Explanation Given Property: Property 5: ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x ( a < c < b ) Verification: The given equation is, ∫ 0 3 ( 1 + x 3 ) d x = ∫ 0 1 ( 1 + x 3 ) d x + ∫ 1 2 ( 1 + x 3 ) d x + ∫ 2 3 ( 1 + x 3 ) d x . First calculate the integration of the left hand side (L.H.S) of the given equation. L .H .S = ∫ 0 3 ( 1 + x 3 ) d x = [ x + x 4 4 ] 0 3 = 3 + 81 4 − 0 = 93 4 Now, evaluate individually each part of the right hand side (R.H.S) of the equation. The value of first integral, I 1 = ∫ 0 1 ( 1 + x 3 ) d x = [ x + x 4 4 ] 0 1 = 1 + 1 4 − 0 = 5 4 The value of second integral, I 2 = ∫ 1 2 ( 1 + x 3 ) d x = [ x + x 4 4

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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Author: Tan

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Chapter 6.5, Problem 76E

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To verify: The equation ∫03(1+x3)dx=∫01(1+x3)dx+∫12(1+x3)dx+∫23(1+x3)dx by direct computation and illustrate that property 5 may be extended.

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Chapter 6.5, Problem 75E

Chapter 6.5, Problem 77E

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Chapter 6 Solutions

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

Ch. 6.1 - What is an antiderivative? Give an example.Ch. 6.1 - Prob. 2CQ Ch. 6.1 - What is the difference between an antiderivative...Ch. 6.1 - Prob. 4CQ Ch. 6.1 - In Exercises 1-4, verify directly that F is an...Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - In Exercises 1-4, verify directly that F is an...Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E

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To verify: The equation ∫ 0 3 ( 1 + x 3 ) d x = ∫ 0 1 ( 1 + x 3 ) d x + ∫ 1 2 ( 1 + x 3 ) d x + ∫ 2 3 ( 1 + x 3 ) d x by direct computation and illustrate that property 5 may be extended.

Solution Summary: The author illustrates that property 5 may be extended by calculating the integration of the left hand side (L.H.S).

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To verify: The equation ∫03(1+x3)dx=∫01(1+x3)dx+∫12(1+x3)dx+∫23(1+x3)dx by direct computation and illustrate that property 5 may be extended.

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Chapter 6 Solutions