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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 
3.
 5cos x
5.
x2  2 x  3
dx 
 x4
7.
1 

  x  2 x  dx 
dx 
1
2.
x
4.
Find the general solution for differential
dr
 .
equation
d
dx 
4
6.
x
8.
  2t
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
 k t . The initial size of the
population size and t is the time in days  0  t  10 . That is,
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:_____________________________________
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)
Sum of cubes of 1st n positive integers =
 i3 
i 1
Example 6
n
(a)

i 1

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- 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
14
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.
15
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 
a
b
f ( x)dx

a
 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

b
f ( x)dx
a

 g ( x)dx.
a
Examples of Use:
4
Example 9: Given
 x dx  60
3
2
4
and
 dx  2,
evaluate
2
2
 x dx
3
2
4
 15dx
2
4
x
3

 4 dx
2
16
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)

6
f ( x)dx 
(b)
0
0
6
(c)

 2  f ( x) dx 
2
f ( x) dx 
0
(d)
 2 f ( x)dx 
0
(e) The average value of f over the interval [0,6] is
17
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.
Look Ma, No long sum!
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 dt
b.
2
e.

 5 x  4 dx

1

t 3  9t dt
1

2
1
1
c.
  3x
d.
x  x2
 3 dx
8 2 x
  2  csc x dx

2
2
4
18
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
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].
19
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.
20
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, 
0
2
].
F (0) 
 6 
F 
 4 
F 
 3 
F 
 2 
F 
21
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
22
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
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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.


2
3
 x x  8 dx
2
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
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AP Calculus
Lesson- Areas and Volumes
Name:____________________________________
Date:_____________________________________
Area Problems
1.
The area of the region bounded between y  x 2  2 x  3 and y  4 x  45 .
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
4.
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.
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5.
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
6.
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.
7.
The volume of the solid generated by rotating the region bounded by y  x  4 , x  0 , and y  3
around the y-axis.
8.
The volume of the solid generated by rotating the region in quadrant I bounded by y  x 2  1 ,
x  1 and y  10 around
a.
The line y  10 .
b.
The line x  3 .
c.
The line y  1 .
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