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§12.5 The Fundamental Theorem of Calculus

The student will learn about:

the definite integral,the fundamental theorem of calculus,and using definite integrals to find average values.

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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

By the fundamental theorem we can evaluate

Easily and exactly.

b

adx)x(f

)a(F)b(F

By the fundamental theorem we can evaluate

Easily and exactly. We simply calculate

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Definite Integral Properties

a

a 0dx)x(f

a

b

b

a dx)x(fdx)x(f

b

a

b

a dx)x(fkdx)x(fk

b

a

b

a

b

a dx)x(gdx)x(fdx])x(g)x(f[

b

c

c

a

b

a dx)x(fdx)x(fdx)x(f

REMINDER !

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Example 1

5 · 3 – 5 · 1 = 15 – 5 = 10

Make a drawing to confirm your answer.

31 dx5

31 dx5

3

1xx5

0 x 4

- 1 y 6

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Example 2

4

31 dxx

31 dxx

3

1x

2

2x

Make a drawing to confirm your answer.

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29

0 x 4

- 1 y 4

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Example 3

1. 9 - 0 =30

2 dxx 3

0x

3

3x

0 x 4

- 2 y 10

9

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Example 4

2. Let u = 2x du = 2 dx

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x2 dxe

1

1x

u

2e

11

x2 dx2e21

11

u due21

2e

2e 22

1

1x

x2

2e

3.6268604

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Example 5

3. 21 dx

x1 2

1xxln

= ln 2 – ln 1 = ln 2

= 0.69314718

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Examples 6

4. This is a combination of the previous three problems

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1x

x23xln

2e

3x

3

1x22 dx

x1ex

= 9 + (e 6)/2 + ln 3 – 1/3 – (e2)/2 – ln 1

= 207.78515

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Examples 7

5. Let u = x3 + 4

5

0x3uln

50 3

2dx

4xx

50 3

2dx

4xx3

31

du = 3x2 dx

50 du

u1

31

50x

3

3)4x(ln

(ln 129)/3 – (ln 4)/3 = 1.1578393

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Examples 7 REVISITED

5. Let u = x3 + 4

5

0x3uln

50 3

2dx

4xx

50 3

2dx

4xx3

31

du = 3x2 dx

50 duu1

31

129

4u3uln

(ln 129)/3 – (ln 4)/3 = 1.1578393

At this point instead of substituting for u we can replace the x value in terms of u. If x = 0 then u = 4 and if x = 5, then u = 129.

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Numerical Integration on a Graphing Calculator

Use some of the examples from above.

21 dx

x1

50 3

2dx

4xx

3.

5.

0 x 3- 1 y 3

-1 x 6- 0.2 y

0.5

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Example 8

From 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 x is the total accumulated cost of maintenance for x years.

dx000,5x907

3

2

Write a definite integral that will give the total maintenance cost from the end of the seventh year to the end of the seventh year. Evaluate the integral.

30 x 3 + 5,000x |7

3x

= 10,290 + 35,000 – 810 – 15,000

= $29,480

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Using Definite Integrals for Average Values

Average Value of a Continuous Function f over [a, b].

dx)x(fab

1 b

a

Note this is the area under the curve divided by the width. Hence, the result is the average height or average value.

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Example

§6.5 # 70. The total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x a. Find the average cost per unit if 1000 dictionaries are produced

Continued on next slide.

Note that the average cost is .x

)x(C)x(C

10x

20000)x(C

)1000(C 30 10100020000

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Example - continued

The total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x

b. Find the average value of the cost function over the interval [0, 1000]

Continued on next slide.

dx)x(f

ab1 b

a dxx)1020000(1000

1 1000

0

1000

0x2 )x5x20000(

10001

20,000 + 5,000 = 25,000

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Example - continued

The total cost (in dollars) of printing x dictionaries is C (x) = 20,000 + 10x

c. Write a description of the difference between part a and part b

From part a the average cost is $30.00From part b the average value of the cost is $25,000.The average cost per dictionary of making 1,000 dictionaries us $30. The average total cost of making between 0 and 1,000 dictionaries is $25,000.

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Summary.

We can find the average value of a function f by:

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

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

ba

dx)x(fab

1 ba

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Practice Problems

§6.5; 1, 3, 5, 9, 13, 17, 21, 23, 27, 31, 35, 39, 41, 45, 49, 53, 55, 61, 65, 69, 73, 81, 85.

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§13.3 Integration by Parts

The student will learn tointegrate by parts.

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Integration by Parts

Integration by parts is based on the product formula for derivatives: )x('f)x(g)x('g)x(f)x(g)x(f

dxd

We rearrange the above equation to get:

)x('f)x(g)x(g)x(fdxd)x('g)x(f

And integrating both sides yields:

)x('f)x(g)x(g)x(fdxd)x('g)x(f

Which reduces to:

)x('f)x(g)x(g)x(f)x('g)x(f

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Integration by Parts Formula

If we let u = f (x) and v = g (x) in the preceding formula the equation transforms into a more convenient form:

Integration by Parts Formula

duvvudvu

The goal is to change the integral from u dv to v du and have the second integral be easier than the first.

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Integration by Parts Formula

This method is useful when the integral on the left side is difficult and changing it into the integral on the right side makes it easier.

Remember, we can easily check our results by differentiating our answers to get the original integrals.

Let’s look at an example.

duvvudvu

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Example

Our previous method of substitution does not work. Examining the left side of the integration by parts formula yields two possibilities.

Let’s try option 1.

duvvudvu

dxexConsider x

Option 2

dxxe x

u dv

Option 1

dxex x

u dv

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Example - continued

We have decided to let u = x and dv = e x dx, (Note: du = dx and v = e x), yielding

Which is easy to integrate= x e x – e x + C

As mentioned this is easy to check by differentiating.

duvvudvu dxex x

dxeexdxex xxx

xxxxxx exeeex C e– ex dxd

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Selecting u and dv

1. The product u dv must equal the original integrand.

2. It must be possible to integrate dv by one of our known methods.

duvvudvu

4. For integrals involving x p e ax , tryu = x p and dv = e ax dx

5. For integrals involving x p ln x q , tryu = (ln x) q and dv = x p dx

3. The new integral should not be more involved than the original integral .

duv dvu

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Example 2

Let u = ln x and dv = x 3 dx, yielding

Which when integrated is:

Check by differentiating.

duvvudvu dxxlnx 3

dxx1

4xxln

4xdxxlnx

443

xnlx16x4xlnx

x1

4xC

16x– x ln

4x

dxd 3

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444

du = 1/x dx and v = x 4/4, and the integral is

C16xxln

4x 44

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Summary.

We have developed integration by parts as a useful technique for integrating more complicated functions.

Integration by Parts Formula

duvvudvu

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Practice Problems

§7.3; 1, 3, 5, 7, 11, 15, 19, 21, 23, 25, 29, 33, 37, 41, 43, 47, 49, 57, 63, 65.