§12.5 the fundamental theorem of calculus
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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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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
33
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.
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