Embed Size (px)
Citation preview
Lin McMullin
Accumulation &
Functions Defined by Integrals
Accumulation & Functions Defined by Integrals
t
af t f a f x dx
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:
” b
af x dx f b f a
t
af t f a f x dx
Final Value = Starting Value + Accumulated Change
t
af t f a f x dx
t
af t f a f x dx
Final Position = Initial Position + Displacement
t
a
v t s t
s t s a v x dx
t
af t f a f x dx
The first time you saw this ….
0
0
x
xy x y mdt
t
af t f a f x dx
The first time you saw this ….
0
0
0
0
x
x
x
x
y x y mdt
y m dt
t
af t f a f x dx
The first time you saw this ….
0
0
0
0
0
0
x
x
x
x
x
x
y x y mdt
y m dt
y m t
t
af t f a f x dx
The first time you saw this ….
0
0
0
0
0
0
0 0
x
x
x
x
x
x
y x y mdt
y m dt
y m t
y y m x x
t
af t f a f x dx
t
af t f a f x dx
y mx b
t
af t f a f x dx
y mx b
AP Example from 1997 BC 89
If f is an antiderivative of such that f (1) = 0
Then f (4) =
2
51
x
x
24
514 1 0.376
1
xf f dx
x
t
af t f a f x dx
24
514 1
1
xf f dx
x
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?
23 6v t t t 0t
11 2 3 2
0 01 2 3 6 2 3 2 4 6x t t dt t t
t
af t f a f x dx
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?
23 1v t t
3 23
03 2 1 2 4.51153 6.512x t dt
t
af t f a f x dx
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
2f t dt
4
27 f t dt
4
27 f t dt
t
af t f a f x dx
A quick look at some free-response questions
2000 AB 4
2008 AB2 / BC2 (d)
2008 AB 3 (c)
2008 AB4 / BC 4 (a)
t
af t f a f x dx
A quick look at some free-response questions
2010 AB 1 (a, c, d)
2010 AB 2 (c)
2010 AB 3 (a,d)
2010 AB 5 (a)
t
af t f a f x dx
2009 AB 6
/3
for 4 0
5 3 for 0 4 x
g x xf x
e x
The x-intercepts are x = - 2 and x = 3ln(5/3) = M With the initial condition f (0) = 5
M
t
af t f a f x dx
f (0) = 5
Find f (4)
4 /3
0
4/3 4/3
0
4 0 5 3
5 15 3 8 15
x
x
f f e dx
e x e
M
t
af t f a f x dx
f (0) = 5
Find f (-4)
4
0
0
4
4 0
0
5 8 2 2 3
f f g x dx
f g x dx
M
t
af t f a f x dx
f (0) = 5
Find f (-4)
0
40 4
5 4 8 2
5 8 2 4
4 2 3
f f g x dx
f
f
f
M
t
af t f a f x dx
M = 3ln(5/3)
Find the x-coordinate of the absolute maximum value and justify your answer.
M
t
af t f a f x dx
and since f ´(x) ≥ 0 on
[-4, M ] it follows that f (M) > f (-4).
4
4M
f M f f x dx
M
t
af t f a f x dx
and since on [M, 4]
£ 0 it follows that f (M) > f (4)
4 /34 5 3x
Mf f M e dx
M
t
af t f a f x dx
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
t
af t f a f x dx
Lin McMullin
E-mail: [email protected]
Blog: TeachingCalculus.wordpress.com
Website: www.LinMcMullin.net
Click on AP Calculus
