AP Calculus Notes: Unit 6 – Definite Integrals
Syllabus Objective: 3.4 – The student will approximate a definite integral using rectangles.
Recall: If a car is traveling at a constant rate (cruise control), then its distance traveled is equal to rate x time.
On a graph, this would look like the one shown, for a car traveling at 60 mph for 2 hours.
Rate
Time
The area of the shaded region = 60∙2 = 120 miles; the total distance traveled in 2 hours.
Finding Distance Traveled When Velocity Varies
To estimate the area of an irregular region, we will estimate using rectangles formed by subintervals.
3
Ex1: A particle starts at x  0 and moves along the x-axis with velocity v  t   t for time t  0 . Where
is the particle at t  2 ? Use 8 subintervals. (We will use “midpoint” rectangles.)
Suberintervals:
Midpoints  mi  :
 1
0, 4 
1
8
3
1 1
 4 , 2 
3
8
3
1 3
 2 , 4  …
5
8
Widths  x  =
1
4
3
1
27
1
 3
 5  125
  
  
  
 8  512
 8  512
 8  512
1 1
1
1 27
27
1 125 125
1
3
Areas =  mi  :






4 512 2048
4 512 2048
4 512 2048
4
1
27
125
343 729 1331 2197 3375 8128
Area 








 3.969
2048 2048 2048 2048 2048 2048 2048 2048 2048
So the particle is close to x  4 when t  2 .
Heights =  mi  :
3
Rectangular Approximation Methods (RAM): Used to approximate the area under a curve
Page 1 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals



LRAM: Left-Hand Rectangles – Use the left endpoints of each interval to draw the heights of each
rectangle.
MRAM: Midpoint Rectangles – Use the midpoints of each interval to draw the heights of each rectangle.
RRAM: Right-Hand Rectangles – Use the right endpoints of each interval to draw the heights of each
rectangle.
Ex2: Approximate the area under the curve y  x 2 from x  1 to x  2 using 6 subintervals. Use all of the
approximation methods listed above.
LRAM
x 
2
Area 
1
2
2
2
1
1 1
1
1 1
1
1 3
 12       0 2     12     2.375 LRAM
2
2 2 2
22 2
22
RRAM
x 
2
Area 
2
1
2
2
1
1 3
1
1 1
1
1 1
 2 2     12      0 2      3.875 RRAM
2
22 2
22 2
2 2
Page 2 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
MRAM
x 
2
2
2
2
1
2
2
2
1 3 1 1 1 1 1 3 1 5 1 7
Area                      2.9375 RRAM
2 4 2 4 24 24 24 24
Note: To get a better approximation of the area, more rectangles (smaller intervals) can be used.
Teacher Note: There are programs for the TI-84 and TI-89 for approximating area using RAM. These programs
will allow you to estimate the area using hundreds of rectangles! Check the technology resource guide given to you
with your textbook to find programs.
Exploration: Which RAM is the biggest?

Estimate the area under the graph of y  e x in the interval 0,3 . Using all three RAM. List
them from smallest to greatest.

Estimate the area under the graph of y  3 x in the interval 0,3 . Using all three RAM. List
them from smallest to greatest.
Draw a conclusion using properties of the graphs above as to which RAM is the biggest. Test your
conclusion with a few graphs that you choose on your own based upon the properties in your conclusion.
Solution: If a curve is increasing, the RRAM is the biggest because the rectangles go above the curve, and
LRAM is the smallest because the rectangles lie below the curve. If a curve is decreasing, the LRAM is the
biggest because the rectangles go above the curve, and the RRAM is the smallest because the rectangles lie
below the curve. In both cases, the MRAM is somewhere in between the RRAM and LRAM. In curves
that are not strictly monotonic, such as y  sin x  3 on the interval 0,3 , it cannot be determined.
Page 3 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
You Try: The table below shows the velocity of a car at 10 second time intervals. Estimate how far the car traveled
using LRAM and RRAM sums.
Time Velocity
(sec) (ft/sec)
0
0
10
15
20
35
30
20
40
43
50
15
60
22
QOD: Give an example of a function (and an interval) for which the RRAM and LRAM are equal on that interval.
Page 4 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Syllabus Objective: 3.5 – The student will evaluate a definite integral as the limit of a Reimann sum. 3.11 –
The student will find the area between two or more curves (between curve and x-axis).
Recall: Sigma Notation  = sum
n
a
k
 a1  a2  a3  ...  an 1  an
k 1
Vocabulary and Notation:
Below is the graph of a function f  x  . The interval  a , b  has been partitioned into n subintervals, called
partitions: P  x0 , x1 , x2 ,..., xn 1 , xn  (Note – Each subinterval is not necessarily the same width.)
The width of the kth rectangle is equal to xk .
Each rectangle is created by choosing a value, cn , within each subinterval and drawing a rectangle with a height of
f  cn  .
cn
a = x0
c1
Notation:
x1 c2
xn-1
x2
xn = b
P = length of subinterval (called the norm)
Riemann Sum (sum of the area of n rectangles): Sn 
n
 f c
k
  xk
k 1
Note: Smaller partitions create more rectangles.
Definition: Let f be a function defined on  a , b  . For any partition P of  a , b  , let the numbers ck be chosen
arbitrarily in the subintervals  xk 1 , xk  .
n
If there exists a number I such that lim
P 0
 f c
k
  xk  I , then f is integrable on  a, b and I is the
k 1
definite integral of f over  a , b  .
Page 5 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Theorem: All continuous functions are integrable.
Definite Integral: Let f be continuous on  a , b  , and let  a , b  be partitioned into n subintervals of equal length
x 
n
ba
f  ck  x .
. Then the definite integral of f over  a , b  is lim
n 
n
k 1

n
Note: The lim
P 0
 f  c   x
k
k
means that the length of the partitions are approaching 0 (getting smaller).
k 1
n
This is the same idea as lim
n 
 f c
k
 x , which means that the number of rectangles are approaching
.
k 1
n
Notation for Definite Integrals: lim
n 

k 1
f  ck  x 
b
 f  x  dx , read “the integral from a to b of f of x dx.”
a
Note that the integral symbol
resembles an S, because an
integral is a sum    .
Vocabulary of Definite Integrals:
Upper Limit
of Integration
b
Integrand
Variable
of Integration
(dummy variable)
 f  x  dx
Integral
Sign
a
Lower Limit
of Integration
Ex1: The interval  4,2 is partitioned into n subintervals of equal length x 
 4  m 
n
midpoint of the kth subinterval. Express the limit lim
n 
The interval  4,2 represents  a , b  . (This is why x 
3
k
6
. Let mk denote the
n

 7mk  2 x as an integral.
k 1
6 b  a 2   4 


.)
n
n
n
mk represents a value of x in each subinterval. So f  x   4 x 3  7 x  2 .
2
Integral:
 4x
3

 7 x  2 dx
4
Page 6 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Area Under a Curve (above the x-axis): If y  f  x  is nonnegative and integrable over  a , b  , then the area
under the curve y  f  x  from a to b is the integral of f from a to b:
b
 f  x  dx
a
Note: We can use integrals to calculate areas, and use areas to calculate integrals!
0
Ex2: Evaluate the integral

9  x 2 dx .
3
Graph the function y  9  x 2 . Shade the region in the interval  3,0 .
Find the area of the shaded region: The figure shaded is ¼ of a circle with a radius of 3.
 r2
9
Area =

.
4
4
0
So,

9  x 2 dx 
3
9
4
Question: What if a function is nonpositive? The value of the function representing the heights of the rectangles
will all be negative. So the heights of the rectangles would be the opposite of the function value.
Area of a Region Between a Curve and the x-Axis (under the x-axis): If y  f  x  is nonpositive and integrable
over  a , b  , then the area between the curve y  f  x  and the x-axis from a to b is the opposite of the integral of f
b
from a to b: Area = 
 f  x  dx
a
Area of a Region Between a Curve and the x-Axis: For any integrable function y  f  x  with both positive and
negative values on an interval  a , b  , the total area between the curve and the x-axis is equal to the Riemann sum of
y  f  x  over the interval the function is above the x-axis minus the Riemann sum of y  f  x  over the
interval the function is below the x-axis.
b
Definite Integral: For any integrable function,
 f  x  dx = (area above the x-axis) – (area below the x-axis).
a
Note: An integral can be negative, an area cannot!
Page 7 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
5
Ex3: Use areas to find the value of
 3dx .
1
Sketch the graph and shade the region.
The area of the shaded region is equal to A  lw  3 4   12 . So the value of
5
 3 dx  12 . (The answer is
1
negative because the function f  x   3 lies below the x-axis. Use this example to explain the following formula.
The Integral of a Constant Function: If f  x   c , where c is a constant, on the interval  a , b  , then
b

b
f  x  dx  cdx  c  b  a  .

a
a
Calculating Definite Integrals: On the TI-84, a definite integral is approximated using numerical integration,
NINT. The keystrokes are fnInt(function, variable of integration, lower limit, upper limit). fnInt can be found in the
Math menu.
7
Ex4: Evaluate
5
 x dx .
2
7
5
 x dx  6.264
2
Now check your answers to the previous examples using the calculator.
Page 8 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Note: All continuous functions are integrable. A discontinuous function MAY be integrable.
2
Ex5: Evaluate the integral:

4
x
dx
x
Sketch the graph and shade the region.
Area above the x-axis = 2(1) = 2
2

4
Area below the x-axis = 4(1) = 4
x
dx = (area above the x-axis) – (area below the x-axis) = 2  4  2
x
8
You Try: Evaluate the integral

0
 25  x dx using areas. Check your answer on the calculator.
2
5 x
QOD: Give an example of a function and interval on which the function is NOT integrable.
Page 9 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Syllabus Objective: 3.7 – The student will solve problems using the properties of definite integrals. 3.10 – The
student will find the average value of a function on an interval.
RULES FOR DEFINITE INTEGRALS
b

a
f  x  dx   f  x  dx

a
b
Explanation: The “width” of the subintervals will be
a b
ba
, which is the opposite of
.
n
n
a
 f  x  dx  0
a
Explanation: Integrating from a to a will create no region under the curve, which will have an area of 0.
b
b
 k  f  x  dx  k  f  x  dx , for any constant k
a
a
Explanation: The area will be multiplied by the constant.
b
b
b
a
a
  f  x   g  x   dx   f  x  dx   g  x  dx
a
Explanation: This follows from the rules of Riemann Sums.
b
c
c
b
a
 f  x  dx   f  x  dx   f  x  dx
a
Explanation: Accumulation of area over two intervals.
min f   b  a  
b
 f  x  dx  max f   b  a 
a
Explanation: The value of the “net” area will be between the signed area of the smallest rectangle and the
signed area of the largest rectangle.
If f  x   g  x  for
b
a  x  b , then

a
Page 10 of 20
b
f  x  dx  g  x  dx .

a
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Ex1: Use the properties of integrals to evaluate the following, given
5

f  x  dx  6,
2

g  x  dx  2,
1
2
2

2.
 f  x  dx  8
1
2
1.
2
h  x  dx  3 ,
1
2
2
2
1
1
1
1
2
 f  x  dx
2
5
 f  x  dx    f  x  dx  6
5
5
2
  g  x   h  x   dx
  g  x   h  x   dx   g  x  dx   h  x  dx   2  3 
5
2
3.
 g  x   h  x  dx
Not possible with the information given; no product rule for integration
1
5
4.

5
f  x  dx
f  x  dx 

1
1
2

f  x  dx 
1
5
 f  x  dx  8  6  2
2
2
5.
 4 f  x   2 g  x  dx
1
2

2
2
 4 f  x   2 g  x   dx  4 f  x  dx  2 g  x  dx  4  8   2  2   28

1

1
1

Ex2: Find the upper and lower bounds of the integral
 1  sin x  dx .
0
Upper bound: max f   b  a   1  0   
Lower bound: min f   b  a   0   0   0

So, 0 
 1  sin x  dx   .
0

Checking on the calculator, we have
 1  sin x  dx    2 , which is within our bounds.
0
Average Value of a Function: If f is integrable on  a , b  , its average (mean) value on  a , b  is
av  f  
1
ba
b
 f  x  dx
a
Think about it: To find the average (mean) of a set of numbers, we add the numbers and divide by how many values
there are. The integral is an infinite sum. So we are “adding” all of the function values and dividing by the length
of the interval.
Page 11 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Ex3: Find the average value of f  x   x 2  1 on the interval 1,4 . Does the function take on this value
at some point in the given interval?
4


1
1
av  f  
x 2  1 dx   24  8
4 1 1
3
Does the function take on this value?
8  x 2  1  x   7 ; Yes,
7 is in the interval 1,4 .
Mean Value Theorem for Definite Integrals: If f is continuous on  a , b  , then at some point c in  a , b  ,
1
f c 
ba
You Try: Let
b
 f  x  dx
a
6
10
10
2
2
6
 f  x  dx  9 and  f  x  dx  5 . What is the value of  f  x  dx ?
QOD: Describe how finding the average value of a function relates to finding the average value of a group of
numbers.
Sample AP Calculus AB Exam Question(s):
y = f (x)
B
C
A
1.
The regions A, B, and C in the figure above are bounded by the graph of the function f and the x-axis. If
3
the area of each region is 2, what is the value of
  f  x   1 dx ?
3
(A) −2
(B) −1
(C) 4
(D) 7
(E) 12
Page 12 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
The velocity, in ft/sec, of a particle moving along the x-axis is given by the function v  t   et  tet .
2.
What is the average velocity of the particle from time t  0 to time t  3 ?
(A)
(B)
(C)
(D)
(E)
3.
20.086 ft/sec
26.447 ft/sec
32.809 ft/sec
40.671 ft/sec
79.342 ft/sec
On the closed interval  2,4  , which of the following could be the graph of a function f with the property
4
1
that
f  t  dt  1 ?
4  2 2
(A)
(B)
(D)
(E)
Page 13 of 20
(C)
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Syllabus Objectives: 3.8 – The student will evaluate definite integrals using the Fundamental Theorem of
Calculus. 3.9 – The student will interpret geometrically the Fundamental Theorem of Calculus. 3.1 – The
student will determine the antiderivative of an elementary function.
The Fundamental Theorem of Calculus (Part 1): If f is continuous on  a , b  , then the function F  x  
x
 f  t  dt
a
has a derivative at every point x in  a , b  , and
x
dF d

f  t  dt  f  x  .
dx dx a

Note: The lower limit is a constant, and the upper limit is x.
Proof: Use the definition of derivative.
F  x  h  F  x
dF
 lim
dx h0
h
x h
 lim


a
dF
1
 lim
h

0
dx
h
 lim
a
h 0
So,
x h
x
f  t  dt  f  t  dt

f  t  dt . Note that
x
1
h
x h

f  t  dt
x
h 0
h
x h

h
f  t  dt is the average value of f from x to x + h. So, by the Mean
x
Value Theorem, there must exist some value c between x and x + h such that f  c  
approaches zero, c must approach x: lim f  c   f  x  . Therefore,
h 0
1
h
x h

f  t  dt . As h
x
F  x  h  F  x
dF
 lim
 f  x .
dx h0
h
Note: A simplified way of thinking about the Fundamental Theorem is that you are taking the derivative of an
antiderivative. Since these are inverses, the result is the original function.
Ex1: Find
d
dx
x
  7t
2

 1 dt .
2
Because the lower limit of integration is a constant, and the upper limit of integration is x, we can use the
Fundamental Theorem. So
Page 14 of 20
d
dx
x
  7t
2

 1 dt  7 x 2  1 .
2
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Special Case #1: What if the upper limit of integration is a variable expression other than x? We must use the chain
rule.
dy
Ex2: Find
if y 
dx
x3
 

1  cos t 2 dt .
d
By the Fundamental Theorem and the chain rule,
dx
Solution:
3
Derivative of x
0
x3

 
 
1  cos t 2 dt  1  cos x 3
2
 3x 2
0
 
dy
 3x 2 1  cos x 6
dx
Special Case #2: What if the variable is the lower limit of integration? We must use the properties of integration to
switch the limits of integration.
1
d 1
dt .
Ex3: Find
dx x t

1
Using the properties of integrals,
x
x
d 1
1
1
1
dt   dt . So 
dt   .
dx 1 t
x
t
t
x
1



Special Case #3: What if there are variables in both the lower and upper limits of integration? Use the properties of
integration to split them into two integrals.
dy
Ex4: Find
for
dx
3x
  2t ln t  dt .
x
We will choose a constant, a, that is in the domain of f  t   2t ln t . (Note: Your choice is arbitrary, but must be
in the domain of the function.) For this example, use a = 1. Split the integral into a sum of two integrals:
3x

 2t ln t  dt =
x
1

 2t ln t  dt 
x
3x

d
So,
dx
 2t ln t  dt
1
3x

 2t ln t  dt =
x
d 
  2t ln t  dt 
dx  x
1

3x

1

  2t ln t  dt 
To use the Fundamental Theorem, we must switch the limits of integration in the first integral, and use the chain rule
in the second integral.
d 
   2t ln t  dt 
dx  1
x

3x

  2t ln t  dt   2 x ln x   2 x ln x   3  4 x ln x
1
The Fundamental Theorem of Calculus (Part 2) – also known as the Integral Evaluation Theorem: If f is
continuous at every point of  a , b  , and if F is any antiderivative of f on  a , b  , then
b
 f  x  dx  F  b   F  a 
a
Page 15 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
x
Derivation: By definition and the first Fundamental Theorem,
 f  t  dt  F  x   C
a
a
We know that
 f  t  dt  F  a   C  0 , so it follows that C   F  a  .
a
b
If we let x  b , we have
 f  t  dt  F  b   C .
a
Substituting C   F  a  , we have
b
 f  t  dt  F  b   F  a  .
a
2
Ex5: Evaluate the integral.
 x
2

 x  1 dx
1
2
Find the antiderivative, F  x  of f  x   x  x  1 .
F  x 
We will use the following notation for F  b   F  a  :
2

1
x3 x2

x
3
2
2
3
2
 9
 23 2 2
   1  1
x3 x2
x  x  1 dx 

 x     2  

  1  
 2
3
2
2
 3 2
  3
1


2
Finding Total Area: To find the area between the graph of y  f  x  and the x-axis over the interval  a , b  ,
1.
Partition  a , b  with the zeros of f.
2.
3.
Integrate f over each subinterval.
Add the absolute values of the integrals.
Ex6: Find the area of the region between the curve y  x 2  9 and the x-axis on the interval 0,5 .
1.
x 2  9  0  x  3
3
2.

0
5

3
3.
Partitions:
0,3 , 3,5
3

x3
33
x  9 dx 
 9 x   9  3   0  0   18
3
3
0
2
3
 53
  33
 44
x3
x  9 dx 
 9 x    9  5      9  3  
3
3
 3
 3
0
2
18 

44
98

3
3
Check your answer on the calculator. Absolute value (abs) can be found in the Math menu.
Page 16 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Business Application: Inventory
Ex7: Suppose a wholesaler receives a shipment of 1300 cases of candy every 30 days. The candy is sold
to retailers at a steady rate, and x days after the shipment arrives, the inventory still on hand is
I  x   1300  50 x . Find the average daily inventory. Then find the average daily holding cost if
holding on to a case costs 3 cents a day.
Average Daily Inventory:
avI  x  
b
1
1
I  x  dx 
ba a
30  0

30
1
 1300  50 x dx  30 1300 x  25x
2 30
  550 cases
0
0
Average daily holding cost: 550  0.03  $16.50 / day

3
You Try: Evaluate the integral
 sec tan  d .
0
QOD: Explain the Fundamental Theorem of Calculus (Part 1) in your own words.
Sample AP Calculus AB Exam Question(s):
1.
Let f be a differentiable function with f  2   3 and f   2   5 , and let g be the function defined by
g  x   x f  x  . Which of the following is an equation of the line tangent to the graph of g at the point
where x  2 ?
(A) y  3x
(B) y  3  5  x  2 
(C) y  6  5  x  2 
(D) y  6  7  x  2 
(E) y  6  10  x  2 
x
2.
Let g be the function given by
g  x    sin  t 2  dt for 1  x  3 . On which of the following
0
intervals is g decreasing?
(A) 1  x  0
(B) 0  x  1.772
(C) 1.253  x  2.171
(D) 1.772  x  2.507
(E) 2.802  x  3
Page 17 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
y  f  x 
3.
The graph of f  , the derivative of f, is the line shown in the figure above. If f  0   5 , then f 1 
(A) 0
(B) 3
(C) 6
(D) 8
(E) 11
x

d 
  sin  t 3  dt  

dx  0

2
4.
 
sin  x 
sin  x 
2 x sin  x 
2 x sin  x 
(A)  cos x 6
(B)
(C)
(D)
(E)
Page 18 of 20
3
6
3
6
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Syllabus Objective: 3.6 – The student will approximate a definite integral using the trapezoidal sum.
Recall: Area of a Trapezoid Formula: A 
1
h  b1  b2 
2
Ex1: Use 4 trapezoids of equal heights to approximate the area under the curve y  x 2 on the interval
0,2 .
Then find the exact value.
Draw the graph and sketch the trapezoids. (Note: Trapezoid I is a “special” trapezoid with one base equal to 0.)
IV.
I.
II.
III.
I.
.
Find the areas of each trapezoid using the formula given above. Note: The bases of the trapezoids are the function
values for each value of x.
I. A 
2
1  1  2  1   1
0


 
   
2  2  
 2   16
2
1  1   2  3   13
III. A     1     
2  2  
 2   16
II. A 
2
1  1  1 
5
2

      1  
2  2  2 
16

2
1  1  3 
25
2
IV. A         2   


2  2  2 
 16
1 5 13 25 44 11
  


16 16 16 16 16
4
Sum of the areas:
Note that all of the trapezoids have the same height. Also, trapezoids I & II, II & III, and III & IV share a common
base. So instead of finding each area individually, we could put them all together:
2
2
1  1 
11
2
2
1
 3
A     0  2    2 1  2     2   

2  2  
2
2
 
 
 4
2
2
x3
8

Exact value: x dx 
The trapezoidal approximation was slightly larger. Did you expect it to be?
3 0 3
0

2
b
Trapezoidal Rule: To approximate
h
 f  x  dx , T  2  y
0
 2 y1  2 y2  ...  2 yn 1  yn  , where h 
a
ba
.
n
Teacher Note: Encourage students NOT to try to memorize this formula. Using the area of a trapezoid formula with
commonality of bases will allow them to recreate the rule. Also, the intervals may not be constant, so the heights of
the trapezoids may differ!
Page 19 of 20
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.
AP Calculus Notes: Unit 6 – Definite Integrals
Ex: The table was created by recording the temperature every hour from noon until midnight. Use the
trapezoidal rule to approximate the average temperature for the 12-hour period.
Time
Temperature
Noon
76
1
78
2
80
3
79
4
85
5
86
6
82
7
80
8
78
9
70
10
68
11
65
Midnight
63
Average Temperature =
1
12  0
12
1 1

 f  x  dx  12  2  76  2  78  2 80   2  79   ...  2  68   2  65   63  76.708
0
You Try: The table below shows the velocity of a car at 10 second time intervals. Estimate how far the car traveled
using a trapezoidal sum. Compare this answer to the answers you obtained in the You Try problem a few days ago
(LRAM and RRAM). Can you draw a conclusion based on this comparison?
Time Velocity
(sec) (ft/sec)
0
0
10
15
20
35
30
20
40
43
50
15
60
22
QOD: What is the relationship between LRAM, RRAM, and Trapezoidal Sums. Prove this relationship
algebraically.
Sample AP Calculus AB Exam Question(s):
4
If a trapezoidal sum overapproximates

f  x  dx , and a right Riemann sum underapproximates
0
which of the following could be the graph of
(A)
(B)
(D)
(E)
Page 20 of 20
4
 f  x  dx ,
0
y  f  x ?
(C)
Pearson Prentice Hall 2007 – Calculus: Graphical, Numerical, Algebraic
4.1 – 4.3
These notes are aligned to the textbook referenced above and to the College Board Calculus AB curriculum.