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4029 u-du: Integrating Composite Functions
AP Calculus
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Find the derivative 5π₯ 5 +4 π₯ 3 +3π₯+2 5 5 5π₯ 5 +4 π₯ 3 +3π₯ π₯ π₯ 2 +3 dx/du-part of the antiderivative
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Integrating Composite Functions
u-du Substitution Integrating Composite Functions (Chain Rule) Revisit the Chain Rule If let u = inside function du = derivative of the inside becomes = π₯ 2 +1 2π₯ππ₯
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A Visual Aid USING u-du Substitution ο a Visual Aid REM: u = inside function du = derivative of the inside let u = becomes now only working with f , the outside function π₯ 2 +1 2π₯ππ₯ π₯ 2 +1 ππ’= 2π₯ππ₯ 3 π’ π π’ 3 +π π₯ π
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Example 1 : du given Ex 1: π’= 5π₯ 2 +1 π’ 3 ππ’ ππ’=10π₯ππ₯ π’ π π₯ π π₯ π proof π₯ π₯ 5π₯ π₯
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Example 2: du given π’= π₯ 3 +1 Ex 2: ππ’= 3π₯ 2 ππ₯ π₯ 3 +1 1 2 3π₯ 2 ππ₯
ππ’= 3π₯ 2 ππ₯ π₯ π₯ 2 ππ₯ π’ ππ’ π¦= π₯ π π’ π 2 3 π’ π
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Example 3: du given π’= π₯ 2 +1 Ex 3: ππ’=2π₯ππ₯ π₯ 2 +1 β 1 2 2π₯ππ₯
π’= π₯ 2 +1 Ex 3: ππ’=2π₯ππ₯ π₯ β π₯ππ₯ 2π’ π π’ β ππ’ 2 π₯ π π’ π
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Example 4: du given Ex 4: π’= tan (π₯ ) π’= sec (π₯) ππ’= π ππ 2 (π₯)ππ₯
π’= tan (π₯ ) π’= sec (π₯) Derivative only ππ’= π ππ 2 (π₯)ππ₯ ππ’= sec π₯ tanβ‘(π₯) π ππ 2 π₯ ππ₯ π’ ππ’ π’ ππ’ π’ π π’ π Function and derivative TWO WAYS! Differ by a constant 1 2 π‘ππ 2 (π₯)+π 1 2 π ππ 2 π₯ +π tan π₯ π ππ 2 π₯ ππ₯ 1+ π‘ππ 2 π₯ = π ππ 2 (π₯) Both ways !
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Example 5: Regular Method
π’= sin (π₯) ππ’= cos π₯ ππ₯ Ex 5: cosβ‘(π₯) sinβ‘(π₯) β2 ππ₯ cos (π₯) sin (π₯) β 1 sin π₯ = cot π₯ csc π₯ ππ₯ sinβ‘(π₯) β2 cos π₯ ππ₯ β csc π₯ +π π’ β2 ππ’ π’ β1 β1 +π βπ’ β1 +π β sin π₯ β1 +π β 1 sin π₯ +π=β csc π₯ +π
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Working with Constants < multiplying by one>
Constant Property of Integration ILL let u = du = and becomes = Or alternately = =
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Example 6 : Introduce a Constant - my method
β2 β2 π₯ 9β π₯ 2 ππ₯ π’=9β π₯ 2 ππ’=β2π₯ππ₯ β 1 2 β2 β β π₯ β2π₯ππ₯ β π’ ππ’ β π’ π β 1 3 π’ π β β π₯ π
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Example 7 : Introduce a Constant
π’=3π₯ ππ’=3ππ₯ π ππ 2 3π₯ ππ₯ π ππ 2 3π₯ 3ππ₯ π ππ 2 π’ ππ’ 1 3 tan π’ +π 1 3 tan 3π₯ +π
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sec π₯ tan π₯ ππ₯ π’= sec π₯ ππ’= sec π₯ tan π₯ sec π₯ sec π₯ sec π₯ tan π₯ ππ₯ 1 sec π₯ sec π₯ tan π₯ sec π₯ ππ₯ 1 sec π₯ π’ ππ’ 1 sec π₯ π’ π 1 sec π₯ π ππ 2 π₯ 2 +π 1 2 sec π₯+π
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Example 8 : Introduce a Constant << triple chain>>
π’= sin (2π₯) ππ’= cos 2π₯ 2ππ₯ π ππ 4 2π₯ cos 2π₯ β2ππ₯ π’ 4 ππ’ π’ π π’ π 1 10 π ππ 5 2π₯ +π
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Example 9 : Introduce a Constant - extra constant
You is what You is inside 5 3π₯+4 5 ππ₯ << extra constant> π₯ ππ₯ π’=3π₯+4 ππ’=3ππ₯ π’ 5 ππ’ π’ π π₯ π
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π’= 3π₯ 2 β2π₯+1 Example 10 : Polynomial ππ’=(6π₯β2)ππ₯ (3π₯β1) 3π₯ 2 β2π₯ ππ₯ π’ β4 ππ’ π’ β3 β3 +π β π₯ 2 β2π₯+1 β3 +π
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Example 11: Separate the numerator
π’= π₯ 2 +1 Example 11: Separate the numerator ππ’=2π₯ππ₯ 2π₯ π₯ 2 +1 ππ₯ π₯ 2 +1 ππ’ π’ π₯+1 2 π’ β1 ππ’ π₯ 2 +1 = π’ 0 0 ln π’ + arctan (π₯) +π
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Formal Change of Variables << the Extra βxβ>>
ILL: Let Solve for x in terms of u then and becomes = π’ β6 π’ ππ’ π’ ππ’β π’ ππ’= π’ β π’ = 1 5 π’ β2 π’ 3 2 π₯ β2 2π₯β π
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Formal Change of Variables << the Extra βxβ>>
Rewrite in terms of u - du π’=π₯+3 ππ’=ππ₯ π₯=π’β3 2π’β7 π’ β 1 2 ππ’ 2π₯=2π’β6 2π₯β1=2π’β7 2π’ β 3 2 β7 π’ β ππ’ 2β 2 5 π’ β7β2 π’ π 4 5 π’ β14 π’ π 4 5 π₯ β14 π₯ π
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Assignment Day 1 Worksheet Larson HW 4029 Day 2 Basic Integration Rules Wksht extra x Larson f anti for tan /cot Text p # (3x)
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Formal Change of Variables << the Extra βxβ>>
Solve for x in terms of u - du <<alt. Method>> - could divide or multiply by
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Complete Change of Variables << Changing du >>
At times it is required to even change the du as the u is changed above. We will solve this later in the course.
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Integrating Composite Functions
(Chain Rule) Remember: Derivatives Rules Remember: Laymanβs Description of Antiderivatives *2nd meaning of βduβ du is the derivative of an implicit βuβ
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Development must have the derivative of the inside in order to find the antiderivative of the outside *2nd meaning of βdxβ dx is the derivative of an implicit βxβ more later if x = f then dx = f /
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Development from the laymanβs idea of antiderivative βThe Family of functions that has the given derivativeβ must have the derivative of the inside in order to find the antiderivative of the outside
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Working With Constants: Constant Property of Integration
With u-du Substitution REM: u = inside function du = derivative of the inside Missing Constant? u = du = Worksheet - Part 1
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