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4.9 β Antiderivatives
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In addition to finding derivatives, there is an important βinverseβ problem
Given the derivative, find the function itself. For example, in physics we may know the velocity π£ π‘ (the derivative) and wish to compute the position π π‘ of an object. Since π β² π‘ =π£ π‘ , this amounts to finding a function whose derivative is π£ π‘ . A function πΉ π₯ whose derivative is π π₯ is called an antiderivative of π π₯ .
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Definition Antiderivatives
A function πΉ π₯ is an antiderivative of π π₯ on π, π if πΉ β² π₯ =π π₯ for all π₯β π, π .
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Examples πΉ π₯ =β cos π₯ is an antiderivative of π π₯ = sin π₯ because πΉ β² π₯ = π ππ₯ β cos π₯ = sin π₯ =π π₯ πΉ π₯ = 1 3 π₯ 3 is an antiderivative of π π₯ = π₯ 2 because πΉ β² π₯ = π ππ₯ π₯ 3 = π₯ 2 =π π₯
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One critical observation is that antiderivatives are not unique.
We are free to add a constant C because the derivative of a constant is zero. If πΉ β² π₯ =π π₯ , then πΉ π₯ +πΆ β² =π π₯ . Each of the following is an antiderivative of π₯ 2 1 3 π₯ 3 1 3 π₯ 3 +5 1 3 π₯ 3 β4 The process of finding an antiderivative is called integration.
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Notation Indefinite Integral
The notation π π₯ ππ₯ =πΉ π₯ +πΆ means that πΉ β² π₯ =π π₯ We say that πΉ π₯ +πΆ is the general antiderivative or indefinite integral of π π₯
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Power Rule for Integrals
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Antiderivative of y = 1/x
The function πΉ π₯ = ln π₯ is an antiderivative of π¦= 1 π₯ in the domain π₯:π₯β 0 ; that is ππ₯ π₯ = ln π₯ +πΆ
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Linearity of the Indefinite Integral
Sum Rule π π₯ +π π₯ ππ₯ = π π₯ ππ₯ + π π₯ ππ₯ Multiples Rule ππ π₯ ππ₯ =π π π₯ ππ₯
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Basic Trigonometric Integrals
sin π₯ ππ₯ =β cos π₯ +πΆ cos π₯ ππ₯ = sin π₯ +πΆ sec 2 π₯ ππ₯ = tan π₯ +πΆ csc 2 π₯ ππ₯ =β cot π₯ +πΆ sec π₯ tan π₯ ππ₯ = sec π₯ +πΆ csc π₯ cot π₯ ππ₯ =β csc π₯ +πΆ
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Integrals Involving ex
π π₯ ππ₯ = π π₯ +πΆ π ππ₯+π ππ₯ = 1 π π ππ₯+π +πΆ
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Examples Evaluate cos π₯ ππ₯ Evaluate 3 π₯ 4 β5 π₯ 2/3 + π₯ β3 ππ₯
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Examples Evaluate sin 8π‘β3 +20 cos 9π‘ ππ‘ Evaluate 3 π π₯ β4 ππ₯
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We can think of an antiderivative as a solution to the differential equation ππ¦ ππ₯ =π π₯
In general, a differential equation is an equation relating an unknown function and its derivatives. A differential equation with an initial condition is called an initial value problem.
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Examples Solve ππ¦ ππ₯ =4 π₯ 7 with an initial condition π¦ 0 =4.
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Examples Solve the differential equation ππ¦ ππ‘ = sin ππ‘ with the initial condition π¦ 2 =2
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Homework Text page 281 #s even, even, all
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Speed Quiz
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5.3 β The Fundamental Theorem of Calculus, Part I
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Fundamental Theorem of Calculus Part I
Assume that π π₯ is continuous on π, π . If πΉ π₯ is an antiderivative of π π₯ on π, π , then π π π π₯ ππ₯ =πΉ π βπΉ π OR π π π β² π₯ ππ₯ =π π βπ π
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Other Important Integral Theorems
π π π π₯ ππ₯ = signed area of region between the graph and x-axis over [a, b] For a < b, π π π π₯ ππ₯ =β π π π π₯ ππ₯ π π π π₯ ππ₯ =0 Let πβ€πβ€π, and assume that π π₯ is integrable. π π π π₯ ππ₯= π π π π₯ ππ₯ + π π π π₯ ππ₯ π π ππ₯ π₯ = ln π β ln π = ln π π For an object in linear motion with velocity π£ π‘ , then Displacement during π‘ 1 , π‘ 2 = π‘1 π‘2 π£ π‘ ππ‘ Distance traveled during π‘ 1 , π‘ 2 = π‘1 π‘2 π£ π‘ ππ‘
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Example Calculate the area under the graph of π π₯ = π₯ 3 over 2, 4
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Example Calculate βπ/4 π/4 sec 2 π₯ ππ₯ and sketch the corresponding region.
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Example Calculate 0 π sin π₯ ππ₯ and sketch the corresponding region.
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Example Calculate 0 2π sin π₯ ππ₯ and sketch the corresponding region.
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Example Calculate 0 2 π π₯ ππ₯ given π π₯ = 3 π₯ for π₯<1 2π₯+1 for π₯β₯1
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Example Calculate β2 5 π π₯ ππ₯ given π π₯ = 2π₯β π₯ 2 for π₯< for π₯β₯1
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Example Calculate β3 2 π₯ ππ₯ by first rewriting the integral without absolute value.
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Example Calculate β1 4 π₯β2 ππ₯ by first rewriting the integral without absolute value.
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Homework Text pages #s 6-14 even, 28, 30, 38, 43, 44, 55, 56
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5.4 β The Fundamental Theorem of Calculus, Part II
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Example Find a formula for the area function π΄ π₯ = 3 π₯ π‘ 2 ππ‘ .
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Example Find the formula for the area function π΄ π₯ = 2 π₯ 2 2π₯+5 ππ₯
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Fundamental Theorem of Calculus Part II
Assume that π π₯ is continuous on an open interval I and let πβπΌ. Then the area function π΄ π₯ = π π₯ π π‘ ππ‘ is an antiderivative of π π₯ on I; that is, π΄ β² π₯ =π π₯ . Equivalently, π ππ₯ π π₯ π π‘ ππ‘ =π π₯ Furthermore, π΄ π₯ satisfies the initial condition π΄ π =0 If the upper bound is a function of x, apply the chain rule and multiply by the derivative of the upper bound.
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Example Find the derivative of π΄ π₯ = 2 π₯ 1+ π‘ 3 ππ‘ . Evalute π΄ β² 2 , π΄ β² 3 , and π΄ 2 .
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Example Find the derivative of πΊ π₯ = β2 π₯ 2 sin π‘ ππ‘
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Homework Text page 320 #s 4-10 even, 28
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Quiz
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5.6 β Substitution Method
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Integration (antidifferentiation) is generally more difficult than differentiation.
There are no sure-fire methods, and many antiderivatives cannot be expressed in terms of elementary functions. However, there are a few important general techniques. One such technique is the Substitution Method, which uses the Chain Rule βin reverse.β Consider the integral 2π₯ cos π₯ 2 ππ₯ . We can evaluate it if we remember the Chain Rule π ππ₯ sin π₯ 2 =2π₯ cos π₯ 2 This tells us that sin π₯ 2 is an antiderivative of 2π₯ cos π₯ 2 , and therefore 2π₯ Derivative of the inside function cos π₯ 2 Inside function ππ₯=sinβ‘ π₯ 2 +πΆ
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The integrand is the product of a composite function and the derivative of the inside function.
The Chain Rule does not help if the derivative of the inside function is missing. For instance, we cannot use the Chain Rule to compute cos π₯+ π₯ 3 ππ₯ because the factor 1+3 π₯ 2 does not appear.
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Theorem β The Substitution Method
If πΉ β² π₯ =π π₯ , then π π’ π₯ π’ β² π₯ ππ₯ =πΉ π’ π₯ +πΆ
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Equivalently, ππ’ and ππ₯ are related by ππ’= π’ β² π₯ ππ₯ For example
Before proceeding to the examples, we discuss the procedure for carrying out substitution using differentials. Differentials are symbols such as ππ’ or ππ₯ that occur in the Leibniz notations. In our calculators, we shall manipulate them as thorugh they are related by an equation in which the ππ₯ βcancelsβ: ππ’= ππ’ ππ₯ ππ₯ Equivalently, ππ’ and ππ₯ are related by ππ’= π’ β² π₯ ππ₯ For example If π’= π₯ 2 then ππ’=2π₯ππ₯ If π’= cos π₯ 3 then ππ’=β3 π₯ 2 sin π₯ 3 ππ₯
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Example Evalaute 3 π₯ 2 sin π₯ 3 ππ₯
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Example Evalaute 3 π₯ 2 sin π₯ 3 ππ₯
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Example Evalaute π₯ π₯ ππ₯
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Example Evaluate π₯ 2 +2π₯ ππ₯ π₯ 3 +3 π₯
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Example Evalaute sin 7π+5 ππ
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Example Evaluate π β9π‘ ππ‘
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Change of Variables for Definite Integrals
π π π π’ π₯ π’ β² π₯ ππ₯ = π’ π π’ π π π’ ππ’
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Example Evaluate π₯ 2 π₯ 3 +1 ππ₯
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example Evaluate π₯+1 π₯ 2 +2π₯ 5 ππ₯
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example Evalaute 0 π/4 tan 3 π sec 2 π ππ
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example Calculate the area under the graph of π¦= π₯ π₯ over 1, 3
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Homework Text pages #s even, 48, 50, 80, 82, 86
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