ANTIDERIVATIVES v.04
d f HxL
For any two functions f(x) and g(x), if d x = gHxL, or what is the same f ¢ HxL = gHxL ,
f HxL is said to be an antiderivative of gHxL.
In front of each function on the first column you find several of its antiderivatives. Double
check that they are indeed antiderivatives of the function on the first column
Function
Some Antiderivatives
f HxL =
g HxL = x
g HxL =
1
x
g HxL = 2x
1
2
x2
f HxL = Ln x
f HxL =
1
LnH2L
h HxL =
1
2
x2 + 3
h HxL = Ln x +10
2x h HxL =
1
LnH2L
r HxL =
1
2
x2 - 5.7
r HxL = Ln x -3
2x + 3 r HxL =
1
LnH2L
2x - 2
As a consequence of the Mean Value Theorem, functions with the same derivative differ
only by a constant. It is, if f HxL is an antiderivative for gHxL, then any other antiderivative
for gHxL is of the form f HxL + C , C a constant.
The graph below shows the function y = x and three different antiderivatives
y = 12 x2 , y = 12 x2 + 2, y = 12 x2 - 4
6
4
2
0
-2
-4
-3
-2
-1
0
1
2
3
2
Integration Antiderivatives.nb
The family of functions gHxL =
gHxL =
1
2
1
2
x2 + C , C a constant, are antiderivatives for f HxL = x.
x2 + C is called the general antiderivative of f HxL = x and is denoted as
Ù x âx =
1
2
x2 + C
Notice that each value of C will yield a different antiderivative. Since these antiderivatives are obtained by vertical translations, they do not intersect. It means that if we choose
a point in the plane, and there is an antiderivative passing through that point, it has to be
unique.
EXAMPLE 1
Find the antiderivative of f HxL = x passing through the point (1, -2).
Since the general antiderivative is of the form gHxL =
1
2
x2 + C, we need to find the value of C
such that (1, -2) is on that function. Need to solve -2 =
antiderivative we are looking for is gHxL =
1
2
x2 -
1
2
H1L2 + C, C = - 52 . Hence, the
5
2
EXERCISE 1
a. Find the antiderivative of f HxL = x + 2 passing through the point (2, -2).
b. Find the antiderivative of gHxL = 3 - x2 passing through the point (0,4)
PROPERTIES OF ANTIDERIVATIVES
Antiderivatives have the same basic properties as derivatives:
1. Ù c f HxL â x = c Ù f HxL â x, where c is a constant
2. Ù H f HxL + gHxLL â x = Ù f HxL â x+Ù gHxL â x
Integration Antiderivatives.nb
Antiderivatives have the same basic properties as derivatives:
1. Ù c f HxL â x = c Ù f HxL â x, where c is a constant
2. Ù H f HxL + gHxLL â x = Ù f HxL â x+Ù gHxL â x
BASIC ANTIDERIVATIVES
3
4
Integration Antiderivatives.nb
This is the list of most common antiderivatives you will
encounter:
1. Ù â x = x + C
2. Ù xn â x =
3. Ù
1
x
xn+1
n+1
+ C, n ¹ -1
â x = Ln x +C
4 Ù ãx â x = ãx + C
5. Ù ax â x =
1
LnHaL
ax + C, a > 0
6. Ù cosHxL â x = sinHxL + C
7. Ù sinHxL â x = -cosHxL + C
8. Ù sec2 HxL â x = tanHxL + C
9. Ù secHxL tanHxL â x = secHxL + C
10. Ù
1
1+x2
11. à
â x = arctanHxL + C
1
â x = arcsinHxL + C
1-x2
EXERCISE 2
Find general antiderivative below. For the ones that have a point after it, also find the particular anti-derivative passing through the given point.
1. Ù 3 â x , (1,2)
2. Ù Hx - 0.5L â x , (-1, 1)
3. Ù Ix2 + x -
1
+ 2M â x,
(1,1)
lar anti-derivative passing through the given point.
1. Ù 3 âAntiderivatives.nb
x , (1,2)
Integration
5
2. Ù Hx - 0.5L â x , (-1, 1)
3. Ù Ix2 + x -
1
x
+ 2M â x,
4. Ù H3 - sinHtLL â t,
(1,1)
(-3, 3)
5. Ù Isec2 HxL - cosHxLM â x
6. Ù Iex + 2x - x2 + xã+3 - ΠM â x
7. à J2 x7.5 +
2
x
- 2N â x