Functions Defined by Integrals Accumulation & Functions Defined by Integrals My Favorite Equation Lin McMullin

Accumulation & Functions Defined by Integrals 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: ” “Accumulation” is not mentioned in the indexes of Anton, Finney, Foerster, Hughes-Hallett, Ostebee, Rogawski or Stewart. There is a single sentence in the current edition of Larson, with no exercises using the idea. Rogawski does have a good section on the rate of change is the amount of change idea. As with some other topics that appear on the AP calculus exams, since the topic does not appear in the current editions of most of the textbooks used in AP Calculus courses it is important that you use examples from recent exams and make up problems of your own if necessary. On the 2008 AP Calculus exams there are no less than 7 questions where this can be used, yet, as mentioned above, this form and this approach is not mentioned in the most textbooks. It appears 10 times on the 2010 AB scoring standards.

Final Value = Starting Value + Accumulated Change Click

Final Position = Initial Position + Displacement Click

The first time you saw this …. 1 of 3

The first time you saw this …. 3 of 3 At this point I will make a slight digression ….

The first time you saw this …. 3 of 3 At this point I will make a slight digression ….

The first time you saw this …. 3 of 4 At this point I will make a slight digression ….

Much more difficult to use Much more difficult to understand Just use point-slope Slope-intercept is a Special Case; Why do the special case first and for several years? But I digress (I told you I was going to) So back to our favorite equation Some Examples from the AP Calculus multiple-choice questions

If f is an antiderivative of such that f (1) = 0 Then f (4) = AP Example from 1997 BC 89 If f is an antiderivative of such that f (1) = 0 Then f (4) = #3 of 4 Euler’s Method

A particle moves along the x-axis with velocity given by 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? Some Examples from the AP Calculus multiple-choice questions #1 of 4

An object traveling in a straight line has position 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? #2 of 4

If G(x) is an antiderivative for f (x) and G(2) = -7, then G(4) = 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) #4 of 4

Leave PPT for One-note 2008 AB2 / BC2 2008 AB 3 2008 AB4 / BC 4 A quick look at some free-response questions 2000 AB 4 2008 AB2 / BC2 (d) 2008 AB 3 (c) 2008 AB4 / BC 4 (a) Leave PPT for One-note 2008 AB2 / BC2 2008 AB 3 2008 AB4 / BC 4 2000 AB 4 2010 AB 1

Leave PPT for One-note 2008 AB2 / BC2 2008 AB 3 2008 AB4 / BC 4 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) Leave PPT for One-note 2008 AB2 / BC2 2008 AB 3 2008 AB4 / BC 4 2000 AB 4 2010 AB 1

The x-intercepts are x = - 2 and x = 3ln(5/3) = M 2009 AB 6 M Click The x-intercepts are x = - 2 and x = 3ln(5/3) = M With the initial condition f (0) = 5

M f (0) = 5 Find f (4) Click

M f (0) = 5 Find f (-4) The second definite integral is most easily found by subtracting the area of the semicircle from the area of a rectangle, 8, drawn around it. Notice that is “backwards”: the upper limit of integration is less than the lower limit. Therefore, its value is the opposite of the area of the region between the semicircle and the x-axis. Integration questions with the lower limit of integration is greater than the upper limit and where the values must be found from a graph are confusing for students. A way of looking at this situation is to reason this way: We are looking for f(-4). If we integrate starting at x = –4 we have accumulated the amount 8 – 2pi by the time we get to x = 0 where the value is 5 (given). So if we subtract the (8 – 2pi) from 5, we will have our starting value . Symbolically, the details look like this and this avoids the integral with the lower limit greater than the upper. You can always start with the left hand side of the interval as the starting value.

M f (0) = 5 Find f (-4) Click

Find the x-coordinate of the absolute maximum value and justify your answer. M = 3ln(5/3) The Easy way The elegant way.

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

and since on [M, 4] 5 𝑒 − 𝑥 3 −3 £ 0 it follows that f (M) > f (4)

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 Click on AP Calculus E-mail: lnmcmullin@aol.com Blog: TeachingCalculus.wordpress.com Website: www.LinMcMullin.net Click on AP Calculus I will post the handout at my website