4032 Fundamental Theorem AP Calculus. Where we have come. Calculus I: Rate of Change Function.

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4032 Fundamental Theorem AP Calculus

description

f’ T T f PDPD D C P : f ( 0 ) = v(t)

Transcript of 4032 Fundamental Theorem AP Calculus. Where we have come. Calculus I: Rate of Change Function.

Page 1: 4032 Fundamental Theorem AP Calculus. Where we have come. Calculus I: Rate of Change Function.

4032 Fundamental Theorem

AP Calculus

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Where we have come.Calculus I:

Rate of Change Function f f

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x

y

f’

T

T

f

PD

DC

P : f ( 0 ) = 0 + -

∪∩∪2.5 6 8

v(t)

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Where we have come.Calculus II:

Accumulation Function

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x

y

Accumulation: Riemann’s Right V

T

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Accumulation (2)Using the Accumulation Model, the Definite Integral representsNET ACCUMULATION -- combining both gains and losses

V

T

D

T

REM: Rate * Time = Distance

x

y

5 8 8 63

-3-4

-3

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Where we have come.Calculus I:

Rate of Change Function

Calculus II:

Accumulation Function

Using DISTANCE model f’ = velocity f = Position

Σ v(t) Δt = Distance traveled

f f

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x

y

Distance Model: How Far have I Gone?V

T

Distance Traveled: a)

b)

If I go 5 mph for one hour and 25mph for 3 hours what is the total distance traveled?

Ending position-beginning position

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B). The Fundamental Theorem

DEFN: THE DEFINITE INTEGRAL

If f is defined on the closed interval [a,b] and

exists , then0 1

lim ( )n

i ix i

f c x

0 1

lim ( ) ( )bn

i ixi a

f c x f x dx

Height base

Rate time

The Definition of the Definite Integral shows the set-up.

Your work must include a Riemann’s sum! (for a representative rectangle)

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( )b

b

aa

f x dx F x

The Fundamental Theorem of Calculus (Part A)

If or F is an antiderivative of f,

then

F b F a

( ) ( )F x f x

𝑣 (𝑡 )𝑑𝑡 ( )b

ad t 𝑑(𝑏)−𝑑 (𝑎)

The Fundamental Theorem of Calculus shows how to solve the problem!

Your work must include an anti-derivative!

REM: The Definite Integral is a NUMBER -- the Net Accumulation

of Area or Distance -- It may be positive, negative, or zero.

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Practice:Evaluate each Definite Integral using the FTC.

1) 1

3xdx

2).

4 2

1( 1)x dx

3).2

2sin( )x dx

The FTC give the METHOD TO SOLVE Definite Integrals.

Top-bottom

¿𝑥2

2 | 1−3

¿𝑥3

3− 𝑥| 4

− 1

¿−𝑐𝑜𝑠𝑥| 𝜋2

− 𝜋2

¿12

− −32

2=− 4

¿ ( 643

−4 )−( − 13

+1)523 − 2

3=503

¿− 0+0=0

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Example: SET UPFind the NET Accumulation represented by the region between

the graph and the x - axis on the

interval [-2,3].

2( ) 2 5f x x x

REQUIRED:

Your work must include a Riemann’s sum! (for a representative rectangle)

𝑏=∆ 𝑥 [− 2,3 ]-0

𝑡𝑜𝑝−𝑏𝑜𝑡𝑡𝑜𝑚𝐴= (𝑥2 −2 𝑥+5 ) ∆ 𝑥

lim𝑛→ ∞

∑ (𝑥2− 2𝑥+5 ) ∆ 𝑥

−2

3

(𝑥2− 2𝑥+5 ) ∆𝑥

∆ 𝑥

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Example: WorkFind the NET Accumulation represented by the region between

the graph and the x - axis on the

interval [-2,3].

2( ) 2 5f x x x

REQUIRED:

Your work must include an antiderivative!

𝑥3

3−𝑥2+5 𝑥| 3

− 2

( 273

−9+15)−( − 83

− 4 − 10)4 53 +

503 =

953

∆ 𝑥

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Method: (Grading)

A). 1.

2.

3.

B) 4.

5.

C). 6.

7.

Graph and rectangle

Base and boundaries

Height (top – bottom) or (right – left) or (big – little)Riemann’s Sum

Definite Integral [must have dx or dy]

antiderivative

answer

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Example:Find the NET Accumulation represented by the region between

the graph and the x - axis on the

interval .

3( ) 27f x x 0,3

0

3

(27 −𝑥3 )𝑑𝑥

27 𝑥− 𝑥4

4 |30(81− 81

4 )− (0− 0 )

2434

𝑏=∆ 𝑥 [ 0,3 ]h=27 −𝑥3 −0𝐴= (27 −𝑥3 ) ∆ 𝑥

lim𝑛→ ∞

∑0

3

(27 −𝑥3 ) ∆𝑥

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Example:Find the NET Accumulation represented by the region between

the graph and the x - axis on the

interval .

( ) sec( ) tan( )f x x x

,4 3

−𝜋

4

𝜋3

( sec (𝑥 ) tan (𝑥 ) )𝑑𝑥

sec (𝑥 ) ⌊𝜋3

−𝜋4

¿¿

(2 − 2√2 )=2√2− 2

√2=

4 − 2√22

𝑏=∆ 𝑥 [− 𝜋4

, 𝜋3 ]

h=sec (𝑥 ) tan (𝑥 )− 0

lim𝑛→ ∞

∑− 𝜋

4

𝜋3

(𝑠𝑒𝑐 (𝑥 ) 𝑡𝑎𝑛 (𝑥 ) ) ∆ 𝑥

√√

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Last Update:

• 1/20/10

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AntiderivativesLayman’s Description:

2x dx cos( )x dx 2sec ( )x dx

2

1 dxx

1 dxx

Assignment: Worksheet

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Accumulating Distance (2)Using the Accumulation Model, the Definite Integral representsNET ACCUMULATION -- combining both gains and losses

V

T

D

T

REM: Rate * Time = Distance

4

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Rectangular Approximationsy = (x+5)(x^2-x+7)*.1

Velocity

Time

V = f (t)

Distance Traveled: a)

b)