This document provides an overview of antiderivatives and integrals. It discusses how when taking the derivative, any constants disappear, but when finding an antiderivative we include an unknown constant C. It also introduces the concept of an initial value problem, where an initial condition is given to determine the value of C. Finally, it distinguishes between definite integrals, where a specific value can be found, and indefinite integrals, where the constant C must be retained.

1. 6.1 Antiderivatives

2. First, a little review:
Consider: y = x2 + 3 y = x2 ! 5
or
then: y! = 2 x y! = 2 x
It doesn’t matter whether the constant was 3 or -5, since
when we take the derivative the constant disappears.
!

3. First, a little review:
Consider: y = x2 + 3 y = x2 ! 5
or
then: y! = 2 x y! = 2 x
It doesn’t matter whether the constant was 3 or -5, since
when we take the derivative the constant disappears.
However, when we try to reverse the operation:
Given: y! = 2 x find y
!

4. First, a little review:
Consider: y = x2 + 3 y = x2 ! 5
or
then: y! = 2 x y! = 2 x
It doesn’t matter whether the constant was 3 or -5, since
when we take the derivative the constant disappears.
However, when we try to reverse the operation:
Given: y! = 2 x find y We don’t know what the
constant is, so we put “C” in
y = x2 + C the answer to remind us that
there might have been a
constant.
!

5. If we have some more information we can find C.
Given: y! = 2 x and y = 4 when x = 1 , find the equation for y .
!

6. If we have some more information we can find C.
Given: y! = 2 x and y = 4 when x = 1 , find the equation for y .
y = x2 + C
!

7. If we have some more information we can find C.
Given: y! = 2 x and y = 4 when x = 1 , find the equation for y .
y = x2 + C
4 = 12 + C
3=C
y = x2 + 3
!

8. If we have some more information we can find C.
Given: y! = 2 x and y = 4 when x = 1 , find the equation for y .
y = x2 + C
2
This is called an initial value
4 =1 +C problem. We need the initial
values to find the constant.
3=C
y = x2 + 3
!

9. If we have some more information we can find C.
Given: y! = 2 x and y = 4 when x = 1 , find the equation for y .
y = x2 + C
2
This is called an initial value
4 =1 +C problem. We need the initial
values to find the constant.
3=C
y = x2 + 3
An equation containing a derivative is called a differential
equation. It becomes an initial value problem when you
are given the initial condition and asked to find the original
equation.
!

10. 4
Integrals such as ! x 2 dx are called definite integrals
1
because we can find a definite value for the answer.
4
2
! x dx
1

11. 4
Integrals such as ! x 2 dx are called definite integrals
1
because we can find a definite value for the answer.
4
2
! x dx
1
4
1 3
x +C
3 1
The constant always cancels
!1 3 quot; !1 3 quot; when finding a definite
% # 4 + C & $ % #1 + C &
'3 ( '3 ( integral, so we leave it out!
64 1 63
+C ! !C = = 21
3 3 3

12. 2
Integrals such as ! x dx are called indefinite integrals
because we can not find a definite value for the answer.
2
! x dx
1 3
x +C
3 When finding indefinite
integrals, we always
include the “plus C”.
!

13. Many of the integral formulas are listed on page 332. The
first ones that we will be using are just the derivative
formulas in reverse.
CW P. 337 (2, 6, 8, 10)