Chapter 4 notes from the Text Book

Section 4.1 Antiderivative and Indefinite Integration

Definition of an Antiderivative:

A function F is an antiderivative of f on an interval I  if F ’(x) = f(x) for all x in I.

Theorem 4.1 Representation of Antiderivative:

If F is an antiderivative of f on an interval I, then G is an antiderivative of f on the interval I if an only if G is of the form

G(x)= F (x) + C, for all x in I    where C is a constant.

Using Theorem 4.1:

G(x) = x^2 + C               ( Family of all antiderivative of f(x) = 2x)

and  G(x) = x^2 + C      is the general antiderivative of the differential equation

G ‘ (x) = 2x                    (Differential equation)

Example 1  Solving a Differential Equation:

Find the general solution of the differential equation y ‘ = 2

Solution: To begin, you need to find a function whose derivative is 2.

One such function is:     y= 2x   (2x is an antiderivative of 2 )

Now, you can use Theorem 4.1 to conclude that the general solution of the differential equation is:

y= 2x + C   (General Solution )

Basic Integration Rules

 Differentiation Formula Integration Formula d/dx [C] = 0 int 0 dx = C d/dx [kx] = k int k dx = kx + C d/dx [kf(x)] = k f ‘ (x) int k f(x)dx = k(int) f(x) dx d/dx [f(x)] +- [g(x)] = f ‘(x) +- g ‘(x) int [f(x)+- g(x)]dx = int f (x)dx +- int g (x)dx d/dx [x^n] = nx^(n-1) int [x^n]dx = ((X^(n+1))/(n+1)) + C,  n not equal –1  (Power Rule) d/dx [sin x] = cos x int cos x dx = sin x + C d/dx [cos x] = (- sin) x int sin x dx = (- cos) x + C d/dx [tan x] = sec^2 x int sec^2 x dx = tan x + C d/dx [sec x] = sec x tan x int sec x tan x dx = sec x + C d/dx [cot x] = (- csc)^2 x int csc^2 x dx = (- cot) x + C d/dx [csc x] = (- csc) x cot x int csc x cot x dx =( - csc) x + C

Example 6 ( integration of trig functions)

Rewrite before Integration

int (sinx /cos^2x)dx

=  int (1/cosx)(sinx/cosx)dx  à  rewrite as a product

= int secx tanx dx                    à  rewrite using trig identities

= sec x + C                              à  Integrate.

Example 8 Solving a Vertical Motion Problem
A ball is thrown upward with an initial velocity of 64 feet per second
from an initial height of 80 feet. (See figure 4.4 page 248) s (t) = (-16)t^2 +64t + 80
(A). Find the position function giving the height s as a function of the time t.
(B). When does the ball hit the ground?

Solution
(A). Let t = 0 represent the initial time. The two given initial conditions can be written as fallows.
s (0) = 80 Initial height is 80 feet.
s‘ (0) = 64 Initial velocity is 64 feet per second.
Using –32 feet per second per second as the acceleration due to gravity, you can write:

s“ (t) = (-32)
s‘(t) = int s”(tP dt = int –32 dt = -32t + C1

Using the initial velocity, you obtain s’(0) = 60 = -32(0) + C1 , which implies that C1 = 64.
Next, by integrating s‘(t), you obtain

s (t) = int s ‘(t)dt = int (-32t + 64) dt = -16t^2 +64t + C2

Using the initial height, you obtain
s (0) = 80 = -16(0)^2 +64(0) + C2 ,which implies that C2 = 80.

Therefore, the position function is
s (t) = (-16)t^2 +64t + 80
------------
(B). Using the position function found in part (a), you can find the time that the ball hits the ground by solving the equation s(t) = 0.

s (t) = (-16)t^2 +64t + 80 = 0
(-16)(t+1)(t-5) = 0
t = -1 and t = 5
Because t must be positive, you can conclude that the ball hits the ground 5 seconds after it was thrown.

Example 8 shows how to use calculus to analyze vertical motion problems in which the acceleration is determined by a gravitational force. You can use a similar strategy to analyze other linear motion problems (vertical or horizontal) in which the acceleration ( or deceleration) is the result of some other force.
Also look at problems # 77 & # 78.

Text book page 251 #77 & # 78
Rectilinear Motion In Exercises 77-80, consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x ‘ (t) is its velocity, and x “(t) is its acceleration.

77. x(t) = t^3 – 6t^2 + 9t – 2, 0 ≤ t ≤ 5

(a) Find the velocity and acceleration of the particle.
v(t) = x’(t) = 3t^2 – 12t + 9 à 3(t^2 – 4t + 3) à 3 (t – 1) ( t – 3)
a(t) = v’(t) = 6t – 12 à 6 (t - 2)

(b) Find the open t-intervals on which the particle is moving to the right.

v(t) > 0 when 0 < t < 1 or 3 < t < 5

(c. ) Find the Velocity of the particle when the acceleration is 0.
a(t) = 6 (t - 2) = 0
t = 2
v(2) = 3 (2 – 1) (2 – 3) = -3

78. Repeat Exercise 77 for the position function
x(t) = (t - 1) (t - 3)^2, 0 ≤ t ≤ 5.

(a) Find the velocity and acceleration of the particle.

x(t) = (t - 1) (t - 3)^2 à t^3 – 7t^2 + 15t – 9
v(t) = x ’ (t)= 3t^2 – 14t + 15 à (x - 3)·(3·x - 5)

a(t) = v ’ (t) = 6t -14

(b) Find the open t-intervals on which the particle is moving to the right.

v(t) > 0 when 0 < t < (5/3) or 3 < t < 5

(c. ) Find the Velocity of the particle when the acceleration is 0.

a(t) = v ’ (t) = 6t -14 = 0
t = 14/6 = 7/3
v(7/3) = ((7/3) – 3)(3*(7/3)-5) = -4/3