Antiderivatives and
Indefinite Integration
Lesson 5.1
Reversing Differentiation
• An antiderivative of function f is
 a function F
 which satisfies
F’ = f
• Consider the following:
F ( x)  2 x
G( x)  2 x 4  17
4
f ( x)  F '( x)  8 x
3
g ( x)  G '( x)  8 x
3
• We note that two antiderivatives of the same
function differ by a constant
Reversing Differentiation
• General antiderivatives
f(x) = 6x2
F(x) = 2x3 + C
 because F’(x) = 6x2
k(x) = sec2(x)
K(x) = tan(x) + C
 because K’(x) = k(x)
Differential Equation
• A differential equation in x and y involves
 x, y, and derivatives of y
• Examples
y '  5x
y' x 4
3
• Solution – find a function whose derivative is
the differential given
Differential Equation
• When
y '  5x
5 2
• Then one such function is y  x
2
5 2
• The general solution is y  x  C
2
Notation for Antiderivatives
dy
 f ( x)
• We are starting with
dx
• Change to differential form
dy  f ( x)  dx
• Then the notation for antiderivatives is
y   f ( x) dx  F ( x)  C
"The antiderivative of f with respect to x"
Basic Integration Rules
• Note the inverse nature of integration and
differentiation
 F '( x) dx  F ( x)  C
d 
  f ( x)
f
(
x
)
dx


dx 
• Note basic rules, pg 286
Practice
• Try these
 4x
3
 6 x  1 dx
2
x2  2 x  3
 x 4 dx
 sec y   tan y  sec y  dy
Finding a Particular Solution
• Given
dy
 2   x  1
dx
• Find the specific equation
from the family of
antiderivatives, which
contains the point (3,2)
• Hint: find the general antiderivative, use the
given point to find the value for C
Assignment A
• Lesson 5.1 A
• Page 291
• Exercises 1 – 55 odd
Slope Fields
• Slope of a function f(x)
 at a point a
 given by f ‘(a)
• Suppose we know f ‘(x)
 substitute different values for a
 draw short slope lines for successive values
of y
• Example
f '( x)  2 x
Slope Fields
• For a large portion of the graph, when
f '( x)  2 x
• We can trace the line for a specific F(x)
 specifically when the C = -3
Finding an Antiderivative Using a Slope
Field
• Given
f '( x)  e
x2
• We can trace the version of the original
F(x) which goes through the origin.
Vertical Motion
• Consider the fact that the acceleration due to
gravity a(t) = -32 fps
• Then v(t) = -32t + v0
Why?
• Also s(t) = -16t2 + v0t + s0
• A balloon, rising vertically with velocity = 8
releases a sandbag at the instant it is 64 feet
above the ground
 How long until the sandbag hits the ground
 What is its velocity when this happens?
Rectilinear Motion
• A particle, initially at rest, moves along the xaxis at velocity of
a(t )  cos t
t 0
• At time t = 0, its position is x = 3
 Find the velocity and position functions for the
particle
 Find all values of t for which the particle is at rest
Assignment B
• Lesson 5.1 B
• Page 292
• Exercises 77 – 93, EOO