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. 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 PresentationTRANSCRIPT
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.
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.
Section 4.8 - AntiderivativesAntiderivative Formulas where k is a constant
(from page 281 of the textbook)
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
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
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
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
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
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
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
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.
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
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
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→
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¿
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 ))