Explanation Given : D is the rectangle 0 ≤ x ≤ 2 , − 1 8 ≤ y ≤ 1 8 , f ( x , y ) = x 3 + 1 Calculation : Step 1: We can write ∫ ∫ D f ( x , y ) d A ≤ M . A r e a ( D ) Where, M is a constant such that f ( x , y ) ≤ M . We can see that A r e a ( D ) = 2 ( 1 / 4 ) = 1 / 2 So, it remains to show that there is some constant M so that f ( x , y ) ≤ M ...

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ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

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Domain

In simple mathematics terms, a domain is defined as the space on which all the possible inputs for the function will lie. It can also be expressed as a set of departures for the required function.

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Chapter 16.2, Problem 59E

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The mathematical proof of the relation $\int \int_{D} f (x, y) d A \leq \frac{3}{2}$ where D is the rectangle bounded by the inequalities $0 \leq x \leq 2, \frac{- 1}{8} \leq y \leq \frac{1}{8}$ and $f (x, y) = \sqrt{x^{3} + 1}$

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Chapter 16.2, Problem 58E

Chapter 16.2, Problem 60E

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CALCULUS (CLOTH)

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The mathematical proof of the relation ∫ ∫ D f ( x , y ) d A ≤ 3 2 where D is the rectangle bounded by the inequalities 0 ≤ x ≤ 2 , − 1 8 ≤ y ≤ 1 8 and f ( x , y ) = x 3 + 1

Solution Summary: The author explains how the mathematical proof of the relation D f ( x, y ) d A = 3 2 is derived.

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The mathematical proof of the relation $\int \int_{D} f (x, y) d A \leq \frac{3}{2}$ where D is the rectangle bounded by the inequalities $0 \leq x \leq 2, \frac{- 1}{8} \leq y \leq \frac{1}{8}$ and $f (x, y) = \sqrt{x^{3} + 1}$

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Chapter 16 Solutions

The mathematical proof of the relation ∫ ∫ D f ( x , y ) d A ≤ 3 2 where D is the rectangle bounded by the inequalities 0 ≤ x ≤ 2 , − 1 8 ≤ y ≤ 1 8 and f ( x , y ) = x 3 + 1

Solution Summary: The author explains how the mathematical proof of the relation D f ( x, y ) d A = 3 2 is derived.

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The mathematical proof of the relation ∫∫Df(x,y)dA≤32 where D is the rectangle bounded by the inequalities 0≤x≤2,−18≤y≤18 and f(x,y)=x3+1

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Chapter 16 Solutions

The mathematical proof of the relation $\int \int_{D} f (x, y) d A \leq \frac{3}{2}$ where D is the rectangle bounded by the inequalities $0 \leq x \leq 2, \frac{- 1}{8} \leq y \leq \frac{1}{8}$ and $f (x, y) = \sqrt{x^{3} + 1}$