Explanation Given: ∫ 0 6 ∫ y 2 3 ( x + y ) d x d y Calculation: Now, evaluate iterated integral, ∫ 0 6 ∫ y 2 3 ( x + y ) d x d y = ∫ 0 6 ( x 2 2 + x y ) y 2 3 d y = ∫ 0 2 ( 9 2 + 3 y − y 2 8 − y 2 2 ) d y

Calculus: Early Transcendental Functions

6th Edition

ISBN: 9781305005303

Author: Ron Larson, Bruce Edwards

Publisher: CENGAGE C

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Chapter 14.2, Problem 7E

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To graph: The iterated integral $∫06∫y23(x+y)dxdy$ and sketch of the region R.

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Chapter 14.2, Problem 6E

Chapter 14.2, Problem 8E

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Chapter 14 Solutions

Calculus: Early Transcendental Functions

Ch. 14.1 - Evaluate the iterated integral: 0433cosrdrd Ch. 14.1 - Prob. 1E Ch. 14.1 - Evaluate the integral: xx2yxdy Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Evaluate the integral: 04x2x2ydy Ch. 14.1 - Evaluate the integral: x3x(x2+3y2)dy Ch. 14.1 - Evaluate the integral: eyyylnxxdx;y0 Ch. 14.1 - Evaluate the integral: 1y21y2(x2+y2)dx Ch. 14.1 - Evaluate the integral: 0x2yeyxdy

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To graph: The iterated integral ∫ 0 6 ∫ y 2 3 ( x + y ) d x d y and sketch of the region R .

Solution Summary: The author analyzes the value of the iterated integral, cc

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To graph: The iterated integral $∫06∫y23(x+y)dxdy$ and sketch of the region R.

Want to see the full answer?

Chapter 14 Solutions