Explanation Given : A r e a ( D ) = ∫ ∫ D 1 d A , D is the region between the two curves y = g 1 ( x ) and y = g 2 ( x ) with g 2 ( x ) ≤ g 1 ( x ) for a ≤ x ≤ b Calculation : Step 1: The region D is defined by the inequalities a ≤ x ≤ b g 1 ( x ) ≤ y ≤ g 2 ( x ) Step 2: We compute the double integral of f ( x , y ) = 1 on D using Theorem 2, by evaluating the following iterated integral: ∫ ∫ D 1 d A = ∫ a b ( ∫ g 1 ( x ) g 2

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.2, Problem 67E

To determine

The mathematical proof of the statement stating that if D is the region between the two curves $y = g_{1} (x)$ and $y = g_{2} (x)$ with $g_{2} (x) \leq g_{1} (x)$ for $a \leq x \leq b$ , then

$\int \int_{D} 1 d A = \int_{a}^{b} (g_{1} (x) - g_{2} (x)) d x$

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Chapter 16.2, Problem 66E

Chapter 16.2, Problem 68E

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Chapter 16 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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The mathematical proof of the statement stating that if D is the region between the two curves y = g 1 ( x ) and y = g 2 ( x ) with g 2 ( x ) ≤ g 1 ( x ) for a ≤ x ≤ b , then ∫ ∫ D 1 d A = ∫ a b ( g 1 ( x ) − g 2 ( x ) ) d x

Solution Summary: The author calculates the double integral of f (x, y ) = 1 on D using Theorem 2.

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The mathematical proof of the statement stating that if D is the region between the two curves $y = g_{1} (x)$ and $y = g_{2} (x)$ with $g_{2} (x) \leq g_{1} (x)$ for $a \leq x \leq b$ , then

$\int \int_{D} 1 d A = \int_{a}^{b} (g_{1} (x) - g_{2} (x)) d x$

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Chapter 16 Solutions

The mathematical proof of the statement stating that if D is the region between the two curves y = g 1 ( x ) and y = g 2 ( x ) with g 2 ( x ) ≤ g 1 ( x ) for a ≤ x ≤ b , then ∫ ∫ D 1 d A = ∫ a b ( g 1 ( x ) − g 2 ( x ) ) d x

Solution Summary: The author calculates the double integral of f (x, y ) = 1 on D using Theorem 2.

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The mathematical proof of the statement stating that if D is the region between the two curves y=g1(x) and y=g2(x)with g2(x)≤g1(x) for a≤x≤b, then ∫∫D1dA=∫ab(g1(x)−g2(x))dx

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Chapter 16 Solutions

The mathematical proof of the statement stating that if D is the region between the two curves $y = g_{1} (x)$ and $y = g_{2} (x)$ with $g_{2} (x) \leq g_{1} (x)$ for $a \leq x \leq b$ , then

$\int \int_{D} 1 d A = \int_{a}^{b} (g_{1} (x) - g_{2} (x)) d x$