AP Calculus
Lesson- Antiderivatives and Indefinite Integrals
Name:____________________________________
Date:_____________________________________
Anti-derivative = Anti-differentiation = Working backwards
x2 is an antiderivative of 2x
Example:
The general form of the antiderivative is given as F(x) + C, where C is any constant, called the constant of
integration. Any 2 antiderivatives of f will differ only by a constant.
Notation:
Let y = x2 + C. Then
dy
 2x
dx
. This is called a differential equation, an equation that involves
x, y, and a derivative of y.
We can re-write this equation as dy  2x dx . This is called the differential form of the equation.
In general,
Let y  F ( x) and F ( x)  f ( x).
dy
 f(x)
dx
Then
dy  f(x) dx
We write this using “integral notation:”

is a differential equation and
is the differential form of the equation.
y   f ( x)dx  F ( x)  C
is the indefinite integral sign; it means “the antiderivative of f with respect to x.”
f(x) is the integrand, the function to be antidifferentiated.
x = the variable of integration
C = the constant of integration
Rules of Integration
0
dx 
 sin x
dx 
k
dx 
 sec
2
x dx 
2
x dx 
 kf ( x)
x
n
dx 
dx 
 cos x
dx 
 csc
 sec x
tan x dx =
 csc x
cot x dx 
1
Steps for Integrating:
(1) Re-write the original integral into a form in the table.
(2) Integrate (take the antiderivative).
(3) Simplify.
Examples
1.
 2x
dx 
1
2.
x
4.
Find the general solution for differential
dr
 .
equation
d
dx 
4
3.
 5cos x
5.
x2  2 x  3
dx 
 x4
6.
x
7.
1 

  x  2 x  dx 
8.
  2t
dx 
2
 sec 2 x
2

1
2

dx 
dt 
2
AP Calculus
Lesson- Rolle’s Theorem, MVT, AVT
Name:____________________________________
Date:_____________________________________
Objective:
To learn about and apply Rolle’s Theorem and the Mean Value Theorem
Rolle’s Theorem:
Given an interval (a, b) if f(a)=f(b) then there is a location in the interval where the first
derivative = 0.
MVT:
On the closed interval [a,b]:
f ' (c ) 
f (b)  f (a )
{used for finding average velocity)
ba
The idea:
The “average” slope is the slope of a line drawn from endpoint to endpoint. No matter how much the function
may change between endpoints, we say that its “average” change is simply the ratio of changes in y to changes
in x.
For average slope, use the slope formula.
The MVT says that there has to be a point somewhere between the endpoints where the instantaneous rate-ofchange (derivative) is the same as the average slope.
The formula
f (b)  f (a )
f ' (c ) 
where c is an x-value between a and b.
ba
Find the values of x that satisfy the MVT for each of the following:
1.
y = x2 on [1,5]
2.
f(x) = sin (x) on [0, ]
Determine if Rolle’s Theorem can be applied. If so, apply it. If not, explain why.
f ( x)  sin x [0,2 ]
3.
4.
f ( x)   x 2  3 x
[0,3]
3
Determine if Mean Value Theorem can be applied. If so, apply it. If not, explain why.
f ( x)  cos x  tan x
5.
6.
f ( x)  x 4  8 x
[0,2]
[0,  ]
AVT:
The “average value” of a function is a number that represents the central tendency of the output. The
calculation of this value is straightforward; it is simply the integral of the function over an interval divided by
the interval.
Average Value =
_
y
b
1
f ( x)dx
ba  a
Practice:
7.
Find the average value of y on [2, 6] for y = 2x + 3.
8.
Find the average value of y on [0, ] for y = cos(x) + x
9.
Find the value of c that makes the average value of x2 equal to 9 on [0,c].
10.
Design a function of your own whose average value is 0 on [-2, 2].
4
AP Calculus
Lesson- Antiderivatives and Initial Conditions
and Particular Solutions
Name:____________________________________
Date:_____________________________________
Objective-
To learn how to find anti-derivatives involving particular values and initial conditions.
dy
 3x 2  1 is
The general solution to the differential equation
dx
dy  (3x 2  1) dx



y   3x 2  1
dx
If we are given the value of y for one value of x – this is called an initial condition – then we can get a
particular solution; that is, we can find a particular constant C and write a specific antiderivative as our
answer.
Examples
1.
Given F ( x)   3x 2  1


dx and F (2)  4. Find the particular solution.
F(x) =
2.
Find g(x) if g ( x)  6 x 2 and g(0) = -1.
3.
A ball is thrown upward with an initial velocity of 64 ft/sec from an initial height of 80 ft. Using the
fact that the acceleration due to gravity is 32 ft/sec²,
(a) find the position function giving the height s as a function of time t.
(b) When does the ball hit the ground?
5
4.
The rate of growth dP/dt of a population of bacteria is proportional to the square root of t, where P is the
dP
population size and t is the time in days  0  t  10 . That is,
 k t . The initial size of the
dt
population is 500. After 1 day the population has grown to 600. Estimate the population after 7 days.
5.
The Grand Canyon is 1800 m deep at its deepest point. A rock is dropped from the rim above this point.
Write the height of the rock as a function of the time t in seconds. How long will it take the rock to hit
the canyon floor?
6
AP Calculus
Lesson- Sigma Notation
Name:____________________________________
Date:_____________________________________
Objective: To review sigma notation and to apply that knowledge to finding anti-derivatives.
Summation Notation
Sigma Notation
 , read “sigma, “ is the uppercase Greek letter S. It is used to denote a sum.
Let a1 be the first term of a sum,
a2 be the second term of a sum,
a3 be the third term of a sum,
.
.
.
an be the nth term of a sum.
Then we write the sum a1 + a2 + a3 +… + an in sigma notation as
n
ai

i 1
where i is called the index of summation.
6
Example 1
i 
Here ai = i.
i 1
5
Example 2
  i  1 
Here ai = i + 1.
i 0
Example 3
Write the sum 42 + 52 + 62 + 72 + 82 + 92 + 102 in sigma notation.
Example 4
Write the sum represented by
n
1
k 
k 1
6
Example 5
Expand
 f ( x )x .
i 1
i
7
Properties of Summation
(1)
n
n
i 1
i 1
 kai  k  ai
Example:
(2)
n
n
n
i 1
i 1
i 1
n
n
n
i 1
i 1
i 1
 (ai  bi )   ai   bi , by the commutative property of addition
and
 (ai  bi )   ai   bi
Summation Formulas
n
(1)
 c  cn
i 1
n(n  1)
2
n
(2)
Sum of 1st n positive integers =
i 
i 1
n
(3)
i
Sum of squares of 1st n positive integers =
2
i 1
n
(4)
st
Sum of cubes of 1 n positive integers =
i
i 1
Example 6
n
(a)

i 1
3


n(n  1)(2n  1)
6
n2 (n  1)2
4
Use the properties of summation and summation formulas to evaluate the sums:
i 1
n2
(b) Evaluate the sum in part (a) when n = 10.
8
Evaluate Each:
6
7.
15
 k (k  2)
8.
k 3
10
  2i  3
9.
i 1
 i (i
2
 1)
i 1
Write in Sigma Notation
10.
5
5
5
5


 ... 
11 1 2 1 3
1  15
12.
  2 2   2 
  2n  2   2 
1    1     ...  1    1    
   n 
  n    n 
  n
11.
  1 2    2 2 
  4 2 
1      1      ...  1     
  4     4  
  4  
How to do Summations on your Calculator
(1) In MODE menu, set 4th line to SEQ.
(2) In LIST menu (2nd STAT) select MATH menu #5, get display:
sum(
(3) In LIST (2nd STAT) menu, select OPS menu #5, press ENTER, get display:
(4) Enter, in turn:
sum(seq(
formula, n, lower bound of summation, upper bound of summation))
(5) Hit ENTER. This will return the sum of the indicated terms in your list.
Try this with Example 7, 8, and 9 above.
9
AP Calculus
Lesson- Area of a Plane Region, Upper and Lower Sums
Name:____________________________________
Date:_____________________________________
Given a continuous, non-negative function f(x) find the area under the graph of y = f(x) between the vertical
lines x = a and x = b.
Pictures:
Method:
Step 1:
Step 2:
Break the interval [a, b] into n equal parts.
Each little sub-interval so formed will have length x 
Label the endpoints of the sub-intervals as follows:
a  x0 ,
x1  x0  x,
x2  x1  x, ...,
xi  xi 1  x,
..., xn 1,
xn  b
Step 3:
Draw vertical lines at each of these endpoints to inscribe rectangles inside the region
Since f is continuous, f has a minimum value on each of these sub-intervals.
Call the x value in the i th interval where the minimum value of f occurs mi .
Step 4:
Compute the area of these inscribed rectangles. This is called a lower sum.
s(n) =
 Area under the curve of f
Step 5:
Repeat Step 3, only this time draw circumscribed rectangles.
Since f is continuous, f has a maximum value on each of these sub-intervals.
Call the x value in the i th interval where the maximum value of f occurs M i .
Step 6:
Repeat Step 4, only this time use circumscribed rectangles.
Compute the area of these circumscribed rectangles. This is called an upper sum.
Area under the curve of f  S(n) =
10
Example 1
Find the upper and lower sums for the region bounded by the graph of f ( x)  x 2 and the x-axis
between x = 0 and x = 2 for n = 4.
Sketch a graph.
Find x.
Lower Sum
Find mi for each subinterval.
Find M i for each subinterval.
Find areas and add.
Find areas and add.
Actual area A is between theses two values.
11
Example 2
Find the upper and lower sums for the region bounded by the graph of f ( x)  x 2 and the x-axis
between x = 0 and x = 2 for n = 8.
x =
mi =
Mi =
s(8) =
S(8)=

Area 
In a similar manner we, can compute s(16) =
So

and S(16) =
,
Area 
12
Example 3
Find the upper and lower sums for the region bounded by the graph of f ( x)  x 2 and the
x-axis between x = 0 and x = 2 for general n.
x =
mi =
Mi =
Theorem: If f is a continuous, nonnegative function on the interval [a, b], then the limits as n   of both the
lower and upper sums exist and are equal to each other. In symbols,
n
lim s(n)
n 
where x 
ba
,
n
 lim  f (mi )x
n 
i 1
n
 lim  f ( M i ) x
n 
i 1
 lim S (n)
n 
f (mi )  the minimum value of f on the i th subinterval, and
f ( M i )  the maximum value of f on the i th subinterval.
Definition: Let f be a continuous, non-negative function on [a,b]. The area of the region bounded by the
graph of f, the x-axis, and the vertical lines x = a and x = b is
n
ba
xi 1  ci  xi , where x 
Area = lim  f (ci ) x,
.
n 
n
i 1
Example 4
Find the area under the graph of y = 1 – x2 over the interval [-1, 1].
13
AP Calculus
Lesson- MRAM and Trapezoidal Method
Name:____________________________________
Date:_____________________________________
Example 5:
Find the Medial Rectangle Approximation for f ( x)  x 2 for 0,2 using 4 subintervals.
Sketch a graph.
Find x
Find mi for each subinterval.
Find f (mi ) for each subinterval
Find areas and add.
14
Example 5:
Find the Trapezoidal Approximation for f ( x)  x 2 for 0,2 using 4 subintervals.
Sketch a graph.
Find x
Find mi for each subinterval.
Find f (mi ) for each subinterval
Find areas and add.
15
Approximating the Area Under a Curve
1. Estimate the area under y = x4 on the interval [-1,2] using 6 subintervals and:
a. LRAM
b. RRAM
c. MRAM
d. Trapezoidal Approximation (is this an under- or over-approximation?)
2. The following table shows the velocity v (in ft/s) of a charging elephant at various intervals of time t (in
sec).
T
5
7
8
12
15
20
23
V
5
12
15
18
20
20
25
Use each method listed below to estimate the distance in feet the elephant has covered from t = 5 sec to t =
23 sec.
a. Left Endpoints
b. Right Endpoints
c. Trapezoidal
Approximation Methods
1. Find RRAM
and LRAM for f(x) =
8
x on the interval [0,1] with
four equal subintervals.
8
6
6
4
4
2
2
-5
-10
-2
5
-5
10
5
10
-2
16
-4
-4
-6
-6
2. Which of the following is true for f(x) = 4 – x2 on the interval [0,2] with6 n equal subintervals?
(a)
(b)
(c)
(d)
(e)
LRAM = RRAM
LRAM > RRAM
LRAM < RRAM
Area under curve > LRAM
Area under curve < RRAM
4
2
-10
-5
5
-2
8
-4
3. Find MRAM for f(x) = x3 on the interval[-1,1] using four equal subintervals.
-10
10
6
-6
4
-8
2
-5
5
-2
-4
2
4. Find the Trapezoidal Approximation for f(x) = x on the interval [-1,3] using 4 equal subintervals.
-6
10
-8
8
6
4
2
-10
-5
How does concavity relate to the trapezoidal approximation method?
5
10
-2
17
-4
-6
Approximating Areas under Curves
If a particle is moving with a constant velocity of 10 m/s for 5 sec, it is easy to figure out how far it has traveled.
10 ∙ 5 = 50 m. If you were to look at this graphically, you would see that this corresponds to the area of the
rectangle formed.
14
12
10
8
6
4
2
-10
-5
5
10
-2
-4
If you have a particle moving with a velocity which is constantly increasing, you can determine the distance
traveled using the same method.
-6
12
-8
10
-10
v(t) = 2t
-12
8
Distance = ½ (5)∙(10) = 25 m
-14
6
4
2
-10
-5
5
10
2
-2
Now what happens if you don’t have an easy geometric
shape to find the area of?
1.5
-4
-6
1
-8
v(t) = sin(πt/5)
-10
0.5
-12
-6
-4
-2
2
4
6
-0.5
-1
-1.5
18
-2
We can approximate the area by using geometric shapes. The most common used are rectangles. So we
partition the interval over which we want the area, and then create a rectangle (or other geometric shape) on
each subinterval. Let’s take the same curve and approximate the area under y = sin(πx/5) using 5 subintervals
and:
1.6
a.
 
x
hx = sin
1.6
5
Rectangles
from the right endpoints
b. Rectangles from the left endpoints
1.4
1.4
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
-6
-0.2
1.6
-0.4
1.4
-4
2
-2
4
2
6
 
hx = sin
4
6
-0.2
1.6
x
5
-0.4
1.4
c. Rectangles using the midpoints for the height
d. Inscribed rectangles
-0.6
1.2
-0.6
1.2
-0.8
1
-0.8
1
-1
0.8
-1
0.8
-1.2
0.6
-1.2
0.6
-1.4
0.4
-1.4
0.4
-1.6
0.2
-1.6
0.2
-1.8
-1.8
-6
2
-4
4
-2
2
6
-0.2
-0.2
1.6
-0.4
-0.4
4
6
e.1.4 Trapezoids
-0.6
-0.6
1.2
-0.8
-0.8
1
-1
-1
0.8
-1.2
-1.2
0.6
-1.4
-1.4
0.4
-1.6
-1.6
0.2
-1.8
-1.8
2
4
6
-0.2
-0.4
-0.6
-0.8
19
-1
-1.2
-1.4
Reimann Sums
Express the following limits as definite integrals:
2 2
4
2n 

 ... 


n  n
n
n 
 n
1. Right Hand Sum: lim
1  n   n  1 
 n  n 1
lim    
  ...  

n  n
n


 n   n 
3
2. Left Hand Sum:
2
lim
3. Right Hand Sum: n e
n
4. Right Hand Sum: lim
n
n2
n
3
e
n4
n
 ...  e
3n2 n
n
3




 e3 

2 3
6
 2n  n  
sin

sin

...

sin






n   n 
n
 n 
2
 2  4n 
 6n  2 
lim 64  
 ...  


n n
 n 
 n 

3
5. Left Hand Sum:
3



2
2
2
3  n  3   n  6 
 n  3n  
6. Right Hand Sum: lim 
 
  ...  
 
n n
 n  
 n   n 
7. Left Hand Sum:
1
1
2
 n 1 
1  2   1  2   1  ...  2
  1

n  n
n
n
 n  

lim
20
2   2n
n4 n2
lim 1  
 ...  

 

n n
 n   n 
  n 
2
8. Left Hand Sum:
2
1  n   n 
 n 
9. Right Hand Sum: lim 
 
  ...  

n  n
 2n  n 
 2n  1   2n  2 
2
2
2
2






Express the following limits as definite integrals:
1. lim
n 
2. lim
n 
3. lim
n 
4. lim
n 
n
 2x
i 1
i
 1x over [1,3]
n
 2  3x x
i 1
n

i 1
i 1
over [-3,2]
25  x i2 x over [0,5]
n
 cos 2x x over [0,∏/2]
i 1
i
21
AP Calculus
Activity- LRAM and RRAM for FUNCY Functions
Name:____________________________________
Date:_____________________________________
Objective-
To apply your knowledge of rectangular approximation toward finding areas involving nonpolynomial functions.
1. Use rectangular approximation to find the area under f(x) = x  1 on the interval [0, 3] with 6
subdivisions. Verify with the graphing calculator.
2. Use rectangular approximation to find the area under f(x) = sinx on the interval 0,2  with 8
subdivisions. Verify with the graphing calculator.
22
 x if  2  x  0

3. Use rectangular approximation to find the area under f(x) =  x 2 if 0  x  2 using 8 subdivisions.
4 if x  2

4. Use rectangular approximation to find the area under f(x) = x 3  1 on the interval 1,2 with n
subdivisions. Apply the Riemann Sum process to determine the actual area.
23
Finding Integrals by Areas

1. Being given that
 sin xdx  2 and using what you know about integrals, graphing curves, and using a
0
little bit of intuition, determine the values of the following integrals:
2
a.
 sin xdx
b.
 (2  sin x )dx
e.
 2
 2sin xdx
f.

x
h.  sin dx
2
0

2

 sin( x  2)dx
2
2
sin udu
 sin xdx
0
0

2. Use areas to evaluate the integral
c.

0
g.
 sin xdx
0

d.
 /2
2
i.
 cos xdx
0
4  x 2 dx .
2
2
x
3. Use areas and your knowledge of discontinuous functions to find  dx
x
1
24
AP Calculus
Lesson- Riemann Sums and Definite Integrals
Name:____________________________________
Date:_____________________________________
Consider the sum whose limit is defined to be the area of the region under the graph of (continuous,
nonnegative function) f(x) between x = a and x = b:
n
ba
f (ci ) x ,
where x 
is the same for every sub-interval.

n
i 1
If we generalize this sum to allow for varying widths in each subinterval, we have the definition of a Riemann
sum.
Definition:
Let f be defined on the closed interval [a, b] and
let  be a partition of the interval [a, b] given by a  x0  x1  x2  ...  xn1  xn  b
where xi is the width of the ith subinterval.
If ci is any point in the ith subinterval, then the sum
n
 f (c )x ,
i
i 1
i
xi 1  ci  xi
is called a Riemann sum for the partition  .
The upper and lower sums we computed in Examples 1 through 4 are Riemann sums.
Notation/Vocabulary: The width of the largest subinterval of a partition  is called the norm of the
partition and is designated by   max(x1 , x2 ,..., xn ) .
A partition where every subinterval is of equal width   x 
Definition of a Definite Integral:
ba
is called a regular partition.
n
If f is defined on the closed interval [a, b] and the limit
n
lim  f (ci )xi exists, then f is said to be integrable on [a,b] and the limit is denoted by:
 0
i 1
n
lim  f (ci )xi =
 0
i 1
b
 f ( x)dx.
a
We call this limit the definite integral of f from a to b.
a is called the lower limit of integration;
b is called the upper limit of integration.
Note: A definite integral is a NUMBER.
Example 5
An indefinite integral is a FAMILY OF FUNCTIONS.
Write these limits as definite integrals:
n
lim  6ci  4  ci  xi
on the interval [0,4]
n
 3 
lim   2 xi
 0
i 1  ci 
on the interval [1,3]
 0
2
i 1
25
3
Example 6 Use the limit definition to evaluate the definite integral:
x
dx
2
Important Theorem: The Definite Integral as the Area of a Region
If f is continuous and nonnegative on the closed interval [a, b], then the area of the region bounded by the graph
b
of f, the x-axis, and the vertical lines x = a and x = b is given by: Area 
 f ( x)dx.
a
This theorem simply states that the area under the graph of f (f continuous and nonnegative) is the limit of the
Riemann sum of any partition of f.
Example 7
Set up the definite integrals that give the area of the regions shown.
Example 8
Sketch the region whose area is described by the definite integral
8
 8  x dx ; then use a
0
geometric formula to evaluate the integral.
26
Important Properties of Definite Integrals
a
(1) If f is defined at x = a, then
 f ( x)dx  0.
a
a
(2) If f is integrable on [a,b], then

b
f ( x)dx   f ( x)dx.
b
a
(Reversing the limits of integration reverses the sign.)
(3) Additive Interval Property: If f is integrable on the closed intervals determined by a, b, and c, with
b
a  c  b ,then:
c
 f ( x)dx 
 f ( x)dx
a
a
b

 f ( x)dx.
c
(4) Constant Multiple Property: If f is integrable on [a, b] and k is a constant, then:
b
b
a
a
 kf ( x)dx  k  f ( x)dx.
(5) Additive Function Property: If f and g are integrable on [a, b], then
b
  f ( x)  g ( x)dx
b
  f ( x)dx
a
a
b

 g ( x)dx.
a
b
(6) Preservation of Sign Property: If f is integrable and nonnegative on [a, b], then 0   f ( x)dx.
a
(7) Preservation of Inequality Property: If f and g are integrable on [a, b], and f ( x)  g ( x) for every x in
b
[a, b], then
 f ( x)dx
a
b

 g ( x)dx.
a
Examples of Use:
4
Example 9: Given
3
 x dx  60
2
4
and
 dx  2,
evaluate
2
2
 x dx
3
2
4
 15dx
2
4
x
3

 4 dx
2
27
3
Example 10: Given
 f ( x)dx  4
6
and
0
 f ( x)dx  1,
evaluate
3
6

a.
3
0
6
3
6
 f ( x)dx
c.
 f ( x)dx
b.
f ( x)dx
 5 f ( x)dx
d.
3
3
Properties of Definite Integrals Practice
6
Given that the shaded region A has an area of 1.5, and
 f ( x)dx  3.5, evaluate the following integrals.
0
2
(a)
(c)

6
f ( x)dx 
(b)
 2  f ( x) dx 
0
0
6
2

f ( x) dx 
0
(d)
 2 f ( x)dx 
0
(e) The average value of f over the interval [0,6] is
28
AP Calculus
Lesson- The Fundamental Theorem of Calculus & MVT
Name:____________________________________
Date:_____________________________________
This is the important theorem that links the 2 branches of calculus--differential and integral.
Fundamental Theorem of Calculus
If the function f is continuous on [a, b] and F is an antiderivative of f on the interval [a, b], then
b
 f ( x)dx  F (b)  F (a)
a
This says that taking a definite integral, the limit of a Riemann sum, the area under a curve, is the same as
taking an antiderivative.
b
 f ( x)dx
Notation:
 F ( x)a
b
 F (b)  F (a)
a
Example 1
Use the Fundamental Theorem of Calculus to evaluate the following definite integrals:
7
a.
3
3
b.
dt
2
 t
3
e.

 5 x  4 dx
1

 9t dt
1

2
1
1
c.
  3x
d.
x  x2
 3 dx
8 2 x
  2  csc x dx

2
2
4
29
1
, the x-axis, x=1 and x=2.
x2
Example 2
Find the area of the region bounded by the graphs of y 
Example 3
Find the area under the graph of y  x  sin x from 0 to  .
2
0
2
Example 4
Evaluate
 2 x 1dx
0
Mean Value Theorem for Integrals (AKA Average Value Theorem)
b
If f is continuous on [a, b], then there exists a number c in [a, b] such that
 f ( x)dx

f (c)  b  a 
a
Example 5
Find the value of c guaranteed by the Mean Value Theorem for Integrals for f ( x) 
9
on the
x3
interval [1,3].
30
The value f(c) guaranteed by the MVT for Integrals is called the average value of f on [a, b].
Definition
b
If f is integrable on [a, b] then the average value of f on the interval [a, b] is
Example 6
1
f ( x)dx.
b  a a
Find the average value of the function f(x) = cos x on the interval [0, 
2
].
Example 7
This graph shows the velocity, in ft/sec, of a car accelerating from rest. Use the graph to
estimate the distance the car travels in 8 seconds.
31
Second Fundamental Theorem of Calculus
x

d 
If f is continuous on an open interval I containing a, then, for every x in the interval,
  f (t )dt   f ( x).
dx  a

x
The function F ( x)   f (t )dt is called an accumulation function.
a
It accumulates the area under the curve y = f(t).
x
Example 8
Let f(x) = cos x. Consider the values for the function F ( x)   cos tdt on the interval [0,  ].
2
0
F (0) 
 6 
F 
 4 
F 
 3 
F 
 2 
F 
32
Note: F(x) is a strictly increasing function.
The First Fundamental Theorem of Calculus says that integration “undoes” differentiation.
The Second Fundamental Theorem of Calculus says that differentiation “undoes” integration.
Taken together, they complete the notion that differentiation and integration are inverse processes.
Examples of Use:
Example 9
Find F as a function of x, then evaluate it at x = 2, 5, and 8.
x
 t
3

 2t  2 dt
2
Example 10
Integrate to find F(x), then differentiate to show the 2nd Fundamental Theorem holds.
x
F ( x)   tdt
4
Example 11
Use the 2nd Fundamental Theorem to find F ( x ).
x
F ( x)   sec3tdt
0
2nd Fundamental Theorem and the Chain Rule
What happens when the upper limit of an accumulation function is not x, but a function of x???
dF dF du


Make a u-substitution, then use the chain rule:
dx du dx
Example 12
Find F ( x) if F ( x) 
x2
1
t
3
dt.
2
x3
Example 13
Find F ( x) if F ( x)   cos t dt.
2
33
Estimating Accumulation Functions
Part I: Evaluate the integrals below:
a) for f(x) = 10 x3, determine F(x).
4   x 
 1    e 5 
 5 

4
b) for f(x) =
c) for f(x) =
 4 

 x 
 1  e 5  




2
, determine F(x).
1
, determine F(x).
x 3
d) for f(x) = -3 (2 x-2), determine F(x)
Part II.
The function v(t) = 0.2(x – 5)2 gives the speed of a rocket in kilometers per second for the interval 5 < t < 20.
i)
How quickly is the rocket moving after 10 seconds?
ii)
How quickly is the rocket accelerating after 10 seconds?
iii)
How far has the rocket traveled after 10 seconds? Estimate your answer using delta x = one second and
both the left and right rectangle methods.
iv)
Estimate this same distance using the trapezoid method.
v)
How far does the rocket travel between 10 and fifteen seconds? Estimate using the same criteria.
vi)
How does your answer compare to the algebraic calculation of the definite integral?
34
Improper Integrals
We can use limits to evaluate integrals that have  as one or both of the limits of integration. Integrals of this
type are call ‘improper’ integrals. Some improper integrals have solutions (they converge), others do not.
Example 1:


3x  2 dx
10
Example 2:
10


4 x 3dx
Example 3:
3

7e 7 x dx

Example 4:


15
5(0.36)t dt
Example 5:
The rate of decay of U-238, an isotope of uranium is found to follow the function r(t) = -[1.55(0.9999999845)t] 
10-6, where time is measured in years and r(t) is mg/yr.
Use this model to determine how many milligrams of U-238 were in the original sample. HINT: Eventually,
all of the isotope will decay. Therefore the original amount will be the same as the amount that decays as t 
.
35
AP Calculus
Lesson- Integration by Substitution
Name:____________________________________
Date:_____________________________________
This is a technique that allows us to integrate a great many functions by putting them in the form of some
function we have in our table of integrals.
d
 F ( g ( x))  F ( g ( x))  g ( x) , so,
dx
by the Fundamental Theorem,  F ( g ( x))  g ( x)dx  F ( g ( x))  C .
By the Chain Rule,
If we consider u = g(x), some function of x, then du  g ( x)dx and
 f (u )du 
F (u )  C
Steps for Integrating by Substitution—Indefinite Integrals
1.
2.
3.
4.
5.
6.
Choose a substitution u = g(x), such as the inner part of a composite function.
Compute du  g ( x)dx .
Re-write the integral in terms of u and du.
Find the resulting integral in terms of u.
Substitute g(x) back in for u, yielding a function in terms of x only.
Check by differentiating.
Examples
2
1.
 2 x x  1 dx
2.
 cos
4.

3.
5.
2

x sin xdx
t  2t 2
 t dt
6.
x

2
x3  1dx
x3
1  x 
4
2
dx
Solve the differential equation:
dy
x4

2
dx
x  8x  1
36
Integration by Substitution – Indefinite Integrals with Initial Conditions
Use the method above to find a general solution, then use the initial condition to find C and a particular
solution.
7.
Find a function f that satisfies f ( x)  sec2 (2 x)
and whose graph passes through the point
 
 ,2 .
2 
8.
Find f if f ( x)  2 x 8  x 2 and f(2) = 7.
Integration by Substitution – Definite Integrals
Change of Variables for Definite Integrals:
If the function u = g(x) has a continuous derivative on [a, b] and f is continuous on the range of g, then
g (b )
b
 f ( g ( x))dx

a

f (u )du
g (a)
Steps for Integrating by Substitution—Definite Integrals
1.
2.
3.
4.
5.
Choose a substitution u = g(x), such as the inner part of a composite function.
Compute du  g ( x)dx . Compute new u-limits of integration g(a) and g(b).
Re-write the integral in terms of u and du, with the u-limits of integration.
Find the resulting integral in terms of u.
Evaluate using the u-limits. No need to switch back to x’s!
4
9.
 x x
2
3

2
 8 dx
2
10.
2
3
4  x 2 dx
0


11.
x
 sin x  cos 2 x dx
0
12.
4


cos xdx 
4
2
13.
 x dx 
3
2
37
AP Calculus
Lesson- Areas and Volumes
Name:____________________________________
Date:_____________________________________
S(t) = 3.7(1.19376)t, where S(t) is the yearly sales of a
European tire manufacturer, in millions of dollars, for t
years since 1974.
X:[0,14] Y:[0,50]
A(t) = 0.04t3 – 0.54tx +2.5t +4.47, where A(t) is the
yearly sales of an American tire manufacturer, under
the conditions.
1. What is the total disparity, in millions of dollar,
between the two companies from 1974 to 1988.
2. What was the average disparity over the 14 years?
Here’s a checklist for answering #1, #2

Find the point of intersection.

Establish which function is on top from 0 (1974) to the point of intersection.

Integrate the difference between the upper function and the lower.

Do the same for the second part.

Add these two positive areas together to get the total area.

Divide your answer by the total number of years being examine (14).
Area Problems
1.
The area of the region bounded between y  x 2  2 x  3 and y  4 x  45 .
38
2.
The area of the region bounded by y  ln x , the x and y axes, and y  3
a.
With respect to x.
b.
With respect to y.
3.
The area of the region in quadrant I bounded by x 2  y 2  8 , x 
a.
With respect to x.
b.
1 2
y , and y  0
2
With respect to y.
Volume Problems
The goal is to find the VOLUME of a shape by adding up the volumes of slices. In general, the formula is

b
a
AREA( x)dx .
Examples (Calculator Intensive)
1. Find the volume of a solid created by revolving the line y = 4x on [2, 4] over the x-axis.
2. Find the volume of a solid created by revolving the lines y = e-x and y = x on [2, 4] over the x-axis.
39
3.
Find the volume of the solid whose base is the region in quadrant I that is bounded by y  x3 , y  0 , and
x  2 . All cross-sections perpendicular to the x-axis are
a.
Rectangles with height twice the width.
b.
Semi-circles.
4.
The volume of the solid generated by rotating the region bounded by y  3x , y  0 , between x  0 and
x  4 around the x-axis
5.
The volume of the solid generated by rotating the region in quadrant I bounded by y  sin x and
y  cos x around the x-axis.
6.
The volume of the solid generated by rotating the region bounded by y  x  4 , x  0 , and y  3
around the y-axis.
7.
The volume of the solid generated by rotating the region in quadrant I bounded by y  x 2  1 ,
x 1
y

10
and
around
a.
The line y  10 .
b.
The line x  3 .
c.
The line y  1 .
40
Using Integration Rules- Class Work
1
1. Suppose

4
f ( x)dx  5,
1

1
f ( x)dx  2,
1
 h( x)dx  7
1
Find each of the following integrals, if possible:
1
4
(a)
 f ( x)dx 
(b)
4
1
1
(c)
1
 [2 f ( x)  3h( x)]dx 
(d)
1
4
(e)  h( x )dx 
(f)
2
2


1
1
1
(b)
5
(d)
5
 [ f ( x)  g ( x)]dx 
1
 f ( x)dx 
2
1
(e)
 g ( x)dx 
5
2
 3 f ( x)dx 
1
 g ( x)dx  8
2
(c)
1
f ( x)dx   h( x)dx 
5
f ( x)dx  6,
2
 g ( x)dx 

1
5
f ( x)dx  4,
1
(a)
 f ( x)dx 
0
2
2. Suppose
 f ( x)dx 
5
(f)
 [4 f ( x)  g ( x)]dx 
1
41
Rules for Definite Integrals
42
Integral Formulas
(Yes, you are going to need to memorize these)
n
 x dx 
1 n1
x  C , n  -1
n 1
1
 x dx  ln x  C
1 x
a C
ln a
x
x
e
dx

e
C

x
a
 dx 
 sin xdx   cos x  C
 cos xdx  sin x  C
2
sec
 xdx  tan x  C
2
csc
 xdx   cot x  C
 sec x tan xdx  sec x  C
 csc x cot xdx   csc x  C
1
1
dx

tan
xC
 1  x2

1
1  x2
dx  sin 1 x  C
43
Integration as Net Change
Ex. A car moving with initial velocity of 5 mph accelerates at the rate of a(t) = 2.4t mph/sec for 8 sec.
(a)
How fast is the car going when the 8 seconds are up?
(b)
How far did the car travel during those 8 seconds?
1. v(t) = 5 cos t
0 < t < 2π
(a) determine when the particle is moving to the left, to the right, and stopped
(b) find the particle’s displacement for the given time interval
(c) find the total distance traveled by the particle
44
2. The following is a graph of velocity. The particle starts at x = 2 when t = 0.
(a) Find where the particle is at the end of the trip
(b) Find the total distance traveled by the particle
6
4
2
5
10
-2
-4
-6
3. Suppose that a tank initially contains 1000 gal of water and that the rate of change of its volume after the
tank drains for 5 min is V’(t) = 0.8t – 40 (in gallons per min). How much water does the tank contain
after it has been draining for a half-hour?
4. Suppose that the population in Juneau in 1970 was 125 (in thousands) and that its rate of growth t years
later was P’(t) = 8 + 0.5t + 0.03t2 (in thousands per year). What was its population in 1990?
45
Net and Total Area
Find the net and total area of each function over the given interval analytically:
1.
y = x3 – 4x , [-2,2]
2.
y = sin x , [0,3π/2]
Find the total area of each shaded region.
3.
4.
5.
6.
46
7.
8.
9.
10.
11.
47
Slope Fields
1. One of the following slope fields is for
dy
dy
 x  y and the other is for
 x  y . Which slope field
dx
dx
goes with which equation?
dy
 1.1x , and the other is for
dx
2
2. One of the following slope fields is for the differential equation
dy
 1 .1 y .
dx
2
Why do both graphs have positive slopes?
Which slope field goes with which differential equation?
3. One of the following slope fields is for the differential equation
dy
 siny 3 .
dx
Which slope field goes with which equation?
dy
 siny 2 , and the other is for the
dx
differential equation
Could you use the same reasoning to distinguish between
dy
dy
 cos y 2  and
 cos y 3  ?
dx
dx
48
4. Create a slope field for the differential equation
Slope
(-2,2)
(-2,1)
(-2,0)
(-2,-1)
(-2,-2)
6
dy
 xy
dx
Slope
(-1,2)
(-1,1)
(-1,0)
(-1,-1)
(-1,-2)
at the given points:
Slope
(0,2)
(0,1)
(0,0) 4
(0,-1)
(0,-2)
Slope
(1,2)
(1,1)
(1,0)
(1,-1)
(1,-2)
Slope
(2,2)
(2,1)
(2,0)
(2,-1)
(2,-2)
2
-5
5
-2
-4
-6
49
AP Calculus
Lesson- Cash Flow
Name:______________________________
Date:_______________________________
The facts:
 10% of your salary goes into an annuity;
 Your monthly salary is currently $2500;
 You will keep the investment for 10 years;
 You have secured a fixed rate of 7.5%; and
 Your salary grows 5% yearly.
Let’s forget the growth of salary for a minute. The setup looks like this:

10
0
(deposits)(interest)dt
The details are:
 Deposits $250 monthly
 Time period: months
 Time interval: 10 years by 12 months yearly = 120 months
 Rate (r) = 7.5% or 0.075
This yields

120
0
 0.075/12120 t 
250 e
dt or approximately $44,680.
It includes interest on the annuity stream for all 10 years, but fails to account for the growth in
your salary.
Let’s add the salary growth part.
nt
120
r
0.05 
Periodic growth is modeled by FV  P 1   . In this case, it is deposit  250 1 
 . This
12 
 n

part replaces the deposit portion of our integrand.
Together, we get

120
0
120
 0.05   0.075/12120t 
250 1 
dt , which is approximately $56,165
 e
12 

50