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The student will learn about:

§4.1 Antiderivatives and Indefinite Integrals.

the properties associated with these functions,

antiderivatives and indefinite integrals,

and applications using indefinite integrals.

2

Introduction

• We have been studying differentiation and its uses. We now consider the reverse process, antidifferentiation, which, for a given derivative, essentially recovers the original function.

• Antidifferentiation has many uses.

For example, differentiation turns a cost function into a marginal cost function, and so antidifferentiation turns marginal cost back into cost.

• Later we will use antidifferentiation for other purposes, such as finding areas.

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Antiderivatives.

Many operations in mathematics have reverses – inverses – such as multiplication and division. The reverse operation of finding a derivative (antiderivative) will now command our attention.

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Examples

Find a function that has a derivative of x.

Find a function that has a derivative of 2x.

x2

Find a function that has a derivative of x2.

x2/2

Oops!

x3/3

x2 Since d/dx (x2) = 2x

Since d/dx [x2/2] = x

Since d/dx [x3/3] = x2

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Examples - continued

The above functions that you found are antiderivatives.

Find a function that has a derivative of 2x.

x2

Note that you can find more than one such function?

x2 + 3 x2 - 5

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Fact If a function has more than one antiderivative then the antiderivatives differ by at most a constant.

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Geometrical Interpretation of the Arbitrary constant

The graph below shows three antidirivatives of f (x) = 2x.

x2 + 2, x2, and x2 – 2, corresponding to C being 2, 0, and –2.

Notice that each curve has the same slope at x = 1

(their tangent lines are parallel).

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Antiderivatives and Indefinite Integrals - Algebraic Forms.

Let f (x) be a function, then the family of all functions that are antiderivatives of f (x) is called the indefinite integral and has the symbol

∫ f (x) dx = g (x) + C

g '(x) = f (x)if and only if

The symbol ∫ is called an integral sign, and the function f (x) is called the integrand. The symbol dx indicates that the antidifferentiation is performed with respect to the variable x. the arbitrary constant C is called the constant of integration.

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Example

Now, let’s get to the basics and make this easy!

since 3

2 xx dx C

3

32d x

C xdx 3

3 + 1 =

- 2 + 1 =

1/2 + 1 =

4

-1

3/2

If y = x 4 then y’ = 4 x 3 .

Subtracted 1 from the exponent.

We add 1 to the exponent and

We multiplied by the exponent and

For integration we do the reverse.

Divide by the new exponent.

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Indefinite Integral Formulas and Properties.

1. 1n,C1n

xdxx

1nn

Power rule.

3. ∫ 1 dx = x + C.

∫ k dx = kx + C. Integral of a constant.2.

4. ∫ k f (x) dx = k ∫ f (x) dx

5. ∫ [f (x) ± g (x)] dx = ∫ f (x) dx ± ∫ g (x) dx.

Integral of a sum or difference.B

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Examples

C2

x5 2

c. ∫ 5 x - 3 dx =

a. ∫ 444 dx =

b. ∫ x3 dx =

e. ∫ (x 4 + x + x ½ + 1 + x – ½ + x – 2) dx =

d. ∫ x2/3 dx =

Power rule.

444x + C Integral of a constant.

Power rule.

Power rule3/5 x 5/3 + C

x5 /5

f. ∫ x - 1 dx = x0 /0 + C = Undefined !!

NO!!! Come back next class for this one.

x5 /5 + x 2/2 x5 /5 + x 2/2 + 2 x 3/2/3 x5 /5 + x 2/2 + 2 x 3/2/3 + x x5 /5 + x 2/2 + 2 x 3/2/3 + x + 2 x 1/2 x5 /5 + x 2/2 + 2 x 3/2/3 + x + 2 x 1/2 – x – 1 + C

C

35x 3

5

4xC

4

A

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Applications In spite of the prediction of a paperless computerized office, paper and paperboard production in the United States has steadily increased. In 1990 the production was 80.3 million short tons, and since 1970 production has been growing at a rate given by

f ’ (x) = 0.048x + 0.95

Where x is years after 1970. Noting that f (20) = 80.3, find f (x).

Continued on next slide.

Note this is a typical application problem for Antiderivatives. The derivative is given and we must find the original function.

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Applications - continued

0.024 x 2 + 0.95 x + c = f (x)

Note, f ’ (x) = 0.048 x + 0.95 Where x is years after 1970.

Noting that f (20) = 80.3, find f (x). There will be three parts.

The result from part 1 gives us a “family” of curves as previously mentioned. In part 2 we will find the particular curve that fits our problem.

Continued on next slide.

We need the integral of f ‘ (x) or

∫ (0.048x + 0.95) dx =

f (x) = 0.024 x 2 + 0.95 x + c

Part 1

Part 2

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Applications - continued Note, f ’ (x) = 0.048 x + 0.95 Where x is years after 1970.

Noting that f (20) = 80.3, find f (x). There will be three parts.

80.3 = (0.024)(202) + (0.95)(20) + c

Noting that f (20) = 80.3

Continued on next slide.

80.3 = 28.6 + c and c = 51.7 so

f (x) = 0.024 x 2 + 0.95 x + 51.7

f (x) = 0.024 x 2 + 0.95 x + c Part 1

Part 2

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Applications - concluded

f (0) = (0.024)(02) + (0.95)(0) + 51.7 =

Note, f ’ (x) = 0.048t + 0.95 Where x is years after 1970.

f (x) = 0.024 x 2 + 0.95 x + 51.7 .

f (30) = (0.024)(302) + (0.95)(30) + 51.7 =

51.7

101.8

Find f (0) and f (30), the production levels for 1970 and 2000.

51.7 short tons in 1970

101.8 short tons in 2000

Part 3

There may also be a third part to the problem.

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Summary.

2.

1. ∫ k dx = kx + C.n 1

n xx dx C, n 1

n 1

5. ∫ 1 dx = x + C.

3. ∫ k f (x) dx = k ∫ f (x) dx

4. ∫ [f (x) ± g (x)] dx = ∫ f (x) dx ± ∫ g (x) dx.

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Summary Continued

Soooooo – What about a product rule or a quotient rule?

Come back in a week or two for the exciting conclusion of indefinite integrals.

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ASSIGNMENT

§4.1 on my website.

12, 13, 14, 15.

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Test Review

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§ 3.1

Be able to do implicit differentiation § 2.5.

Know all the formula for cost, revenue, profit (margins, averages, marginal averages, etc.).

Related rate problems using implicit differentiation.

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Introduction to Related Rates

Related rate problems. Find the rate of change of

both variables (implicitly) with respect to time and solve for one rate with respect to the other.

There will be two variables and thus two related rates. You will know three of the four parts and find the fourth!

x y dx/dt dy/dt

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Test Review

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§ 3.2

§ 3.31. Know the basics of logarithmic functions.

2. Know the application problems associated with logarithmic functions.

1. Know the basics of exponential functions.

2. Know all the interest formula.

3. Know how to calculate present value, future value and depreciation.

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Test Review

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§ 3.4

1. Know the three general derivative formula.

2. Know consumer demand and expenditure.

'uunudx

d 1nn

u ude e u'

dx

d 1lnu u'

dx u

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Test Review

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§ 3.5

1. Know relative rate of change.

2. Know elasticity of demand and how it affects revenue.