Accumulation
&
Functions Defined by Integrals
Lin McMullin
Accumulation & Functions Defined by Integrals
Or Thoughts on
f  t   f  a    f   x  dx
t
a
My Favorite Equation!
f  t   f  a    f   x  dx
t
a
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
a
f   x  dx  f  b   f  a  ”
f  t   f  a    f   x  dx
t
a
Final Value = Starting Value + Accumulated Change
f  t   f  a    f   x  dx
t
a
f  t   f  a    f   x  dx
t
a
Final Position = Initial Position + Displacement
v  t   s  t 
s  t   s  a    v  x  dx
t
a
f  t   f  a    f   x  dx
t
a
The first time you saw this ….
y  x   y0   m dt
x
x0
f  t   f  a    f   x  dx
t
a
The first time you saw this ….
y  x   y0   m dt
x
x0
x
 y0  m  dt
x0
f  t   f  a    f   x  dx
t
a
The first time you saw this ….
y  x   y0   m dt
x
x0
x
 y0  m  dt
x0
 
 y0  m t x
x
0
f  t   f  a    f   x  dx
t
a
The first time you saw this ….
y  x   y0   m dt
x
x0
x
 y0  m  dt
x0
 
 y0  m t x
x
0
y  y0  m  x  x0 
f  t   f  a    f   x  dx
t
a
y  mx  b
f  t   f  a    f   x  dx
t
a
y  mx  b
f  t   f  a    f   x  dx
t
a
AP Example from 1997 BC 89
x2
If f is an antiderivative of
5 such that f (1) = 0
1 x
Then f (4) =
x2
f  4   f 1  
dx  0.376
5
1 1 x
4
x2
f  4   f 1  
dx
5
1 1 x
4
f  t   f  a    f   x  dx
t
a
AP Example from 2008 AB 7
A particle moves along the x-axis with velocity given by
v  t   3t 2  6t for time t  0 . If the particle is at the position
x = 2 at time t = 0, what is the position of the particle
at time t = 1?
x 1  2   3t  6t dt  2   t  3t
1
0
2
3
2

1
0
 24 6
f  t   f  a    f   x  dx
t
a
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 v  t   1  t , what is
3
2
the position of the object at t = 3?
x  3  2  
3
0
3
1  t 2 dt  2  4.51153  6.512
f  t   f  a    f   x  dx
t
a
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)
(D)
(B) -7 + f ´(4)
  7  f  t   dt
4
2
 f  t  dt
4
(C)
2
(E) 7  2 f  t  dt
4
f  t   f  a    f   x  dx
t
a
A quick look at some free-response questions
2000 AB 4
2008 AB2 / BC2 (d)
2008 AB 3 (c)
2008 AB4 / BC 4 (a)
f  t   f  a    f   x  dx
t
a
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)
f  t   f  a    f   x  dx
t
a
2009 AB 6
for  4  x  0
 g  x
f   x     x /3
5e  3 for 0  x  4



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M


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


The x-intercepts are x = - 2 and x = 3ln(5/3) = M
With the initial condition f (0) = 5

f  t   f  a    f   x  dx
t

a




f (0) = 5
M









Find f (4)


f  4   f  0     5e  x /3  3 dx
4

0
 5   15e
 x /3
 3x   8  15e 4/3
4
0

f  t   f  a    f   x  dx
t
a





f (0) = 5
M

Find f (-4)



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





4
f  4   f  0    g  x  dx

0
 f  0    g  x  dx
0
4
 5  8  2   2  3

f  t   f  a    f   x  dx
t

a




f (0) = 5
M








Find f (-4)


0

4
f  0   f  4    g  x  dx
5  f  4   8  2 
5   8  2   f  4 
f  4   2  3


f  t   f  a    f   x  dx
t
a
Find the x-coordinate of the absolute
maximum value and justify your answer.




M = 3ln(5/3)

M













f  t   f  a    f   x  dx
t
a
f  M   f  4   
M
4
f   x  dx and since f ´(x) ≥ 0 on

[-4, M ] it follows that f (M) > f (-4).




M












f  t   f  a    f   x  dx
t
a
f 4  f  M   
5𝑒 −𝑥 3
4
M
5e
 x /3
 3 dx and since on [M, 4]
− 3 0 it follows that f (M) > f (4)





M












f  t   f  a    f   x  dx
t
a
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