Sec' 4'1 Antiderivatives and Indefinite Integration

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Sec. 4.1 Antiderivatives and Indefinite
Integration

• By
• Dr. Julia Arnold

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Antidifferentiation is the art of finding the
function F(x) whose derivative would be the given
function F /(x).
Antidifferentiation is also called Indefinite
Integration
Just as represents finding a derivative,
the symbol represents finding
the antiderivative of the function f(x). The
symbol is called the integral symbol and
resembles an elongated s which as we will see in
the next section is associated with finding sums.
The dx shows what variable we are integrating.
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Read your text book. I will supplement the
information there.
For most of the examples, you will be using the
exponent formula which is like the opposite
of the power formula. Example 4 a, b, and c. use
this formula. Example 5 shows that you always
rewrite radicals and use laws of exponents. You
can not integrate numerator and then denominator.
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In Example 5 some of the missing steps are
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Initial Conditions and Particular Solutions
When we find antiderivatives and add the constant
C, we are creating a family of curves for each
value of C.
C0
C1
C3
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Finding a Particular Solution
Find the general solution of
General Solution means the antiderivative C.
A particular solution means that given some
helpful information we will find the value of C.
The following information is referred to as the
initial condition F(1) 0
C1
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Setting Up A Vertical Motion Problem
When you must set up the motion of a falling
object, S is the position function, is
the velocity function, and is the
acceleration function. Acceleration due to
gravity is -32 ft. per second per second. Begin
most problems with
Lets look at a specific problem.
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A ball is thrown upward with an initial velocity
of 64 feet per second from an initial height of
80 feet.
A. Find the position function giving the height
S as a function of the time t. B. When does the
ball hit the ground?
Begin by integrating
The initial velocity is 64 feet per second.
This means at t 0, S 64
The initial height means at t 0, S 80
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C 80
Thus, the position function is
B. When will the ball hit the ground?
When S 0
Since t must be positive, the ball hit the ground
after 5 seconds.
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Page 243 provides a nice summary of Integration
rules.
Sample Homework Problems 8.
Rewrite as
Using the sum and difference formula (4th from
the top on page 243), we integrate each term
using the power formula.
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Evaluate the definite integral and check the
result by differentiation.
18.
Rewrite
Integrate
Simplify
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Checking

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Find the equation for y, given the derivative
and the indicated point on the curve.
47.
Multiply both sides by dx, then integrate both
sides.
Of the family of curves the correct one goes
through the point (0,4)
Substitute x 0 and y 4. Thus c4 and the
solution is
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Find f(x)
Solve the differential equation
52.
Integrate
Substitute x0 and 6 for f(0)
Thus c 6
Integrate again
Substitute x 0 and f(0)3
C3
Answer
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Vertical Motion Use a(t)-32 feet per sec per
sec as the acceleration due to gravity. (Neglect
air resistance.)
60. A balloon rising vertically with a velocity
of 16 feet per second, releases a sandbag at the
instant it is 64 feet above the ground. (a) How
many seconds after its release will the bag
strike the ground? (b) At what velocity will it
hit the ground?
Solution Begin with a(t) -32 and integrate.
Remember v(t)a(t), thus v(t) -32t c. At t
0 (the moment the sandbag is released, the
balloon has a velocity of 16, thus 16 c. v(t)
-32t 16 Continued on next page
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60 continued
v(t) -32t 16 Now integrate v(t) which is
s(t) the position function. S(t) -16t2 16t c
A balloon rising vertically with a velocity of 16
feet per second, releases a sandbag at the
instant it is 64 feet above the ground.
At t 0, s(t) is 64 feet above the ground. Thus
c 64 and s(t) -16t2 16t 64
(a) How many seconds after its release will the
bag strike the ground?
The position of the bag indicates that s 0 on
ground.
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s(t) -16t2 16t 64
0 -16t2 16t 64 0 -16(t2 - t - 4) 0 t2 -
t - 4 Since this doesnt factor we will need to
use the quadratic formula to solve.
One of these ts is

negative.
One is
positive.
We want the positive answer.
t 2.56 approximately Thus the sandbag is on
the ground in about 2.6 seconds
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(b) At what velocity will it hit the ground?
v(t) -32t 16 _at_ t 2.56 v(2.56) -32(2.56)
16 -65.92 The velocity of the sandbag when it
hits the ground is approximately -65.92 feet per
second.