This document discusses the indefinite integral and antiderivatives. It defines an antiderivative as a function whose derivative is the original function, and notes that there are infinitely many antiderivatives that differ by a constant. The process of finding antiderivatives is called indefinite integration or antidifferentiation. Initial conditions can be used to determine a unique particular solution by solving for the constant of integration.

1. The Indefinite Integral

2. Explore:
Find a function with the given derivative.
1) f’(x) = 2x
2) y’ = 3x2
1
3) f’(x) = − x 2

3. Antiderivatives
x2 is an antiderivative of 2x.
Why “an”?
Because the antiderivative of 2x could also be x2 + 5
Let f(x) be a differential equation (a derivative)
Call F(x) an antiderivative of f(x)
Then G(x) will be the antiderivative where
G(x) = F(x) + C

4. Terms & Notation
G(x)
=
general antiderivative
F(x)
antiderivative
+
C
constant of integration
Process is called antidifferentiation → indefinite integration
integral
F ( x) + C = ∫ f ( x)dx
Integrand
Variable of integration

5. In book…
Know the
power rule!

6. ∫ F ′( x)dx = F ( x) + C
d
dx
[∫ f ( x)dx] = f ( x)
Integration is the “inverse” of differentiation
Differentiation is the “inverse” of integration
4
Example: Describe the antiderivatives of 2
x
∫
4
1
dx = 4 ∫ 2 dx =
2
x
x
 1
4
4  − ÷+ C = − + C
x
 x

7. Practice Time !!!
1
1.
∫
2.
∫
3.
∫ 4 cos x dx =
4.
∫ ( x + x ) dx =
5.
( 2x )
3
dx =
x dx =
2
∫ ( 3x
6
− 2x 2 + 7x +1) dx =

8. Just a Few More !!!
1.
2.
3.
∫
cos x
dx =
2
sin x
∫
t 2 − 2t 4
dt =
4
t
∫
x2
dx =
2
x +1
1
t
4. ∫  − 2e  dt =
 2t

5.
∫
dy
=
csc y

9. Initial Conditions & Particular Solutions
(
)
y = ∫ 4 x 3 − 2 x dx = x 4 − x 2 + C
F(x) → general solution
There are infinitely many solutions until you are told an initial
condition about F(x) → F(2) = 5
Plug in the initial condition and solve for C
5 = (2)4 – (2)2 + C
C = -7
F(x) = x4 – x2 - 7
F(x) → particular solution

10. Initial Conditions & Particular Solutions
(
)
y = ∫ 4 x 3 − 2 x dx = x 4 − x 2 + C
F(x) → general solution
There are infinitely many solutions until you are told an initial
condition about F(x) → F(2) = 5
Plug in the initial condition and solve for C
5 = (2)4 – (2)2 + C
C = -7
F(x) = x4 – x2 - 7
F(x) → particular solution