Section 4.4 β Properties of Definite Integrals Theorem: 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. π π π₯ ππ₯ = π π π π π₯ ππ₯ π π₯ ππ₯ + π π Section 4.4 β Properties of Definite Integrals Example: Calculate the area under the given curve between π₯ = β3 and π₯ = 3. βπ₯ + 6 πππ π π₯ = π₯2 πππ 3 π₯β€2 π₯>2 3 2 π π₯ ππ₯ = π₯ 2 ππ₯ βπ₯ + 6 ππ₯ + β3 β3 3 2 π₯3 3 π₯2 2 π π₯ ππ₯ = β + 6π₯ + 3 2 β3 2 β3 = 32.5 + 6.3333 3 π π₯ ππ₯ = 38.8333 β3 Section 4.4 β Properties of Definite Integrals Example: Calculate the area under the given curve between π₯ = β1 and π₯ = 6. π π₯ = 2π₯ β 4 2π₯ β 4 = 0 π₯=2 β2π₯ + 4 π π₯ = 2π₯ β 4 πππ πππ π₯β€2 π₯>2 6 π π₯ ππ₯ = β1 6 2 2π₯ β 4 ππ₯ β2π₯ + 4 ππ₯ + 2 β1 2π₯ 2 2 β + 4π₯ + β1 2 2π₯ 2 6 β 4π₯ 2 2 2 βπ₯ 2 + 4π₯ + π₯ 2 β 4π₯ 6 β1 2 6 = 9 + 16 β π π₯ ππ₯ = 25 β1 Section 4.4 β Properties of Definite Integrals π¨πππ π©ππππππ πͺπππππ 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 β lim βπ₯β0 Copyright ο 2010 Pearson Education, Inc. π π₯π βπ₯ = π=1 π π₯ ππ₯ = πΉ(π) β πΉ(π) π Section 4.4 β Properties of Definite Integrals π¨πππ π©ππππππ πͺπππππ If a continuous function, f(x), has an antiderivative, F(x), on the interval [a, b], then π β lim βπ₯β0 π π₯π βπ₯ = π=1 π π₯ ππ₯ = πΉ(π) β πΉ(π) π π€πππ‘β ππ ππππ‘πππππ = ππ₯ πππππ‘β ππ ππππ‘πππππ = π’ππππ ππ’πππ‘πππ β πππ€ππ ππ’πππ‘πππ πππππ‘β ππ ππππ‘πππππ = π π₯ β π(π₯) Copyright ο 2010 Pearson Education, Inc. Section 4.4 β Properties of Definite Integrals Example: Calculate the area bounded by the graphs of π π₯ = π₯ 2 + 1, π(π₯) = π₯, π₯ = β1 and π₯ = 1. π π΄πππ = π π₯ β π(π₯) ππ₯ π 1 π₯ 2 + 1 β π₯ ππ₯ π΄πππ = β1 π₯3 π₯2 1 +π₯β 3 2 β1 0.8333 β (β1.8333) 2.6667 Section 4.4 β Properties of Definite Integrals Example: Calculate the area bounded by the graphs of π π₯ = π₯ 2 πππ π(π₯) = 4π₯. π π΄πππ = π π₯ β π(π₯) ππ₯ π Find the points of intersection π π₯ = π(π₯) π₯ 2 = 4π₯ π₯ 2 β 4π₯ = 0 4π₯ 2 π₯ 3 4 β 2 3 0 π₯(π₯ β 4) = 0 3 π₯ 4 2π₯ 2 β 3 0 π₯ = 0, 4 4 4π₯ β π₯ 2 ππ₯ π΄πππ = 10.6667 β 0 0 10.6667 Section 4.4 β Properties of Definite Integrals Average Value of a Continuous Function 1 π΄π£πππππ ππππ’π = πβπ Copyright ο 2010 Pearson Education, Inc. π π π₯ ππ₯ π Section 4.4 β Properties of Definite Integrals Average Value of a Continuous Function Find the average value of the function π π₯ = π₯ 2 + 2 over the interval 1, 3 1 π΄π = 3β1 3 π₯ 2 + 2 ππ₯ 1 1 π₯3 π΄π = + 2π₯ 2 3 1 π΄π = 2 3 1 27 1 +6 β +2 3 3 1 7 π΄π = 15 β 2 3 π΄π = 19 = 6.3333 3 6.3333 Section 4.4 β Properties of Definite Integrals A companyβs marginal revenue and marginal cost functions are as follows: πΉβ² π = ππππ β ππ πΉ π = π, πͺβ² π = ππ β ππ πͺ π = π. 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 β πͺ(π) π π»ππππ π¨ππππππππππ π·πππππ = πΉβ² π β πͺβ²(π) π ππ ππππ β ππ β (ππ β ππ) π π a) π»ππππ π¨ππππππππππ π·πππππ = π ππ ππππ = π π π + π β 75 π π = ππππ + + πππ ππ π π = $π, πππ, πππ. ππ Section 4.4 β Properties of Definite Integrals A companyβs marginal revenue and marginal cost functions are as follows: πΉβ² π = ππππ β ππ πΉ π = π, πͺβ² π = ππ β ππ πͺ π = π. a) Find the total profit from the first 10 days. b) Find the average daily profit from the first 10 days. Reminder: 1 π΄π£πππππ ππππ’π = πβπ π π π₯ ππ₯ π 1 π΄π£πππππ π·ππππ¦ ππππππ‘ = πβπ π b) π¨πππππππ«ππππ π·πππππ = ππ β π π = ππ ππ π π π β² π‘ β πΆβ²(π‘) π ππ ππππ β ππ β (ππ β ππ) π π π π π π ππππ + π β 75 π π = ππππ + + πππ ππ π ππ π = $πππ, πππ. ππ Section 4.4 β Properties of Definite Integrals Section 4.5 β Integration Techniques: Substitution Differentiation Review: π = ππ + π π π = π ππ + π Integration: π π = π π π ππ ππ + π π π π π π π πΊπππππππππππ: π = π(π) π π = πβ²(π) π π Copyright ο 2010 Pearson Education, Inc. π π Section 4.5 β Integration Techniques: Substitution π π = Integration: ππ πππ π + π π π π π π πΊπππππππππππ: π = πππ + π π π = ππ π π π π = π ππ + π π πππ π ππ π+π= +π π π π π= π +πͺ π π π = (πππ + π)π + πͺ π Copyright ο 2010 Pearson Education, Inc. π π = ππ π π Section 4.5 β Integration Techniques: Substitution π π = Integrate: ππ ππ + π π π π π π π πΊπππππππππππ: π = ππ + π π π = π π π π π π = ππ ππ + π π π = ππ π β π ππ + π π π π π = ππ β π π π = π Copyright ο 2010 Pearson Education, Inc. π π π ππ + π ππ + π π ππ π π π = π π π π π ππ π π ππ π+π=π +π π π = ππ + πͺ π π π = (ππ + π)π +πͺ Section 4.5 β Integration Techniques: Substitution Integrate: π π π πΊπππππππππππ: ππ π π π π+π π π = π = π + ππ π π = ππ π π π π = π π π π π π + ππ π π = π π π π π = ππ π + πͺ π = ππ π + ππ + πͺ Section 4.5 β Integration Techniques: Substitution Integrate: πππ π π = π + πππ π π π π = πππ π π β π π π + πππ π π π πΊπππππππππππ: π = π + πππ π π = πππ π π π π = π+ π π β π ππ π π = π βπ π π π π= π ππ π π +π Copyright ο 2010 Pearson Education, Inc. β π= πππ π π π πππ +πͺ β π=π π π + πππ π +πͺ Section 4.5 β Integration Techniques: Substitution Integrate: π π π πΊπππππππππππ: ππ π π π π = π π π π = πππ π π π = π π π π π = ππ π π π = π π π π π ππ π π ππ π= +π π π β π = πππ π π +π Section 4.5 β Integration Techniques: Substitution Integrate: π π = π π ππ π π π π π π π πΊπππππππππππ: π = πππ π π = ππππ π π π π = π π π ππ β πππ π π π ππ π π π = ππ π π π = ππ π π ππ πππ π ππ π π π π π= π +πͺ ππ Copyright ο 2010 Pearson Education, Inc. π π β π πππ π= π +πͺ ππ Section 4.5 β Integration Techniques: Substitution π π = Integrate: π π π πΊπππππππππππ: π πβπ π π π π=πβπ π π = π π π π = π π = π π = π π = π π π π π π+π=π π+π π π π π π π + π π π π π π πβπ + πβπ π π πβπ πβπ π= + +πͺ βπ βπ π=β πβπ βπ π β πβπ π βπ +πͺ Section 4.4 β Properties of Definite Integrals
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