|
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
= 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
|
|
|