1 the student will learn about: §4.4 definite integrals and areas. the fundamental theorem of...
TRANSCRIPT
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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
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Example 2
4
3
1dxx
3
1dxx
3
1x
2
2
x
Make a drawing to confirm your answer.
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1
2
9
0 x 4
- 1 y 4
2
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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
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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.?
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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.