if a < b < c, then for any number b between a and c, the integral from a to c is the integral...
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If a < b < c, then for any number b between a and c, the integral from a to c is the integral from a to b plus the integral from b to c.
Theorem:
β«π
π
π (π₯ )ππ₯=ΒΏΒΏβ«π
π
π (π₯ )ππ₯+ΒΏΒΏβ«π
π
π (π₯ )ππ₯
Section 4.4 β Properties of Definite Integrals
Section 4.4 β Properties of Definite Integrals
π (π₯ )={β π₯+6 πππ π₯ β€ 2π₯2 πππ π₯>2
Example:
Calculate the area under the given curve between and
β«β3
3
π (π₯ ) ππ₯=ΒΏΒΏβ«β3
2
(βπ₯+6 )ππ₯+ΒΏΒΏβ«2
3
π₯2ππ₯
βπ₯2
2+6 π₯ΒΏ
π₯3
3 |32β«β3
3
π (π₯ ) ππ₯=ΒΏΒΏ
ΒΏ32.5+ΒΏ6.3333
β«β3
3
π (π₯ ) ππ₯=38.8333
Section 4.4 β Properties of Definite Integrals
π (π₯ )={β 2π₯+4 πππ π₯β€ 22 π₯β 4 πππ π₯>2
Example:
Calculate the area under the given curve between and
β«β1
6
π (π₯ )ππ₯=ΒΏΒΏ
β«β1
2
(β2 π₯+4 )ππ₯+ΒΏΒΏβ«2
6
(2π₯β 4 )ππ₯
β2π₯2
2+4 π₯ ΒΏ 2π₯2
2β 4 π₯|62
ΒΏ9+ΒΏ16 β«β1
6
π (π₯ )ππ₯=25
π (π₯ )=|2 π₯β 4| 2 π₯β 4=0π₯=2
βπ₯2+4 π₯ ΒΏπ₯2β 4 π₯|62 β
Section 4.4 β Properties of Definite Integrals
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As the number of rectangles increased, the approximation of the area under the curve approaches a value.
If a continuous function, f(x), has an antiderivative, F(x), on the interval [a, b], then
π¨πππ π©πππππππͺπππππ
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π¨πππ π©πππππππͺπππππ
hπππππ‘ ππ ππππ‘πππππ= π (π₯ )βπ (π₯)
hπ€πππ‘ ππ ππππ‘πππππ=ππ₯hπππππ‘ ππ ππππ‘πππππ=π’ππππ ππ’πππ‘πππβ πππ€ππ ππ’πππ‘πππ
If a continuous function, f(x), has an antiderivative, F(x), on the interval [a, b], then
Section 4.4 β Properties of Definite Integrals
Example:Calculate the area bounded by the graphs of and
π΄πππ=β«π
π
[ π (π₯ ) βπ (π₯)]ππ₯
0.8333 β(β1.8333)
Section 4.4 β Properties of Definite Integrals
π΄πππ=β«β1
1
[ (π₯2+1 ) β (π₯ ) ] ππ₯
π₯3
3+π₯β
π₯2
2 | 1β1
2.6667
Example:Calculate the area bounded by the graphs of
π΄πππ=β«π
π
[ π (π₯ ) βπ (π₯)]ππ₯
10.6667β 0
Section 4.4 β Properties of Definite Integrals
4 π₯2
2βπ₯3
3 |40
10.6667
Find the points of intersection
π (π₯ )=π (π₯ )π₯2=4 π₯
π₯2β 4 π₯=0π₯ (π₯β 4)=0π₯=0 , 4
π΄πππ=β«0
4
[ ( 4 π₯ )β (π₯2) ]ππ₯
2 π₯2 βπ₯3
3 |40
Average Value of a Continuous Function
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π΄π£πππππππππ’π= 1πβπβ«π
π
π (π₯ )ππ₯
Section 4.4 β Properties of Definite Integrals
Average Value of a Continuous FunctionFind the average value of the function over the interval
π΄π= 13β1
β«1
3
(π₯2+2 )ππ₯
π΄π=12 ( π₯
3
3+2π₯)|31
π΄π=12 [( 27
3+6)β( 1
3+2)]
π΄π=12 [15 β
73 ]
π΄π=193
=6.3333
Section 4.4 β Properties of Definite Integrals
6.3333
Section 4.4 β Properties of Definite IntegralsA companyβs marginal revenue and marginal cost functions are as follows:
a) Find the total profit from the first 10 days.
b) Find the average daily profit from the first 10 days.
Reminder:
π·πππππ=πΉ (π )βπͺ (π)
π»ππππ π¨ππππππππππ π·πππππ=β«π
π
(πΉ β² (π )βπͺ β²(π ) )
a) π»ππππ π¨ππππππππππ π·πππππ=β«π
ππ
(ππππβππβ(ππβππ) )π π
ΒΏβ«π
ππ
(ππππ+πβ75 )π πΒΏππππ+ ππ
π+ππ π|πππ ΒΏ $π ,πππ ,πππ .ππ
Section 4.4 β Properties of Definite IntegralsA companyβs marginal revenue and marginal cost functions are as follows:
a) Find the total profit from the first 10 days.
b) Find the average daily profit from the first 10 days.
Reminder:
b) π¨πππππππ«ππππ π·πππππ= πππβπβ«
π
ππ
(ππππ βππβ(ππβππ) )π π
ΒΏ πππβ«
π
ππ
(ππππ+πβ 75 )π πΒΏ πππ (ππππ+
ππ
π+ππ π)|πππ ΒΏ $πππ ,πππ .ππ
π΄π£πππππππππ’π= 1πβπβ«π
π
π (π₯ )ππ₯
π΄π£ππππππ·ππππ¦ ππππππ‘= 1πβπβ«π
π
π β² (π‘ ) βπΆ β² (π‘)
Section 4.4 β Properties of Definite Integrals
Differentiation Review:
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π=(π π+π )π
π π=π (ππ+π )π (π )π π
Integration:
: π=π(π)π π=π β² (π)π π
β«π π=β«ππ (π π+π )ππ π
Section 4.5 β Integration Techniques: Substitution
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Integration:
: π=π ππ+ππ π=π π π π
β«π π=β«π π (π ππ+π)ππ π
β«π π=β« (π ππ+π)ππ ππ π β«π π=β«πππ ππ+π=ππ
π+π
π=ππππ
+πͺ
π=ππ
(π ππ+π)π
+πͺ
Section 4.5 β Integration Techniques: Substitution
Copyright 2010 Pearson Education, Inc.
Integrate:
: π=π π+ππ π=ππ π
β«π π=β«ππ (π π+π )ππ π
β«π π=ππβ« (π π+π )ππ πβ«π π=ππβ« π
πβπ (π π+π )ππ π π+π=ππ
π
π+π
π=ππ+πͺ
π=(π π+π)π+πͺβ«π π=ππ βππβ«π (π π+π )ππ π
β«π π=πβ« (π π+π )πππ π
β«π π=πβ«πππ π
Section 4.5 β Integration Techniques: Substitution
Integrate:
: π=π+ππ
π π=π ππ π
β«π π=β« ππ
π+ππ π π
π=πππ+πͺ
β«π π=β« ππ+ππ π
ππ π
β«π π=β« πππ π
π=ππ (π+π π )+πͺ
Section 4.5 β Integration Techniques: Substitution
Copyright 2010 Pearson Education, Inc.
Integrate:
: π=π+πππ
π π=π πππ π
β«π π=β« π ππ
βπ+π πππ π
π=π
ππ
ππ
+π π=ππππ+πͺ
β«π π=β«πβππ π π
β«π π=β«π ππ (π+π ππ )βππ π π
β«π π=β« (π+πππ )βππ π πππ π
π=π (π+π ππ)ππ+πͺβ β
Section 4.5 β Integration Techniques: Substitution
Integrate:
: π=ππππ π=
πππ π
β«π π=β« (πππ )π
ππ π
β«π π=β«πππ πβ«π π=β« (ππ π )π π
ππ π
π=ππ
π+πβπ=
ππ
(πππ )π+π
Section 4.5 β Integration Techniques: Substitution
Copyright 2010 Pearson Education, Inc.
Integrate:
: π=π ππ
π π=πππππ π
β«π π=β«πππππ π
π π
β«π π=β« πππ
βπππππππ π
π π
π= πππ
ππ
+πͺ
β«π π=πππβ«ππππππ ππ
π π
β«π π=πππβ«πππ π
π= πππ
πππ π
+πͺβ
Section 4.5 β Integration Techniques: Substitution
Integrate:
: π=π βππ π=π π
β«π π=β« π(πβπ )π
π π
π=πβπ
βπ+π
βπ
βπ+πͺ
β«π π=β« πππ π π
β«π π=β« πππ+
πππ π π
π+π=π
β«π π=β« π+πππ π π
β«π π=β« (πβπ+πβπ )π π
π=β (πβπ )βπβππ
(πβπ )βπ+πͺ
Section 4.5 β Integration Techniques: Substitution
Section 4.4 β Properties of Definite Integrals