In this handout,
4.7 Antiderivatives
5.3 Evaluating Definite Integrals
5.4 The Fundamental Theorem of
Calculus
Antiderivatives
Definition: A function F is called an antiderivative of f on an
interval I if F’(x) = f(x) for all x in I.
Example: Let f(x)=x3. If F(x) =1/4 * x4 then F’(x) = f(x)
Theorem: If F is an antiderivative of f
on an interval I, then the most general
antiderivative of f on I is
F(x) + C
where C is an arbitrary constant.
F is an antiderivative of f
Table of particular antiderivatives
Function Antiderivative Function Antiderivative
xr, r  -1
xr+1/(r+1)
1/x
ln |x|
sin x
- cos x
ex
ex
cos x
sin x
ax
ax / ln a
sec2 x
tan x
1
1 x 2
sin-1 x
1
1 x 2
tan-1 x
af + bg
aF + bG
In the last and highlighted formula above we assume that F’ = f and G’ = g.
The coefficients a and b are numbers.
These rules give particular antiderivatives of the listed functions. A general
antiderivative can be obtained by adding a constant .
Computing Antiderivatives
Problem
Let f  x  
cos  x 
2

3
. Find antiderivatives of the function f.
2
1 x
Solution
An antiderivative of the function cos  x  is sin  x 
1
is arctan  x .
2
1 x
sin  x 
Hence an antiderivative of the given function is
 3arctan  x  .
2
and an antiderivative of
Use here the formula
(aF + bG)’ = aF’ + bG’
Analyzing the motion of an object
using antiderivatives
Problem
A particle is moving with the given data. Find the position of the
particle.
a(t) = 10 + 3t -3t2, v(0) = 2, s(0) = 5
Solution
v(t) is the antiderivative of a(t): v’(t) = a(t) = 10 + 3t -3t2
Antidifferentiation gives v(t) = 10t + 1.5t2 – t3 + C
v(0) = 2 implies that C=2; thus, v(t) = 10t + 1.5t2 – t3 + 2
s(t) is the antiderivative of v(t): s’(t) = v(t) = 10t + 1.5t2 – t3 + 2
Antidifferentiation gives s(t) = 5t2 + 0.5t3 – 0.25t4 + 2t + D
s(0) = 5 implies that D=5; thus, s(t) = 5t2 + 0.5t3 – 0.25t4 + 2t + 5
Evaluation Theorem
Theorem: If f is continuous on the interval [a, b], then

b
a
f ( x)dx  F (b)  F (a)
where F is any antiderivative of f, that is, F’=f.
Example:
1 3 1 3 1
0 x dx  F (1)  F (0)  3 1  3  0  3
1
2
since F(x)=1/3 * x3 is an antiderivative of f(x)=x2
Indefinite Integral
Indefinite integral is a traditional notation for antiderivatives
 f ( x)dx  F ( x) means
F ( x)  f ( x)
Note: Distinguish carefully between definite and indefinite
integrals. A definite integral is a number, whereas an
indefinite integral is a function (or family of functions).
The connection between them is given by the Evaluation
Theorem:

b
a
f ( x)dx  f ( x)dx

b
a
Some indefinite integrals
n 1
x
n
x
 dx  n  1 + C , if n  -1.
x
x
e
dx

e
 C,

1
 x dx  ln(| x |)  C
x
a
x
a
 dx  ln(a)  C
 sin( x)dx   cos( x)  C,
 cos( x)dx  sin( x)  C,
 sec ( x)dx  tan( x)  C ,
 csc ( x)dx   cot( x)  C ,
 sec( x) tan( x)dx  sec( x)  C,  csc( x) cot( x)dx   csc( x)  C,
2
1
1
dx

tan
( x)  C ,
 x2  1
2

1
1  x2
dx  sin 1 ( x)  C ,
The Fundamental Theorem of
Calculus
The theorem establishes a connection between the two
branches of calculus: differential calculus and integral
calculus:
Theorem: Suppose f is continuous on [a, b].
x
1. If g ( x)   f (t )dt then g ( x)  f ( x)
a
b
2.  f ( x)dx  F (b)  F (a), where F is any antideriva tive of f , that is F   f
a
Example of Part 1: The derivative of g ( x) 
2
is g ( x)  x
x
 t dt
0
2
Average value of a function
The average value of function f on the interval [a, b] is
defined as
b
f ave
1

f ( x)dx

ba a
Note: For a positive function, we can think of this
definition as saying area/width = average height
Example: Find the average value of f(x)=x3 on [0,2].
f ave
2
1 2 3
1 x 
1 24

x dx       2

20 0
2  4 0 2 4
4
The Mean Value Theorem
for Integrals
Theorem: If f is continuous on [a, b], then there exists
a number c in [a, b] such that
f (c)  f ave
that is,

b
a
1 b

f ( x)dx

ba a
f ( x)dx  f (c)(b  a )
Example: Find c such that fave=f(c) for f(x)=x3 on [0,2].
From previous slide, f(c)=fave=2.
Thus, c3=2, so c  3 2  1.26