- 93 Views
- Uploaded on
- Presentation posted in: General

Lin McMullin

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Accumulation

&

Functions Defined by Integrals

Lin McMullin

- Or Thoughts on

My Favorite Equation!

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:

”

Final Value = Starting Value + Accumulated Change

Final Position = Initial Position + Displacement

The first time you saw this ….

The first time you saw this ….

The first time you saw this ….

The first time you saw this ….

AP Example from 1997 BC 89

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

Then f (4) =

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?

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?

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)

- A quick look at some free-response questions
- 2000 AB 4
- 2008 AB2 / BC2 (d)
- 2008 AB 3 (c)
- 2008 AB4 / BC 4 (a)

- 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)

- AB 1 (a, c, d)

2009 AB 6

M

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

With the initial condition f (0) = 5

f (0) = 5

Find f (4)

M

f (0) = 5

Find f (-4)

M

f (0) = 5

Find f (-4)

M

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

M = 3ln(5/3)

M

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

M

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

M

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

E-mail: lnmcmullin@aol.com

Blog: TeachingCalculus.wordpress.com

Website: www.LinMcMullin.net

Click on AP Calculus