Explanation Given: ∫ 0 1 ∫ 2 3 1 ( x + 4 y ) 3 d x d y Calculation: Apply the power Rule, ∫ x a d x = x a + 1 a + 1 , a ≠ -1 Step 1: ∫ 2 3 1 ( x + 4 y ) 3 d x Compute the indefinite integral ∫ 1 ( x + 4 y ) 3 d x = − 1 2 ( x + 4 y ) 2 + C Apply u-substitution: Let u = x + 4 y ⇒ d u = d x ⇒ ∫ 1 ( x + 4 y ) 3 d x = ∫ 1 u 3 d u = ∫ u − 3 d u = u − 2 − 2 + C ⇒ ∫ 1 ( x + 4 y ) 3 d x = ( x + 4 y ) − 2 − 2 + C = − 1 2 ( x + 4 y ) 2 + C Step 2: Compute the boundaries ∫ 2 3 1 ( x + 4 y ) 3 d x Formula used: ∫ a b f ( x ) d x = F ( b ) − F ( a ) = lim x → b − ( F ( x ) ) − lim x → a + ( F ( x ) ) lim x → 2 + ( − 1 2 ( x + 4 y ) 2 ) = − 1 2 ( 2 + 4 y ) 2 [ ∵ x = 2 ] lim x → 3 − ( − 1 2 ( x + 4 y ) 2 ) = lim x → 3 − ( − 1 2 ( 3 + 4 y ) 2 ) [ ∵ x = 3 ] ∴ ∫ 2 3 1 ( x + 4

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ISBN: 9781319323394

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 16.1, Problem 36E

To determine

The value of the iterated integral ∫01∫231 (x+4y) 3 dxdy

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Chapter 16.1, Problem 35E

Chapter 16.1, Problem 37E

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Chapter 16 Solutions

CALCULUS W/SAPLING ACCESS >IC<

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The value of the iterated integral ∫ 0 1 ∫ 2 3 1 ( x + 4 y ) 3 d x d y

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The value of the iterated integral ∫01∫231 (x+4y) 3 dxdy

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Chapter 16 Solutions