1. Chapter 6 The Definite Integral

2. Chapter Outline Antidifferentiation Areas and Riemann Sums
Definite Integrals and the Fundamental Theorem
Areas in the xy-Plane
Applications of the Definite Integral

3. § 6.1
Antidifferentiation

4. Section Outline Antidifferentiation Finding Antiderivatives
Theorems of Antidifferentiation
The Indefinite Integral
Rules of Integration
Antiderivatives in Application

5. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5
Antidifferentiation
Definition
Example
Antidifferentiation: The process of determining f (x) given f ΄(x)
If , then
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #5

6. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #6
Finding Antiderivatives
EXAMPLE
Find all antiderivatives of the given function.
SOLUTION
The derivative of x9 is exactly 9x8. Therefore, x9 is an antiderivative of 9x8. So is x9 + 5 and x It turns out that all antiderivatives of f (x) are of the form x9 + C (where C is any constant) as we will see next.
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #6

7. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7
Theorems of Antidifferentiation
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #7

8. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #8
The Indefinite Integral
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #8

9. Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9
Rules of Integration
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #9

10. Finding Antiderivatives
EXAMPLE
Determine the following.
SOLUTION
Using the rules of indefinite integrals, we have
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #10

11. Finding Antiderivatives
EXAMPLE
Find the function f (x) for which and f (1) = 3.
SOLUTION
The unknown function f (x) is an antiderivative of One
antiderivative is Therefore, by Theorem I,
Now, we want the function f (x) for which f (1) = 3. So, we must use that information in our antiderivative to determine C. This is done below.
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #11

12. Finding Antiderivatives
CONTINUED
So, 3 = 1 + C and therefore, C = 2. Therefore, our function is
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #12

13. Antiderivatives in Application
EXAMPLE
A rock is dropped from the top of a 400-foot cliff. Its velocity at time t seconds is v(t) = -32t feet per second.
(a) Find s(t), the height of the rock above the ground at time t.
(b) How long will the rock take to reach the ground?
(c) What will be its velocity when it hits the ground?
SOLUTION
(a) We know that s΄(t) = v(t) = -32t and we also know that s(0) = We can now use this information to find an antiderivative of v(t) for which s(0) = 400.
The antiderivative of v(t) is To determine C,
Therefore, C = So, our antiderivative is s(t) = -16t
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #13

14. Antiderivatives in Application
CONTINUED
(b) To determine how long it will take for the rock to reach the ground, we simply need to find the value of t for which the position of the rock is at height 0. In other words, we will find t for when s(t) = 0.
s(t) = -16t
This is the function s(t).
0 = -16t
Replace s(t) with 0.
-400 = -16t2
Subtract.
25 = t2
Divide.
5 = t
Take the positive square root since t ≥ 0.
So, it will take 5 seconds for the rock to reach the ground.
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #14

15. Antiderivatives in Application
CONTINUED
(c) To determine the velocity of the rock when it hits the ground, we will need to evaluate v(5).
v(t) = -32t
This is the function v(t).
v(5) = -32(5) = -160
Replace t with 5 and solve.
So, the velocity of the rock, as it hits the ground, is 160 feet per second in the downward direction (because of the minus sign).
Goldstein/SCHNIEDER/LAY, CALCULUS AND ITS APPLICATIONS, 11e – Slide #15