jcalc6.1

Barnett/Ziegler/Byleen Business Calculus 11e 1

Learning Objectives for Section 6.1 Antiderivatives and Indefinite Integrals

The student will be able to formulate problems involving antiderivatives.

The student will be able to use the formulas and properties of antiderivatives and indefinite integrals.

The student will be able to solve applications using antiderivatives and indefinite integrals.


The Antiderivative

Many operations in mathematics have inverses. For example, division is the inverse of multiplication. The inverse operation of finding a derivative, called the antiderivative, will now command our attention. A function F is an antiderivative of a function f if F ’(x) = f (x).


Examples

Find a function that has a derivative of 2x.

Find a function that has a derivative of x.

Find a function that has a derivative of x2.


Examples


Answer: x2, since d/dx (x2) = 2x.

Find a function that has a derivative of x.

Answer: x2/2, since d/dx (x2/2) = x.

Find a function that has a derivative of x2.

Answer: x3/3, since d/dx (x3/3) = x2.


Examples(continued)


Answer: We already know that x2 is such a function.

Other answers are x2 + 2 or x2 – 5.

The above functions are all antiderivatives of 2x.

Note that the antiderivative is not unique.


Uniqueness of Antiderivatives

The following theorem says that antiderivatives are almost unique.

Theorem 1:

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


The symbol is called an integral sign, and the function f (x) is called the integrand. The symbol dx indicates that anti-differentiation is performed with respect to the variable x.By the previous theorem, if F(x) is any antiderivative of f, then

The arbitrary constant C is called the constant of integration.

Indefinite Integrals

CxFdxxf )()(

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


Find the indefinite integral of x2.

Example


Find the indefinite integral of x2.

Answer: , because

Example

Cx

dxx 3

32

23

3xC

x

dx

d


Indefinite Integral Formulas and Properties

dxxgdxxfdxxgxf

dxxfkdxxfk

Cxdxx

Cedxe

nCn

xdxx

xx

nn

)()()()(.5

)()(.4

||ln1

.3

.2

1,1

.11

(power rule)

It is important to note that property 4 states that a constant factor can be moved across an integral sign. A variable factor cannot be moved across an integral sign.


Examples for Power Rule

444 dx = 444x + C (power rule with n = 0) x3 dx = x4/4 + C (n = 3) 5 x-3 dx = -(5/2) x-2 + C (n = -3) x2/3 dx = (3/5) x5/3 + C (n = 2/3) (x4 + x + x1/2 + 1 + x -1/2) dx =

x5/5 + x2/2 + (2/3) x3/2 + x + 2x1/2 + C

But you cannot apply the power rule for n = -1: x-1 dx is not x0/0 + C (which is undefined).

The integral of x-1 is the natural logarithm.


More Examples

4 ex dx = 4 ex + C 2 x-1 dx = 2 ln |x| + C


Application

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.

Find f (x), and the production levels in 1970 and 2000.


Application(continued )

Noting that f (20) = 80.3, we calculate

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

80.3 = 28.6 + C

C = 51.7

f (x) = 0.024 x2 + 0.95 x + 51.7

We need the integral of f ’(x), or

Cxxdxxdxxfxf 95.0024.0)95.0048.0()(')( 2


Application(continued )

The years 1970 and 2000 correspond to x = 0 and x = 30.

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

f (30) = (0.024)(302) + (0.95)(30) + 51.7 = 101.8

The production was 51.7 short tons in 1970, and 101.8 short tons in 2000.

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