1 The student will learn about: §4.4 Definite Integrals and Areas. the fundamental theorem of calculus, and the history of integral calculus, some applications.

1

The student will learn about:

§4.4 Definite Integrals and Areas.

the fundamental theorem of calculus, and

the history of integral calculus,

some applications.

the history of integral calculus, the definition of the definite integral,

Introduction

• We begin this section by calculating areas under curves, leading to a definition of the definite integral of a function.

• The Fundamental Theorem of Integral Calculus then provides an easier way to calculate definite integrals using indefinite integrals.

• Finally, we will illustrate the wide variety of applications of definite integrals.

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Definite Integral as a Limit of a Sum.

The Definite Integral may be viewed as the area between the function and the x-axis.

APPROXIMATING AREA BY RECTANGLES

We may approximate the area under a curve Inscribing rectangles under it. Use rectangles with equal bases and with heights equal to the height of the curve at the left-hand edge of the rectangles.

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Area Under a Curve

The following table gives the “rectangular approximation” for the area under the curve y = x 2 for 1 ≤ x ≤ 2, with a larger

numbers of rectangles.

The calculations were done on

a graphing calculator, rounding

answers to three decimal places.

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# Rectangles Sum of Areas

4 2.71875

8 2.523438

16 2.427734

32 2.380371

64 2.356812

128 2.345062

256 2.339195

512 2.336264

1024 2.334798

2048 2.334066

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Definite Integral as a Limit of a Sum. Definition. Let f be a continuous function defined on the closed interval [a, b], and let

a. a = x0 < x1 < x2 , … < xn – 1 < xn = b

b. ∆ x = (b – a)/n

c. ∆ xk → 0 as n → ∞

d. xk – 1 ≤ ck ≤ xk Then

is called a definite integral of f from a to b. The integrand is f (x), the lower limit is a, and the upper limit is b.

n

1kkk

n

b

ax)c(flimdx)x(f

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Those Responsible.

Isaac Newton 1642 -1727

Gottfried Leibniz 1646 - 1716

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Example 1

5 · 3 – 5 · 1 = 15 – 5 = 10

Make a drawing to confirm your answer.

3

1dx5

3

1dx5

3

1xx5

0 x 4

- 1 y 6

9

Example 2

4

3

1dxx

3

1dxx

3

1x

2

2

x

Make a drawing to confirm your answer.

2

1

2

9

0 x 4

- 1 y 4

2

8

Nice red box?

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Fundamental Theorem of Calculus

If f is a continuous function on the closed interval [a, b] and F is any antiderivative of f, then

)a(F)b(F)x(Fdx)x(f ba

b

a

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Evaluating Definite Integrals

b

adx)x(f

)a(F)b(F

By the fundamental theorem we can evaluate

Easily and exactly. We simply calculate

No red box?

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Definite Integral Properties

a

a0dx)x(f

a

b

b

adx)x(fdx)x(f

b

a

b

adx)x(fkdx)x(fk

b

a

b

a

b

adx)x(gdx)x(fdx])x(g)x(f[

b c b

a a cf (x)dx f (x)dx f (x)dx

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Example 3

9 - 0 =30

2 dxx 3

0x

3

3

x

0 x 4

- 2 y 10

9

Do you see the red box?

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Example 4

1

1

x2 dxe

2

e

2

e 22

x22

1 e1

1x

3.6268604

There is that red box again?

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Examples 5This is a combination of the previous two problems

3 2x 3

x 1

x e

3 2

3 2 2x1

x e dx

= 9 + (e 6)/2 – 1/3 – (e2)/2

What red box?

= 206.68654

So, what’s with the red box!

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Numerical Integration on a Graphing Calculator

21 dx

x

1

50 3

2dx

4x

x

0 x 3

- 1 y 3

-1 x 6- 0.2 y

0.5

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ApplicationFrom past records a management services determined that the rate of increase in maintenance cost for an apartment building (in dollars per year) is given by M ’ (x) = 90x 2 + 5,000 where M is the total accumulated cost of maintenance for x years.

7 2

090x 5,000 dx

Write a definite integral that will give the total maintenance cost through the seventh year. Evaluate the integral.

30 x 3 + 5,000x7

x 0|

= 10,290 + 35,000 – 0 – 0

= $45,290

Total Cost of a Succession of Units

The following diagrams illustrate this idea. In each case, the curve represents a rate, and the area under the curve, given by the definite integral, gives the total accumulation at that rate.

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FINDING TOTAL PRODUCTIVITY FROM A RATE

A technician can test computer chips at the rate of –3x2 + 18x + 15 chips per hour (for 0 ≤ x ≤ 6), where x is the number of hours after 9:00 a.m. How many chips can be tested between 10:00 a.m. and 1:00 p.m.?

4 21

3x 18x 15 dx 20

Solution - N (t) = –3t 2 + 18t + 15 The total work accomplished is the integral of this rate from t = 1 (10 a.m.) to t = 4 (1 p.m.):

Use your calculator

4 2

13x 18x 15 dx

43 2

x 1x 9 x 15x

= ( - 64 + 144 + 60) – (-1 + 9 + 15) = 117

That is, between 10 a.m. and 1 p.m., 117 chips can be tested.

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Summary.

We can evaluate a definite integral by the fundamental theorem of calculus:

)a(F)b(F)x(Fdx)x(f ba

b

a

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ASSIGNMENT

§4.4 on my website.

8, 9, 10, 11, 12, 13.