Explanation Given: For each continuous function f : [ 0 , 1 ] → ℝ and I ( f ) = ∫ 0 1 x 2 f ( x ) d x and J ( f ) = ∫ 0 1 x ( f ( x ) ) 2 d x Formula Used: The integration formula is ∫ x n d x = x n + 1 n + 1 . Calculation: The integrals given are I ( f ) = ∫ 0 1 x 2 f ( x ) d x and J ( x ) = ∫ 0 1 x ( f ( x ) ) 2 d x Subtract I ( f ) − J ( f ) , I ( f ) − J ( f ) = ∫ 0 1 x 2 f ( x ) d x − ∫ 0 1 x ( f ( x ) ) 2 d x = ∫ 0 1 x 2 f ( x ) − x ( f ( x ) ) 2 d x …… (1) Add and subtract x 3 4 to get, x 2 f ( x ) − x ( f ( x ) ) 2 = x 3 4 − x f ( x ) 2 + x 2 f ( x ) − x 3 4 = x 3 4 − x ( f ( x ) 2 − x f ( x ) + x 2

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Limits of Multivariable Functions

Multivariable calculus is an extension of single variable calculus. It involves several variables instead of just one. Assume there is an open set containing points (x0, y0), let f be a function defined in that open interval except for the points (x0, y0…

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Chapter 4.3, Problem 78E

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To calculate: The maximum value of I(f)−J(f) over all such functions f if for each continuous function f:[0,1]→ℝ and I(f)=∫01x2f(x)dx and J(f)=∫01x(f(x))2dx.

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Chapter 4.3, Problem 77E

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Question 1: If a barometer were built using oil (p = 0.92 g/cm³) instead of mercury (p = 13.6 g/cm³), would the column of oil be higher than, lower than, or the same as the column of mercury at 1.00 atm? If the level is different, by what factor? Explain. (5 pts) Solution: A barometer works based on the principle that the pressure exerted by the liquid column balances atmospheric pressure. The pressure is given by: P = pgh Since the atmospheric pressure remains constant (P = 1.00 atm), the height of the liquid column is inversely proportional to its density: Step 1: Given Data PHg hol=hgx Poil • Density of mercury: PHg = 13.6 g/cm³ Density of oil: Poil = 0.92 g/cm³ • Standard height of mercury at 1.00 atm: hμg Step 2: Compute Height of Oil = 760 mm = 0.760 m 13.6 hoil = 0.760 x 0.92 hoil = 0.760 × 14.78 hoil = 11.23 m Step 3: Compare Heights Since oil is less dense than mercury, the column of oil must be much taller than that of mercury. The factor by which it is taller is: Final…

Question 3: A sealed flask at room temperature contains a mixture of neon (Ne) and nitrogen (N2) gases. Ne has a mass of 3.25 g and exerts a pressure of 48.2 torr. . N2 contributes a pressure of 142 torr. • What is the mass of the N2 in the flask? • Atomic mass of Ne = 20.1797 g/mol • Atomic mass of N = 14.0067 g/mol Solution: We will use the Ideal Gas Law to determine the number of moles of each gas and calculate the mass of N2. PV = nRT where: • P = total pressure • V volume of the flask (same for both gases) n = number of moles of gas • R 0.0821 L atm/mol K • T = Room temperature (assume 298 K) Since both gases are in the same flask, their partial pressures correspond to their mole fractions. Step 1: Convert Pressures to Atmospheres 48.2 PNe = 0.0634 atm 760 142 PN2 = = 0.1868 atm 760 Step 2: Determine Moles of Ne nNe = mass molar mass 3.25 nNe 20.1797 nne 0.1611 mol Step 3: Use Partial Pressure Ratio to Find n

Chapter 4 Solutions

EBK CALCULUS

Ch. 4.1 - Integration and Differentiation In Exercises 5 and...Ch. 4.1 - Integration and Differentiation In Exercises 5 and...Ch. 4.1 - Solving a Differential Equation In Exercises 7-10,...Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Solving a Differential Equation In Exercises 7-10,...Ch. 4.1 - Prob. 7E Ch. 4.1 - Rewriting Before Integrating In Exercises 11-14,...Ch. 4.1 - Rewriting Before Integrating In Exercises 11-14,...Ch. 4.1 - Prob. 10E

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The maximum value of I ( f ) − J ( f ) over all such functions f if for each continuous function f : [ 0 , 1 ] → ℝ and I ( f ) = ∫ 0 1 x 2 f ( x ) d x and J ( f ) = ∫ 0 1 x ( f ( x ) ) 2 d x .

Solution Summary: The author explains the formula used to calculate the maximum value of I(f)-J

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To calculate: The maximum value of I(f)−J(f) over all such functions f if for each continuous function f:[0,1]→ℝ and I(f)=∫01x2f(x)dx and J(f)=∫01x(f(x))2dx.

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