section 4.8 - antiderivatives

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Section 4.8 - Antiderivatives If the following functions represent the derivative of the original function, find the original function. ( ) = 3 ( ) =3 2 ( ) = 1 3 3 + 2 ( ) = 2 + 2 ( ) = 3 ( ) =3 2 1 ( ) = 2 + 4 ( ) =2 3 + 4 ¿ 2 3 + 4 Antiderivative – If F’(x) = f(x) on an interval, then F(x) is the antiderivative of f(x) for every value of x on the interval.

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Section 4.8 - Antiderivatives. If the following functions represent the derivative of the original function, find the original function. Antiderivative – If F’(x) = f(x) on an interval, then F(x) is the antiderivative of f(x) for every value of x on the interval. . - PowerPoint PPT Presentation

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Page 1: Section 4.8 - Antiderivatives

Section 4.8 - AntiderivativesIf the following functions represent the derivative of the original function, find the original function.

𝐹 (𝑥 )=𝑥3𝑓 (𝑥 )=3 𝑥2

𝐹 (𝑥 )=13 𝑥

3+𝑥2𝑓 (𝑥 )=𝑥2+2 𝑥

𝐹 (𝑥 )=𝑥3−𝑥𝑓 (𝑥 )=3 𝑥2 −1

𝐹 (𝑥 )=𝑥−2+4 𝑥𝑓 (𝑥 )=− 2𝑥3 +4¿− 2𝑥−3+4

Antiderivative – If F’(x) = f(x) on an interval, then F(x) is the antiderivative of f(x) for every value of x on the interval.

Page 2: Section 4.8 - Antiderivatives

Section 4.8 - Antiderivatives

𝑓 (𝑥 )=𝑥3𝑓 ′ (𝑥 )=3 𝑥2

𝑓 (𝑥 )=𝑥3+2𝑓 ′ (𝑥 )=3 𝑥2

𝑓 (𝑥 )=𝑥3 −1𝑓 ′ (𝑥 )=3 𝑥2

𝑓 (𝑥 )=𝑥3+4𝑓 ′ (𝑥 )=3 𝑥2

Theorem: – If F(x) is an antiderivative of f(x) on an interval I, then the general antiderivative of f(x) is:

𝑓 (𝑥 )=3 𝑥2→ 𝐹 (𝑥 )=𝑥3+𝐶

State the derivative of each function.

Page 3: Section 4.8 - Antiderivatives

Section 4.8 - AntiderivativesAntiderivative Formulas where k is a constant

(from page 281 of the textbook)

Page 4: Section 4.8 - Antiderivatives

Section 4.8 - Antiderivatives

𝑓 (𝑥 )=𝑥5 𝐹 (𝑥 )= 𝑥5+1

5+1+𝐶

𝑓 (𝑥 )=sin(2 x )𝐹 (𝑥 )=− 12 cos (2 𝑥 )+𝐶

𝑓 (𝑥 )=𝑒− 3𝑥

𝐹 (𝑥 )=− 13𝑒

−3𝑥+𝐶

Write the general antiderivative of each of the following functions.

¿𝑥6

6+𝐶

Page 5: Section 4.8 - Antiderivatives

Section 4.8 - AntiderivativesIndefinite Integrals

∫ (5− 6 𝑥 ) 𝑑𝑥=¿¿5 𝑥− 6 𝑥1+1

1+1 +𝐶¿5 𝑥− 6 𝑥2

2+𝐶¿5 𝑥− 3𝑥2+𝐶

∫− 5𝑠𝑖𝑛𝑡 𝑑𝑡=¿¿−5 (−𝑐𝑜𝑠𝑡 )+𝐶¿5𝑐𝑜𝑠𝑡+𝐶

∫ (2𝑒𝑥− 3𝑒− 2𝑥 )𝑑𝑥=¿¿2𝑒𝑥 − 3𝑒−2𝑥

− 2+𝐶¿2𝑒𝑥+

32 𝑒

− 2𝑥+𝐶

Page 6: Section 4.8 - Antiderivatives

Section 4.8 - AntiderivativesInitial Value Problems

Solve for the original equation if given and .

∫ 𝑑2 𝑦𝑑 𝑥2 =∫ 2− 6 𝑥  

𝑑𝑦𝑑𝑥 =2𝑥− 6𝑥2

2+𝐶

𝑑𝑦𝑑𝑥 =2𝑥−3 𝑥2+𝐶

4=2 (0 ) −3 (0)2+𝐶

4=𝐶𝑑𝑦𝑑𝑥 =2𝑥−3 𝑥2+4

∫ 𝑑𝑦𝑑𝑥=∫2 𝑥−3 𝑥2+4  

𝑦=2𝑥2

2− 3 𝑥3

3+4 𝑥+C

𝑦=𝑥2−𝑥3+4 𝑥+C

1=(0)2− (0 )3+4 (0)+C

1=𝐶𝑦=𝑥2−𝑥3+4 𝑥+1

Page 7: Section 4.8 - Antiderivatives

Section 5.1 – Area and Estimating Finite SumsEstimating Area Under a Curve

Approximate the area under the curve from to using 2 rectangles.

.

Left-hand endpoints

1 2

Right-hand endpoints Midpoints

1 2 1 2

𝐴𝑟𝑒𝑎=h h𝑒𝑖𝑔 𝑡 ∙ h𝑙𝑒𝑛𝑔𝑡 = 𝑓 (𝑥) ∙ ∆ 𝑥

𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝐴𝑟𝑒𝑎𝑈𝑛𝑑𝑒𝑟 h𝑡 𝑒𝐶𝑢𝑟𝑣𝑒= h𝑡 𝑒𝑠𝑢𝑚𝑜𝑓 h𝑡 𝑒𝑎𝑟𝑒𝑎𝑜𝑓 𝑎𝑙𝑙 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠𝐴= 𝑓 (0 ) ∙1+ 𝑓 (1) ∙1𝐴=1 ∙ 1+2 ∙1𝐴=3

𝐴= 𝑓 (1 ) ∙1+ 𝑓 (2)∙ 1𝐴=2∙ 1+5 ∙1𝐴=7

𝐴= 𝑓 ( .5 ) ∙1+ 𝑓 (1.5) ∙1𝐴=1.25 ∙ 1+3.25 ∙ 1𝐴=4.5

Page 8: Section 4.8 - Antiderivatives

Section 5.1 – Area and Estimating Finite SumsEstimating Area Under a Curve

Approximate the area under the curve from to using 4 rectangles.

.

Left-hand endpoints Right-hand endpoints Midpoints

𝐴𝑟𝑒𝑎=h h𝑒𝑖𝑔 𝑡 ∙ h𝑙𝑒𝑛𝑔𝑡 = 𝑓 (𝑥) ∙ ∆ 𝑥

𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝐴𝑟𝑒𝑎𝑈𝑛𝑑𝑒𝑟 h𝑡 𝑒𝐶𝑢𝑟𝑣𝑒= h𝑡 𝑒𝑠𝑢𝑚𝑜𝑓 h𝑡 𝑒𝑎𝑟𝑒𝑎𝑜𝑓 𝑎𝑙𝑙 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒𝑠𝐴= 𝑓 (0 ) ∙ .5+ 𝑓 ( .5 ) .5+ f (1 ) .5+ 𝑓 (1.5 ) .5 ¿3.75

1 2 1 2 1 2

𝐴= 𝑓 ( .5 ) ∙ .5+ 𝑓 (1 ) .5+f (1.5 ) .5+ 𝑓 ( 2 ) .5 ¿5.75𝐴= 𝑓 ( .25 ) ∙ .5+ 𝑓 ( .75 ) .5+ f (1.25 ) .5+ 𝑓 (1.75 ) .5 ¿ 4.625

LH

RHMid

Page 9: Section 4.8 - Antiderivatives

Section 5.1 – Area and Estimating Finite SumsAverage Value of an Integral

Average Value: Given a closed interval for a continuous function, the average value is the function value that when multiplied by the length of the interval produces the same area as that under the curve.

𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑉𝑎𝑙𝑢𝑒(𝐴𝑉 )=𝑎𝑟𝑒𝑎𝑢𝑛𝑑𝑒𝑟 h𝑡 𝑒𝑐𝑢𝑟𝑣𝑒

h𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 h𝑡 𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙

AV AVAV

Page 10: Section 4.8 - Antiderivatives

Section 5.1 – Area and Estimating Finite SumsAverage Value of an Integral

Estimate the average value for the function on the interval using four midpoint subintervals (rectangles) on equal width.

𝑨𝒗𝒆𝒓𝒂𝒈𝒆𝑽𝒂𝒍𝒖𝒆(𝑨𝑽 )=𝒂𝒓𝒆𝒂𝒖𝒏𝒅𝒆𝒓 𝒕𝒉𝒆𝒄𝒖𝒓𝒗𝒆𝒍𝒆𝒏𝒈𝒕𝒉𝒐𝒇 𝒕𝒉𝒆𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍

𝑬𝒔𝒕𝒊𝒎𝒂𝒕𝒆𝒅 𝑨𝒓𝒆𝒂𝑼𝒏𝒅𝒆𝒓 𝒕𝒉𝒆𝑪𝒖𝒓𝒗𝒆

𝐴=21

𝐴𝑉=214 =5.25

𝐴= 𝑓 ( .5 ) ∙1+ 𝑓 (1.5 ) ∙ 1+ f (2.5 ) ∙1+ 𝑓 (3.5 ) ∙ 1

1 2 3 4

.

𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑉𝑎𝑙𝑢𝑒(𝐴𝑉 )=21

4 − 0

Page 11: Section 4.8 - Antiderivatives

Section 5.2 – Sigma Notation and Limits of Finite Sums

Sequence – a function whose domain is positive integers.

Sigma Notation

𝑓 (𝑥 )=2𝑥+1𝑓 (𝑥 ) ,𝑔 (𝑥 ) , h(𝑥 ) 𝑎𝑛 ,𝑏𝑖 ,𝑐𝑘

g

h (𝑥 )= 𝑥+6𝑥2+2 𝑥+3

𝑎𝑛=2𝑛+1𝑏𝑖=𝑖2 −3 𝑖+7

𝑐𝑘=𝑘+6

𝑘2+2𝑘+3

Sigma Notation – A mathematical notation that represents the sum of many terms using a formula.

Page 12: Section 4.8 - Antiderivatives

Section 5.2 – Sigma Notation and Limits of Finite Sums

Examples

∑𝒏=𝟏

𝟒𝟐𝒏

2 (1 )+2 (2 )+2 (3 )+2(4)2+4+6+8

2+2+2+2+2+212

Sigma Notation

20

∑𝒌=𝟏

𝟔𝟐

∑𝒊=𝟏

𝟑(𝒊¿¿𝟐−𝟑 𝒊+𝟕)¿

)

𝟓+𝟓+𝟕𝟏𝟕

Page 13: Section 4.8 - Antiderivatives

Section 5.2 – Sigma Notation and Limits of Finite Sums

Express the sums in sigma notation.

∑𝒊=𝟏

𝟗𝟖𝒊

1+2+3+4+…+98

1+12+

13 +

14 +…+

170

Sigma Notation

∑𝒌=𝟏

𝟕𝟎 𝟏𝒌

1 −2+3 − 4+…− 98

∑𝒊=𝟏

𝟗𝟖(−𝟏)𝒊+𝟏𝒊

1 − 14 +

19 − 1

16 +…− 149

∑𝒊=𝟏

𝟕(−𝟏)𝒊+𝟏 𝟏

𝒊𝟐

Page 14: Section 4.8 - Antiderivatives

Section 5.2 – Sigma Notation and Limits of Finite Sums

Linearity of Sigma

∑𝒊=𝟏

𝒏𝒄 𝒂𝒊=𝒄∑

𝒊=𝟏

𝒏𝒂𝒊

Sigma Notation

∑𝒌=𝟏

𝒏(𝟑𝒌𝟐+𝟐𝒌−𝟕)

∑𝒊=𝟏

𝒏(𝒂𝒊±𝒃𝒊)=¿∑

𝒊=𝟏

𝒏𝒂𝒊±∑

𝒊=𝟏

𝒏𝒃𝒊 ¿

∑𝒌=𝟏

𝒏𝟑𝒌𝟐+∑

𝒌=𝟏

𝒏𝟐𝒌− ∑

𝒌=𝟏

𝒏𝟕

Example

𝟑∑𝒌=𝟏

𝒏𝒌𝟐+𝟐∑

𝒌=𝟏

𝒏𝒌− ∑

𝒌=𝟏

𝒏𝟕→

Page 15: Section 4.8 - Antiderivatives

Section 5.2 – Sigma Notation and Limits of Finite Sums

∑𝒌=𝟏

𝒏𝒄=𝒏 ∙𝒄

Summation Rules

∑𝒌=𝟏

𝒏𝒌𝟐=

𝒏(𝒏+𝟏)(𝟐𝒏+𝟏)𝟔

∑𝒌=𝟏

𝒏𝒌=

𝒏(𝒏+𝟏)𝟐

∑𝒌=𝟏

𝒏𝒌𝟑=¿(𝒏(𝒏+𝟏)

𝟐 )𝟐¿

Page 16: Section 4.8 - Antiderivatives

Section 5.2 – Sigma Notation and Limits of Finite Sums

∑𝒌=𝟏

𝟓𝟏𝟒=¿¿

Summation Rules Examples

∑𝒌=𝟏

𝟏𝟓𝒌𝟐=¿¿

∑𝒌=𝟏

𝟑𝟐𝒌=¿¿

∑𝒌=𝟏

𝟗𝒌𝟑=¿¿

𝟓𝟏 ∙𝟒=¿𝟐𝟎𝟒

𝟑𝟐(𝟑𝟐+𝟏)𝟐 =¿𝟓𝟐𝟖

𝟏𝟓(𝟏𝟓+𝟏)(𝟐 ∙𝟏𝟓+𝟏)𝟔 =¿𝟏𝟐𝟒𝟎

(𝟗(𝟗+𝟏)𝟐 )

𝟐=¿𝟐𝟎𝟐𝟓

𝒔𝒖𝒎(𝒔𝒆𝒒 (𝟒 , 𝒙 ,𝟏 ,𝟓𝟒 ,𝟏 ))

𝒔𝒖𝒎(𝒔𝒆𝒒 (𝒙 ,𝒙 ,𝟏 ,𝟑𝟐 ,𝟏 ))

𝒔𝒖𝒎(𝒔𝒆𝒒 ( 𝒙𝟐 ,𝒙 ,𝟏 ,𝟏𝟓 ,𝟏 ))

𝒔𝒖𝒎(𝒔𝒆𝒒 ( 𝒙𝟑 ,𝒙 ,𝟏 ,𝟗 ,𝟏 ))