AP Problems Involving the
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus:
1. If F ' ( x)  f ( x) ,then
d x
2. dx a f t  dt  f x  .
b
a f  xdx  Fb  Fa .
One of the hardest calculus topics to teach in the
old days was Riemann sums.
They were hard to draw, hard to compute, and
(many felt) totally unnecessary.
Then along came the TI graphing calculators. Using
the integral utility in the CALC menu, students could
actually see an integral accumulating value from left
to right along the x-axis, just as a limit of Riemann
sums would do:
So now we can do all kinds of summing problems before we
even mention an antiderivative.
Historically, that’s what scientists had to do before calculus.
Here’s why it mattered to them:
mi/hr
60
v(t) = 40
40
d = 120 mi
20
1
4
hr
The calculus pioneers knew that the area would still
yield distance, but what was the connection to
tangent lines?
And was there an easy way to find these irregularlyshaped areas?
mi/hr
60
v(t)
40
20
d = 120 mi
1
4
hr
Since the time of Archimedes, scientists had
been finding areas of irregularly-shaped regions
by dividing them into regularly-shaped regions.
That is what Riemann sums are all about.
2.033281
2.008248
2.000329
With graphing calculators, students can find
these sums without the tedium. They can also
imagine the tedium of doing these sums by
hand!
Best of all, they can actually see the limiting
case:
And the calculator shows the thin rectangles
accumulating from left to right – ideal for
understanding the FTC!
Let us consider a positive continuous
function f defined on [a, b].
Choose an arbitrary x in [a, b].
y=f(t)
a
x
b
Each choice of x determines a unique area from a to x, denoted as usual by

x
a
f (t )dt
y=f(t)
a
x
b
So
d x
f (t )dt  f ( x).

dx a
But that is only half the story.

x
Now that we know that a f (t ) dt is an
antiderivative of f,
we know that it differs from any antiderivative
of f by a constant.
That is, if F is any antiderivative of f,

x
a
f (t )dt  F ( x)  C.
To find C, we can plug in a:

a
a
f (t )dt  F (a )  C
0  F (a)  C
C   F (a )
So

x
a
f (t )dt  F ( x)  F (a ).
Now plug in b:

b
a
f (t )dt  F (b)  F (a).
This was the FTC. This was the result that changed the
world.
2.000329
Now, instead of wasting a full afternoon just to get an
approximation of the area under one arch of the sine curve, you
could find one antiderivative, plug in two numbers, and subtract!


0
sin( x)dx   cos( x)

0
  cos( )  cos(0)  2.
Since 2000, the AP Calculus Test Development
Committee has been emphasizing a conceptual
understanding of the definite integral, resulting
in these “new” problem types:
Functions defined as integrals
Accumulation Problems
Integrals from Tables
Finding f (b), given f (a ) and f ( x)
Interpreting the Definite Integral
Problem of the:
Suppose

x
1
f (t )dt  x 3  2 x  k .
(a) Find f(x).
(b) Find k.
(a) By the Fundamental Theorem,
d x
d 3
f
(
t
)
dt

x

2
x

k
)



dx 1
dx
2
f ( x)  3x  2
(b) Plug in x = 1:

1
1
f (t )dt  1  2(1)  k
3
0  1  k
k 1
Here was the problem (1987):
If f ( x)  x2 , which of the following could be the graph of
x
y   f (t )dt ?
1
(A)
(B)
(D)
(E)
(C)
This problem had been checked:
1. by the author who had written it;
2. by the committee that had okayed it;
3. by the committee that had okayed it for a pre-test;
4. by the ETS test development specialists;
5. The committee, reviewing the final form of the college pretest.
The proposed key was (B). That is,
x
If f ( x)  x2 , the graph of y  1 f (t )dt could be
(B)
While everyone was concentrating on the Fundamental
Theorem application, they had missed the hidden “initial
condition” that y must equal zero when x = 1!
Here’s 1995 / BC-6:
2
1
0
-1
-2
-3
1
2
3
4
5
Let f be a function whose domain is the closed interval [0, 5].
The graph of f is shown above.
Let h( x)  
(a)
(b)
(c)
x
3
2
0
f (t )dt.
Find the domain of h.
Find h(2).
At what x is h(x) a minimum? Show the analysis that
leads to your conclusion.
(a) The domain of h is all x for which

x
3
2
0
is defined:
f (t )dt
x
0 35  6 x  4
2
(b) A little Chain Rule:
x
 1
f (t )dt  f   3  
2
 2
1
3
h(2)  f (4)   
2
2
d
h( x)  
dx
x
3
2
0
2
1
0
-1
-2
-3
1
2
3
4
5
Let f be a function whose domain is the closed interval [0, 5].
The graph of f is shown above.
Let h( x)  
(a)
(b)
(c)
x
3
2
0
f (t )dt.
Find the domain of h.
Find h(2).
At what x is h(x) a minimum? Show the analysis that
leads to your conclusion.
(c) Since 
x
3
2
0
f (t )dt is positive from
-6 to -1 and negative from -1 to 4,
the minimum occurs at an endpoint. By
comparing areas, h(4) < h(-6) = 0, so the
minimum occurs at x = 4.
This “area comparison” genre of problem was
pretty common in the early graphing calculator
days.
2500
2000
2007 / AB-2 BC-2
1500
1000
500
1
2
3
4
5
6
7
The amount of water in a storage tank, in gallons, is modeled by a
continuous function on the time interval 0  t  7 , where t is measured in
hours. In this model, rates are given as follows:
(i)
The rate at which water enters the tank is f (t )  100t 2 sin t
 
gallons per hour for 0  t  7 .
(ii) The rate at which water leaves the tank is
250 for 0  t  3
g (t )  
gallons per hour.
2000 for 3  t  7
The graphs of f and g, which intersect at t = 1.617 and t = 5.076, are
shwn in the figure above. At time t = 0, the amount of water in the tank
is 5000 gallons.
2500
2000
1500
1000
500
1
2
3
4
5
6
7
(a) How many gallons of water enter the tank during the time interval
0  t  7 ? Round your answer to the nearest gallon.
(b) For 0  t  7 , find the time intervals during which the amount of
water in the tank is decreasing. Give a reason for each answer.
(c) For 0  t  7 , at what time t is the amount of water in the tank
greatest? To the nearest gallon, compute the amount of water at this
time. Justify your answer.
2500
2000
1500
1000
500
1
2
3
4
5
6
7
The standard description of the FTC is that
“The two central operations of calculus, differentiation and integration,
are inverses of each other.” —Wikipedia
A more useful description is that the two definitions of the
definite integral:
•The difference of the values of an anti-derivative taken at the
endpoints, [definition used by Granville (1941) and earlier
authors]
•The limit of a Riemann sum, [definition used by Courant
(1931) and later authors]
yield the same value.
2004 AB3(d)
A particle moves along the y-axis so that its velocity v at time t ≥ 0 is
given by v(t) = 1 – tan–1(et). At time t = 0, the particle is at y = –1. Find the
position of the particle at time t = 2.
y '(t) = v(t) = 1 – tan–1(et)
y(t) = ?
Velocity  Time = Distance
velocity
time
The Fundamental Theorem of Calculus (part 1):
If F ' ( x)  f ( x) ,then
b
a f  xdx  Fb  Fa .
Change in y-value equals
 vt dt   1  tan e dt  0.3607,
2
2
0
0
1
t
Since we know that y(0) = –1:
y2   y 0  0.3607  1.3607
If we know an anti-derivative, we can use it to find the
value of the definite integral.
All students should know how to interpret the
following applications as accumulations:
Areas (sums of rectangles)
Volumes (sums of regular-shaped slices)
Displacements (sums of v(t)∙∆t)
Average values (Integrals/intervals)
BC: Arclengths (sums of hypotenuses)
BC: Polar areas (sums of sectors)
Problem of the Day :
The population density of the city of Washerton decreases as you move
away from the city center. In fact, it can be approximated (in people per
square mile) by the function 10,000(2 – r) at a distance r miles from the
city center.
(a) What is the radius of the populated portion of the city?
(b) A thin ring around the center of the city has thickness r and
radius r. What is its area? [Hint: Imagine straightening it out to
make a thin rectangular strip.]
(c) What is the population of the strip in part (b)?
(d) Estimate the total population of Washerton by setting up and
evaluating a definite integral.
The population density of the city of Washerton decreases as you move
away from the city center. In fact, it can be approximated (in people per
square mile) by the function 10,000(2 – r) at a distance r miles from the
city center.
(a)
What is the radius of the populated portion of the city?
(a) r = 2 miles.
(b) A thin ring around the center of the city has thickness r and
radius r. What is its area? [Hint: Imagine straightening it out to
make a thin rectangular strip.]
(b) A = 2πr Δr
(d) Estimate the total population of Washerton by setting up and
evaluating a definite integral.
(d)

2
0
10,000(2  r )(2 r )dr
people per sq. mile
 83,776 people
sq. miles
Another implication of the Fundamental
Theorem (and a source of several recent
problems that have caused trouble for
students):

b
a
f ( x)dx  f (b)  f (a )
b
 f (b)  f (a)   f ( x)dx
a
Thus, given f(a) and the rate of change of f on [a, b],
you can find f(b).
The Kicker in 2003 / AB-4 BC-4:
y
2
(–3, 1)
–4
–2
2
4
x
–2
(4, –2)
Let f be a function defined on the closed interval 3  x  4 with f(0) = 3.
The graph of f  , the derivative of f, consists of one line segment and a
semicircle, as shown above.
(d) Find f(–3) and f(4). Show the work that leads to your answers.
f (3)  f (0)  
3
0
f ( x)dx
1
1
 3  (2)(2)  (1)(1)
2
2
9

2
4
f (4)  f (0)   f ( x)dx
0
 3  (8)  (2 )
 2  5
Problem of the Day #35:
If F ( x)  sin ( x) and F(2) = 5, find F(7).
2
Solve this just like the last one was solved.