Accumulation
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Accumulation & Functions Defined by Integrals. Lin McMullin. Accumulation & Functions Defined by Integrals. Or Thoughts on . My Favorite Equation!.

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Lin mcmullin

Accumulation

&

Functions Defined by Integrals

Lin McMullin


Accumulation functions defined by integrals

Accumulation & Functions Defined by Integrals

  • Or Thoughts on

My Favorite Equation!


Lin mcmullin

The goals of the AP Calculus program include the statement, “Students should understand the definite integral … as the net accumulation of change….”[1] The Topical Outline includes the topic the “definite integral of the rate of change of a quantity over an interval interpreted as the [net] change of the quantity over the interval:


Lin mcmullin

Final Value = Starting Value + Accumulated Change


Lin mcmullin

Final Position = Initial Position + Displacement


Lin mcmullin

The first time you saw this ….


Lin mcmullin

The first time you saw this ….


Lin mcmullin

The first time you saw this ….


Lin mcmullin

The first time you saw this ….


Lin mcmullin

AP Example from 1997 BC 89

If f is an antiderivative of such that f (1) = 0

Then f (4) =


Lin mcmullin

AP Example from 2008 AB 7

  • A particle moves along the x-axis with velocity given by

  • for time . If the particle is at the position

  • x = 2 at time t = 0, what is the position of the particle

  • at time t = 1?


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AP Example from 2008 AB 87

An object traveling in a straight line has position

x(t) at time t. If the initial position is x(0) = 2 and

the velocity of the object is , what is

the position of the object at t = 3?


Lin mcmullin

AP Example from 2008 AB 81

  • If G(x) is an antiderivative for f (x) and G(2) = -7,

  • then G(4) =

  • (A) f ´(4) (B) -7 + f ´(4) (C)

  • (D) (E)


Lin mcmullin

  • A quick look at some free-response questions

    • 2000 AB 4

    • 2008 AB2 / BC2 (d)

    • 2008 AB 3 (c)

    • 2008 AB4 / BC 4 (a)


Lin mcmullin

  • A quick look at some free-response questions

    • AB 1 (a, c, d)

      • AB 2 (c)

      • 2010 AB 3 (a,d)

      • 2010 AB 5 (a)


Lin mcmullin

2009 AB 6

M

The x-intercepts are x = - 2 and x= 3ln(5/3) = M

With the initial condition f (0) = 5


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f (0) = 5

Find f (4)

M


Lin mcmullin

f (0) = 5

Find f (-4)

M


Lin mcmullin

f (0) = 5

Find f (-4)

M


Lin mcmullin

Find the x-coordinate of the absolute maximum value and justify your answer.

M = 3ln(5/3)

M


Lin mcmullin

and since f ´(x) ≥ 0 on [-4, M ] it follows that f (M) > f (-4).

M


Lin mcmullin

and since on [M, 4] £ 0 it follows that f (M) > f (4)

M


Lin mcmullin

Since M is the only critical number in the interval [-4, 4] and f (M) > f (-4) and f (M) > f (4), x = M is the location of the absolute maximum value by the Candidates’ Test.

M


Lin mcmullin

Lin McMullin

E-mail: [email protected]

Blog: TeachingCalculus.wordpress.com

Website: www.LinMcMullin.net

Click on AP Calculus


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