4.1 Antiderivatives
The process of recovering a function F(x) from its derivative f(x) is called
antidifferentiation.
Example: Find an antiderivative for each of the following functions.
(a) f(x)=2x
(b) g(x)= cos x
Thus the most general antiderivative of f on I is a family of functions F(x)+C
Whose graphs are vertical translates of one another.
Examples
Find an antiderivative of f(x)=3x2 that satisfies F(1)=-1.
Antiderivative Formulas
Examples
Find the general antiderivative of each of the following functions.
(a) f(x)=x4,
(b) g ( x)  1 , (c) h(x)=sin 2x,
(d) k(x)= e2x
x
3
 sin 2 x .
Example: Find the general antiderivative of f ( x) 
x
Initial Value Problems and Differential Equations
Finding an antidervative for a function f(x) is the same problem as
dy
 f ( x ). This is
finding a function y(x) that satisfies the equation
dx
called a differential equation.
This function is found by taking the antiderivative of f(x). We fix the
arbitrary constant arising in the antidifferentiation process by
specifying an initial condition y(x0)=y0.
This condition means the function y(x) has the value y0 when x=x0.
The combination of a differential equation and an initial condition is
called an initial value problem.
We solve the differential equation by finding its general solution. We
then solve the initial value problem by finding the particular solution
that satisfies the initial condition y(x0)=y0.
Indefinite Integrals
The process of finding antiderivatives is called antidifferentiation or integration.
d
[ F ( x)]  f ( x) , then it can be recast using integral notations
Thus, if
dx
.
 f ( x)dx  F ( x)  C
where C is understood to represent an arbitrary constant.
d
[  f ( x)dx]  f ( x)
Note:
dx
 f ( x)dx

is called an indefinite integral.
Is called an integral sign, the function
f(x) is called the integrand, and the constant C is called the constant of integration.
dx in
d
[ ] and  [ ]dx
dx
serves to identify the independent variable. If it is different from x,
then the notation must be adjusted appropriately. Thus,
d
[ F (t )]  f (t ) and
dt
 f (t )dt  F (t )  C
are equivalent statements.
For example:
Note:
d 3
[ x ]  3x 2 is equivalent to  3x 2 dx x3  C
dx
d
[tan t ]  sec 2 t is equivalent to  sec 2 tdt  tan t  C
dt
 1 dx can be written as  dx
1
dx
dx
can
be
written
as
 x2
 x2
Example:
x5
 x dx  5  C
4
 cos xdx  sin x  C

1
2
xdx   x dx 
x
1
1
2
1
1
2
2 32
C  x C
3
Example: Evaluate (a)
(b)
Solution (a):
Solution (b):
 5sin xdx
4
(
x
  3x)dx
sin x
dx
Example: (a) 
2
cos dx
3
(b) t  4t  5 dt
 t
Solution (a):
Solution (b):
4.2 Area and Estimating with Finite Sums
While we do not yet have method for determining the exact area of R, we can
approximate it in a simple way.
Estimating Areas by Finite Sums
The area under the graph of a positive function can be approximated by finite sums.
If the interval [a, b] is divided into n equal subintervals of equal width Δx=(b-a)/n,
and if f(ck) is the value of f at the chosen point ck in the kth subinterval, the
process gives a finite sum of the form
f(c1)Δx+ f(c2)Δx+ f(c3)Δx+…+ f(cn)Δx.
The choices for the ck could maximized or minimize the value of f in the kth
Subinterval, or give some value in between. The true value lies somewhere
Between the approximations given by upper sums and lower sums.
Example
Suppose we want to find the area of the shaded
Region R that lies above the x-axis, below the graph
of y=1-x2 , and between the vertical lines x=0 and x=1.
4.3 Sigma Notation and Limits of Finite Sums
Sigma notation enables us to write a sum with many terms in the compact form
n
a
k 1
k
 a1  a2  a3
 an 1  an
• The Greek Letter  stands for “sum”.
• The index of summation k tells us where the sum begins and where it ends.
• Any letter can be used to denote the index, but the letters i, j, and k are customary.
Example
Consider the sum1  2  3  4  5
K is one of the integers from 1 to 5.
2
2
2
2
2
in which each term is of the formk 2
, where
5
k
In sigma notation, this sum can be written as 
k 1
2
,
2
k
which is read “the summation of , where k runs from 1 to 5. ”
More generally, if f(k) is a function of k, and if m and n are integers such that m  n,
Then
n
 f (k )
k m
Denotes the sum of the terms that result when we substitute successive integers for
K, starting from k=m and ending with k=n.
Examples
Note: The lower limit of summation does not have to be 1; it can be any integer.
n
f (k )
The numbers m and n in k
are called, respectively, the lower and upper limits
m
Of summation; and the letter k is called the index of summation.
Note that
•It is not essential to use k as the index of summation;
6
1 6 1 6 1
1 1 1 1 1



1

   



i
j
j
2
3 4 5 6
i 1
j 1
j 1
n
•If the upper and lower limits of summation are the same, then the “sum” in
Reduces to a single term.
3
 f (k )
k m
2
2
2

2
i

1
2

1
i 2
k  3 ,
3
k 3
3
•If the expression to the right of the  does not involve the index of summation,
we take all the terms in the sum to be the same, with one term for each allowable
value of the summation index.
2
 3  3  3  3,
i 0
3
2
2
2
2
x

x

x

x

j 1
Changing the limits of summation
A sum can be written in more than one way using sigma notation with different limits
Of summation and correspondingly different summands.
5
4
7
i 1
j 0
k 3
For example  2i  2  4  6  8  10   (2 j  2)   (2 j  4)
6
Example: Express
is 0 rather than 3.
6
4
k 1
4
k 1
in sigma notation so that the lower limit of the summation
3
 42  43  44  45
k 3
 40 2  41 2  42 2  43 2
3
 4
i 0
i2
3
  4k  2
k 0
Properties of Sums
Example:
n
(a)
 (3k  k
2
)
k 1
3
(b)
 (k  4) 
k 1
Summation Formulas
In closed form
In open form
10
Example: Evaluate
 k (k  2)
k 1
Solution:
n
Example: Evaluate
 (4  k )
k 1
Solution:
2
Limits of Finite Sums
The finite sum approximations we considered in section 4.2 got more
accurate as the number of terms increased and the subinterval
widths became thinner.
Example. Find the limiting value of lower sum approximations to the
area of the region R below the graph y=1-x2 and above the interval
[0, 1] on the x-axis using equal width rectangles whose widths
approach zero and whose number approaches infinity.
Riemann Sums
The theory of limits of finite approximations was made precise by Riemann.
We now introduce the notion of a Riemann sum, which underlies the theory of the
Definite integral studied in the next section.
The sum Sp is called a Riemann sum for f on the interval [a, b].
4.4 The Definite Integral
Notation and Existence of the Definite Integral
Properties of Definite Integrals
Area Under the Graph of a Nonnegative Function
A Definition of Area
Suppose f is continuous and nonnegative on [a, b], and let R be the region bounded
below by the x-axis, bounded on the sides by x=a and x=b, and bounded above by
The curve y = f(x). Using the rectangle method, we can motivate the definition for
The area of R as follows:
• Divide [a, b] into n equal subintervals by inserting n-1 equally spaced points
Between a and b, and denote those points by x1 , x2 , , xn1 . Each of these
Subintervals has width x  b  a
n
•Over each subinterval construct a rectangle
whose height is the value of f at an arbitrary
selected point in the interval.
x*1 , x*2 , , x*n denote the points selected,
If
then the rectangle will have heights
f ( x*1 ), f ( x*2 ), , f ( x*n )
, and areas f ( x*1 )x, f ( x*2 )x, , f ( x*n )x
•The union of the rectangles forms a region whose
area can be regarded as an approximation to the
area A of the region R.
A  area(R)  area(Rn )  f ( x1* )x  f ( x2* )x 
n
A   f ( xk* )x
k 1
 f ( xn* )x
•Repeat the process using more and more subdivisions, and define the area A as
n
A  lim  f ( xk* )x
n 
k 1
In summary, we make the following definition
The Choice of xk*
n
f ( xk )x

The values of xk* in A  nlim

k 1
Chosen in some systematic fashion.
*
can be chosen arbitrarily. In practice they are
Suppose that the interval [a, b] is divided into n equal parts of length x=(b-a)/n by
the points x1, x2, …, xn-1, and let x0=a and xn=b. Then xk=a+kx for k=0, 1, 2, …, n.
Some common choices of xk* are
•The left endpoint of each subinterval: xk*=xk-1=a+(k-1)x
•The right endpoint of each subinterval: xk*=xk=a+kx
•The midpoint of each subinterval: xk*=1/2(xk-1+xk)=a+(k-1/2)x
a
a+x a+2x
a+(n-1)x
b=a+nx
x
x
x0
x1
x2
xn-1
xn
.......
a+3x
x
x
x3
x
x
.......
x
Example
Find the area between the graph of f(x)=x2 and the interval [0, 2] with xk* as
the right endpoint of each subinterval.
Solution:
Average Value of a Continuous Function Revisited
4.5 The Fundamental Theorem of Calculus
The Mean Value Theorem for Definite Integrals
Fundamental Theorem of Calculus, Part I
Examples
Use the Fundamental Theorem to find dy/dx if
(a) y  x cos tdt
a
5
(b) y   3t sin tdt
x
2
(c) y  x cos tdt

1
The Fundamental Theorem of Calculus, Part II
(The Evaluation Theorem)
The theorem says that to calculate the definite integral of f over [a, b] all we need to
do is
1. Find an antiderivative F of f, and
2. Calculate the nubmer

b
a
f ( x )  F ( b)  F ( a )
The usual notation for F(b)-F(a) is F ( x )ba or [ F ( x )ba depending on whether F has one
or more terms.
Examples

(a)

0
(b)

1/2
(c) 
1
dx
1  x2
0
4
cos xdx
(
3
x
x  )dx
2
2
Examples
The FTC can be applied to definite integrals in which the lower limit of integration is
Greater than or equal to the upper limit of integration.
Example:

Example:
2
2

x 3 dx
1
2
xdx
Total Area
To compute the value of the area of the region bounded by the graph of a
Function y=f(x) and the x-axis when the function takes both positive and
negative values, we must be careful, otherwise we might get cancellation
between positive and negative signed area.
The correct total area is obtained by adding the absolute value of the definite
integral over each subinterval where f(x) does not change sign.
Example
Figure below shows that graph of the function f(x)=sinx between x=0 and x=2.
Compute
(a) The definite integral of f(x) over [0, 2].
(b) The area between the graph of f(x) and the
x-axis over [0, 2]
Summary
Example
Find the area of the region between the x-axis and the graph of f(x)=x3-x2-2x, -1≤x ≤ 2.
4.6 Indefinite Integrals and Substitution Rule
The Fundamental theorem of Calculus says that a definite integral of a
Continuous function can be computed directly if we can find an antiderivative of
the function.
In section 4.1, we defined the indefinite integral of the function f with respect to x.
We must distinguish carefully between definite and indefinite integrals. A definite
integral is a number, where an indefinite integral is a function plus an arbitrary
constant C.
Suppose that F is an antiderivative of f and that g is a differentiable function. The
Chain rule implies that the derivative of F(g(x)) can be expressed as
d
[ F ( g ( x))]  F '( g ( x)) g '( x)
dx
Which we can write in integral form as
 F '( g ( x)) g '( x)dx  F ( g ( x))  C
Or since F is an antiderivative of f,
 f ( g ( x)) g '( x)dx  F ( g ( x))  C
Let u  g ( x) and write
Then we have
du
 g '( x) as du  g '( x)dx
dx
 f (u )du  F (u )  C
Then process of evaluating an integral of
By converting it into
 F '( g ( x)) g '( x)dx  F ( g ( x))  C
 f (u )du  F (u )  C with the substitution
u  g ( x) and du  g '( x)dx
is called the method of u-substitution.
Substitution: Running the Chain Rule Backwards
The substitution Rule provides the following method to evaluate the integral
 f ( g ( x))g '( x)dx
When f and g’ are continuous functions:
1. Substitute u=g(x) and du=g’(x)dx to obtain the integral
2. Integrate with respect to u.
3. Replace u by g(x) in the result.
 f (u)du .
Examples
Example: Evaluate
Solution:
2
60
(
x

3)
 2 xdx

Example:
 cos( x  4)dx
Solution:
Example:
Solution:
78
(
x

100)
dx

Example: Evaluate
 sin 4xdx
Solution:
Example: Evaluate
Solution:

dx
1
( x  4)6
2
dx
Example: Evaluate
Solution:

dx
1  5x
2
dx
Example: Evaluate
Solution:
x
2
(
e

csc
 x) dx

Example: Evaluate
Solution:
2
cos
 x sin xdx
Example: Evaluate

e
x
x
dx
Solution:
Example: Evaluate
Solution:
5
6
t
2

7
t
dt

4.7 Substitution and Area Between Curves
There are two methods for evaluating a definite integral by substitution.
•The first method is to find an antiderivative using substitution, and then to evaluate
the definite integral by applying the Fundamental Theorem.
•The second method extends the process of substitution directly to definite integrals
by changing the limits of integration.
Examples
Ex1. Evaluate
Ex2. Evaluate

1

ln 2
1
0
3x 2 x 3  1dx
e3 x dx
Definite Integrals of Symmetric Functions
Example
Evaluate

2
2
( x 4  4 x 2  6)dx
Area Between Curves
Examples
Find the area of the region enclosed by the parabola
Y=2-x2 and the line y=-x.
Examples
Find the area of the region enclosed by the parabola Y=2-x2 and the line y=-x.