Notes 4.1 Antiderivatives and Indefinite Integration
Taking the antiderivative of a function is the inverse of taking the derivative.
differentiation
x3
3x2
3x2 is the derivative, x3 is an antiderivative
otherwise known as an indefinite integral
antidifferentiation
More examples of Antidifferentiation (otherwise known as Integration)
F = x4
Differentiation
Integration
I
f’(x) = 4x3
F = x2
f’(x) = 2x
F = 3x2
f’(x) = 6x
x4
4
f’(x) = x3
x3
F=
3
f’(x) = x2
F=
Power Rule for Integration
For f (x) = x n , the antiderivative, F(x) =
x n+1
+C
n +1
What’s up with the “C” ?
Consider several expressions having the same derivative….
4x2 + 1
4x2 ! 7
4x2 + 8
The derivative of each of the above expressions is 8x because the derivative of any constant is 0. Therefore,
when we take the antiderivative, or indefinite integral, of an expression such as 8x, we must write the
antiderivative as 4x2 + C where C is a constant called the constant of integration.
Page 1
8/4/2009
Rules for Integration:
Power Rule
!x
4
! x dx =
dx =
!x
3
6
dx =
Multiplied by a Constant
! 6 x dx = 6! x dx =
Sum or Difference of Functions
! (15 x
2
+ 4 x " 1)dx =
! 15 x
2
dx +
! 4 xdx
"
! 1dx =
Trigonometric Functions
! cos x dx =
! sin x dx =
! sec
2
x dx =
! sec x tan x dx =
! csc
2
x dx = =
! csc x cot x dx =
Examples: Rewrite the expression in a more convenient form before you integrate. Be sure to include the
constant of integration, C, in your answer. Check your answers by taking the derivative.
1
1.
!x
dx
2.
!
3.
tan x cos x dx
4.
' x3 + 2x 2 + 7x $
""dx
! %%&
x
#
5
Page 2
3
x dx
8/4/2009
Differential Equations
What’s a differential equation? It’s an equation containing a derivative, possibly in the form:
dy
= f (x)
dx
Solving differential equations to get a general solution.
Example:
dy
= 3x
dx
Solve this differential equation, i.e. determine y.
dy = 3x dx
Rewrite as a differential equation.
! dy = ! 3x dx
Take the integral of both sides of the equation.
y+C =
y=
3x 2
+C
2
Integrate with respect to x.
3x 2
+C
2
Combine the two constants to make a new constant C.
A general solution to a differential equation consists of a “family” of curves that simply differ by a constant.
The family of curves for the differential equation solved above would look sort of like this… but there would
actually be infinitely many curves since there are infinitely many constants.
10
8
q(x) = fx
( )+ 3
h(x) = fx
( )+ 2
6
4
g(x) = fx
( )+ 1
2
f(x) =
-5
()
3 2
!x
2
5
r(x) = fx
( )- 1
s(x) = fx
( )- 2
t(x) = fx
( )- 3
-2
-4
10
Determine the general solution of the following differential equations.
1.
dy
= x2 + 1
dx
2.
Page 3
y ' = cos x (Rewrite y' as
dy
)
dx
8/4/2009
Using initial conditions to find a particular solution of a differential equation.
Sometimes we want to focus on one particular solution of a differential equation. In that case, we are usually
given a particular point that the solution includes. This is sometimes called an “initial condition” and makes the
most sense when we are solving a differential equation in an application.
Example:
Given f ' ( x ) =
1
x2
for x > 0 and f(1) = 0 is our initial condition, determine the equation for f(x).
First, set up the appropriate integral and solve.
dy 1
=
dx x 2
1
dy = 2 dx
x
1
! dy = ! x 2 dx
y = ! x "2 dx
x "1
y=
+C
"1
1
y= " +C
x
Now, substitute the initial condition, f(1) = 0 to solve for C.
1
0= ! +C
1
C =1
Substitute the value that you found for C into the equation, f(x), and you have a particular solution for the
differential equation.
f ( x) = !
1
+ 1 for x > 0
x
Page 4
8/4/2009
Using initial conditions makes sense when solving certain physics problems. For example, you can derive the
position function for vertical motion under constant acceleration. Assume acceleration due to gravity is –9.8
meters/sec2 and that the initial velocity is vo and the initial position is so.
s ' ' (t ) = !9.8 meters/sec 2
Integrate the differential equation to get...
s ' (t ) = " ! 9.8dt = !9.8t + C
At time t = 0, the velocity is v o , so...
v0 = !9.8(0) + C , and C = v0
Substitute for C in the function s'(t) to get
s ' (t ) = !9.8t + v0 meters/sec
Integrate again to get the position function...
9.8 2
s (t ) = " (!9.8t + v0 )dt = !
t + v0 t + C meters
2
At time t = 0, the position is s 0 , so...
9.8 2
(0) + v0 (0) + C
2
C = s0 ,
s0 = !
and substitute again to get...
9.8 2
s(t) = t + v0 t + s0
2

Page 5
8/4/2009