝑥 )− (𝑥2) ]𝑑𝑥

2 𝑥2 −𝑥3

3 |40

Average Value of a Continuous Function

Copyright 2010 Pearson Education, Inc.

𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑉𝑎𝑙𝑢𝑒= 1𝑏−𝑎∫𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥

Section 4.4 – Properties of Definite Integrals

Average Value of a Continuous FunctionFind the average value of the function over the interval

𝐴𝑉= 13−1

∫1

3

(𝑥2+2   )𝑑𝑥

𝐴𝑉=12 ( 𝑥

3

3+2𝑥)|31

𝐴𝑉=12 [( 27

3+6)−( 1

3+2)]

𝐴𝑉=12 [15 −

73 ]

𝐴𝑉=193

=6.3333

Section 4.4 – Properties of Definite Integrals

6.3333

Section 4.4 – Properties of Definite IntegralsA company’s marginal revenue and marginal cost functions are as follows:

a) Find the total profit from the first 10 days.

b) Find the average daily profit from the first 10 days.

Reminder:

𝑷𝒓𝒐𝒇𝒊𝒕=𝑹 (𝒕 )−𝑪 (𝒕)

𝑻𝒐𝒕𝒂𝒍 𝑨𝒄𝒄𝒖𝒎𝒖𝒍𝒂𝒕𝒆𝒅 𝑷𝒓𝒐𝒇𝒊𝒕=∫𝒂

𝒃

(𝑹 ′ (𝒕 )−𝑪 ′(𝒕 )   )

a) 𝑻𝒐𝒕𝒂𝒍 𝑨𝒄𝒄𝒖𝒎𝒖𝒍𝒂𝒕𝒆𝒅 𝑷𝒓𝒐𝒇𝒊𝒕=∫𝟎

𝟏𝟎

(𝟕𝟓𝒆𝒕−𝟐𝒕−(𝟕𝟓−𝟑𝒕)   )𝒅𝒕

¿∫𝟎

𝟏𝟎

(𝟕𝟓𝒆𝒕+𝒕−75   )𝒅𝒕¿𝟕𝟓𝒆𝒕+ 𝒕𝟐

𝟐+𝟕𝟓 𝒕|𝟏𝟎𝟐 ¿ $𝟏 ,𝟔𝟓𝟏 ,𝟐𝟎𝟗 .𝟗𝟒

Section 4.4 – Properties of Definite IntegralsA company’s marginal revenue and marginal cost functions are as follows:

a) Find the total profit from the first 10 days.

b) Find the average daily profit from the first 10 days.

Reminder:

b) 𝑨𝒗𝒆𝒓𝒂𝒈𝒆𝑫𝒂𝒊𝒍𝒚 𝑷𝒓𝒐𝒇𝒊𝒕= 𝟏𝟏𝟎−𝟎∫

𝟎

𝟏𝟎

(𝟕𝟓𝒆𝒕 −𝟐𝒕−(𝟕𝟓−𝟑𝒕)   )𝒅𝒕

¿ 𝟏𝟏𝟎∫

𝟎

𝟏𝟎

(𝟕𝟓𝒆𝒕+𝒕− 75   )𝒅𝒕¿ 𝟏𝟏𝟎 (𝟕𝟓𝒆𝒕+

𝒕𝟐

𝟐+𝟕𝟓 𝒕)|𝟏𝟎𝟐 ¿ $𝟏𝟔𝟓 ,𝟏𝟐𝟎 .𝟗𝟗

𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑉𝑎𝑙𝑢𝑒= 1𝑏−𝑎∫𝑎

𝑏

𝑓 (𝑥 )𝑑𝑥

𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝐷𝑎𝑖𝑙𝑦 𝑃𝑟𝑜𝑓𝑖𝑡= 1𝑏−𝑎∫𝑎

𝑏

𝑅 ′ (𝑡 ) −𝐶 ′ (𝑡)

Section 4.4 – Properties of Definite Integrals

Differentiation Review:

Copyright 2010 Pearson Education, Inc.

𝒚=(𝟓 𝒙+𝟔 )𝟒

𝒅𝒚=𝟒 (𝟓𝒙+𝟔 )𝟑 (𝟓 )𝒅𝒙

Integration:

: 𝒖=𝒈(𝒙)𝒅𝒖=𝒈 ′ (𝒙)𝒅𝒙

∫𝒅𝒚=∫𝟐𝟎 (𝟓 𝒙+𝟔 )𝟑𝒅𝒙

Section 4.5 – Integration Techniques: Substitution

Copyright 2010 Pearson Education, Inc.

Integration:

: 𝒖=𝟐 𝒙𝟐+𝟑𝒅𝒖=𝟒 𝒙 𝒅𝒙

∫𝒅𝒚=∫𝟒 𝒙 (𝟐 𝒙𝟐+𝟑)𝟐𝒅𝒙

∫𝒅𝒚=∫ (𝟐 𝒙𝟐+𝟑)𝟐𝟒 𝒙𝒅𝒙 ∫𝒅𝒚=∫𝒖𝟐𝒅𝒖𝒚+𝒄=𝒖𝟑

𝟑+𝒄

𝒚=𝟏𝟑𝒖𝟑

+𝑪

𝒚=𝟏𝟑

(𝟐 𝒙𝟐+𝟑)𝟑

+𝑪

Section 4.5 – Integration Techniques: Substitution

Copyright 2010 Pearson Education, Inc.

Integrate:

: 𝒖=𝟓 𝒙+𝟔𝒅𝒖=𝟓𝒅𝒙

∫𝒅𝒚=∫𝟐𝟎 (𝟓 𝒙+𝟔 )𝟑𝒅𝒙

∫𝒅𝒚=𝟐𝟎∫ (𝟓 𝒙+𝟔 )𝟑𝒅𝒙∫𝒅𝒚=𝟐𝟎∫ 𝟏

𝟓∙𝟓 (𝟓 𝒙+𝟔   )𝟑𝒅𝒙 𝒚+𝒄=𝟒𝒖

𝟒

𝟒+𝒄

𝒚=𝒖𝟒+𝑪

𝒚=(𝟓 𝒙+𝟔)𝟒+𝑪∫𝒅𝒚=𝟐𝟎 ∙𝟏𝟓∫𝟓 (𝟓 𝒙+𝟔   )𝟑𝒅𝒙

∫𝒅𝒚=𝟒∫ (𝟓 𝒙+𝟔   )𝟑𝟓𝒅𝒙

∫𝒅𝒚=𝟒∫𝒖𝟑𝒅𝒖

Section 4.5 – Integration Techniques: Substitution

Integrate:

: 𝒖=𝟏+𝒆𝒙

𝒅𝒖=𝒆 𝒙𝒅𝒙

∫𝒅𝒚=∫ 𝒆𝒙

𝟏+𝒆𝒙 𝒅𝒙

𝒚=𝒍𝒏𝒖+𝑪

∫𝒅𝒚=∫ 𝟏𝟏+𝒆𝒙 𝒆

𝒙𝒅𝒙

∫𝒅𝒚=∫ 𝟏𝒖𝒅𝒖

𝒚=𝒍𝒏 (𝟏+𝒆 𝒙 )+𝑪

Section 4.5 – Integration Techniques: Substitution

Copyright 2010 Pearson Education, Inc.

Integrate:

: 𝒖=𝟑+𝟐𝒙𝟑

𝒅𝒖=𝟔 𝒙𝟐𝒅𝒙

∫𝒅𝒚=∫ 𝟔 𝒙𝟐

√𝟑+𝟐 𝒙𝟑𝒅𝒙

𝒚=𝒖

𝟏𝟐

𝟏𝟐

+𝒄 𝒚=𝟐𝒖𝟏𝟐+𝑪

∫𝒅𝒚=∫𝒖−𝟏𝟐 𝒅𝒖

∫𝒅𝒚=∫𝟔 𝒙𝟐 (𝟑+𝟐 𝒙𝟑 )−𝟏𝟐 𝒅𝒙

∫𝒅𝒚=∫ (𝟑+𝟐𝒙𝟑 )−𝟏𝟐 𝟔 𝒙𝟐𝒅𝒙

𝒚=𝟐 (𝟑+𝟐 𝒙𝟑)𝟏𝟐+𝑪→ →

Section 4.5 – Integration Techniques: Substitution

Integrate:

: 𝒖=𝒍𝒏𝒙𝒅𝒖=

𝟏𝒙𝒅𝒙

∫𝒅𝒚=∫ (𝒍𝒏𝒙 )𝟐

𝒙𝒅𝒙

∫𝒅𝒚=∫𝒖𝟐𝒅𝒖∫𝒅𝒚=∫ (𝒍𝒏 𝒙 )𝟐 𝟏

𝒙𝒅𝒙

𝒚=𝒖𝟑

𝟑+𝒄→𝒚=

𝟏𝟑

(𝒍𝒏𝒙 )𝟑+𝒄

Section 4.5 – Integration Techniques: Substitution

Copyright 2010 Pearson Education, Inc.

Integrate:

: 𝒖=𝟒 𝒙𝟑

𝒅𝒖=𝟏𝟐𝒙𝟐𝒅𝒙

∫𝒅𝒚=∫𝒙𝟐𝒆𝟒𝒙 𝟑

𝒅𝒙

∫𝒅𝒚=∫ 𝟏𝟏𝟐

∙𝟏𝟐𝒙𝟐𝒆𝟒𝒙 𝟑

𝒅𝒙

𝒚= 𝟏𝟏𝟐

𝒆𝒖

+𝑪

∫𝒅𝒚=𝟏𝟏𝟐∫𝟏𝟐𝒙𝟐𝒆𝟒 𝒙𝟑

𝒅𝒙

∫𝒅𝒚=𝟏𝟏𝟐∫𝒆𝒖𝒅𝒖

𝒚= 𝟏𝟏𝟐

𝒆𝟒𝒙 𝟑

+𝑪→

Section 4.5 – Integration Techniques: Substitution

Integrate:

: 𝒖=𝒙 −𝟏𝒅𝒖=𝒅𝒙

∫𝒅𝒚=∫ 𝒙(𝒙−𝟏 )𝟑

𝒅𝒙

𝒚=𝒖−𝟏

−𝟏+𝒖

−𝟐

−𝟐+𝑪

∫𝒅𝒚=∫ 𝒙𝒖𝟑 𝒅𝒖

∫𝒅𝒚=∫ 𝒖𝒖𝟑+

𝟏𝒖𝟑 𝒅𝒖

𝒖+𝟏=𝒙

∫𝒅𝒚=∫ 𝒖+𝟏𝒖𝟑 𝒅𝒖

∫𝒅𝒚=∫ (𝒖−𝟐+𝒖−𝟑 )𝒅𝒖

𝒚=− (𝒙−𝟏 )−𝟏−𝟏𝟐

(𝒙−𝟏 )−𝟐+𝑪

Section 4.5 – Integration Techniques: Substitution

Section 4.4 – Properties of Definite Integrals