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6 The Integral

Copyright © Cengage Learning. All rights reserved.

6.4The Definite Integral: Algebraic Viewpoint and the Fundamental Theorem of Calculus

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Example 1 – Finding Cost from Marginal Cost

The marginal cost of producing baseball caps at a production level of x caps is 4 – 0.001x dollars per cap. Find the total change of cost if production is increased from 100 to 200 caps.

Solution:

Method 1: Using an Antiderivative: Let C(x) be

the cost function.

Because the marginal cost function is the derivative of the cost function, we have C(x) = 4 – 0.001x and so

C(x) = ∫ (4 – 0.001x) dx

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Example 1 – Solution

= 4x – 0.001 + K

= 4x – 0.0005x2 + K.

Although we do not know what to use for the value of the constant K, we can say:

Cost at production level of 100 caps

= C(100)

= 4(100) – 0.0005(100)2 + K

= $395 + K

cont’d

K is the constant of integration.

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Example 1 – Solution

Cost at production level of 200 caps

= C(200)

= 4(200) – 0.0005(200)2 + K

= $780 + K.

Therefore,

Total change in cost = C(200) – C(100)

= ($780 + K) – ($395 + K)

= $385.

cont’d

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Example 1 – Solution

Notice how the constant of integration simply canceled out! So, we could choose any value for K that we wanted (such as K = 0) and still come out with the correct total change.

Put another way, we could use any antiderivative of C(x), such as

F(x) = 4x – 0.0005x2

or

F(x) = 4x – 0.0005x2 + 4

compute F(200) – F(100), and obtain the total change, $385.

cont’d

F(x) is any antiderivative of C(x)whereas C(x) is the actual cost function.

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Example 1 – Solution

Summarizing this method: To compute the total change of C(x) over the interval [100, 200], use any antiderivative F(x) of C(x), and compute F(200) – F(100).

Method 2: Using a Definite Integral: Because the marginal cost C(x) is the rate of change of the total cost function C(x), the total change in C(x) over the interval [100, 200] is given by

Total change in cost = Area under the marginal cost function curve

cont’d

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= $385.

Putting these two methods together gives us the following surprising result:

where F(x) is any antiderivative of C(x).

Example 1 – Solution cont’d

Figure 20

See Figure 20.

Using geometry or Riemann sums

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The Definite Integral: Algebraic Viewpoint and the Fundamental Theorem of Calculus

In Example 1, if we replace C(x) by a general continuous

function f (x), we can write

where F(x) is any antiderivative of f (x). This result is known

as the Fundamental Theorem of Calculus.

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The Definite Integral: Algebraic Viewpoint and the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC)

Let f be a continuous function defined on the interval [a, b] and let F be any antiderivative of f defined on [a, b]. Then

Moreover, an antiderivative of f is guaranteed to exist.

In Words: Every continuous function has an antiderivative. To compute the definite integral of f (x) over [a, b], first find an antiderivative F(x), then evaluate it at x = b, evaluate it at x = a, and subtract the two answers.

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The Definite Integral: Algebraic Approach and the Fundamental Theorem of Calculus

Quick Example

Because F(x) = x2 is an antiderivative of f (x) = 2x,

= F(1) – F(0) = 12 – 02 = 1.

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Example 2 – Using the FTC to Calculate a Definite Integral

Calculate

Solution:

To use the FTC, we need to find an antiderivative of 1 – x2. But we know that

We need only one antiderivative, so let’s take F(x) = x – x3/3. The FTC tells us that

.

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Applications

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Example 5 – Oil Spill

Your deep ocean oil rig has suffered a catastrophic failure, and oil is leaking from the ocean floor wellhead at a rate of

v(t) = 0.08t2 – 4t + 60 thousand barrels per day (0 t 20),

where t is time in days since the failure. Compute the total volume of oil released during the first 20 days.

Solution:

We calculate

Total volume =

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Example 5 – Solution cont’d