Explanation Given information: The integrand of ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x is f ( x , y , z ) = x 2 e y + 3 z Formula used: The integral formulae: ∫ x n d x = x n + 1 n + 1 , ∫ e x d x = e x Calculation: The integrand of ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x can be written as f ( x , y , z ) = x 2 e y e 3 z = g ( x ) h ( y ) k ( z ) , where g ( x ) = x 2 , h ( y ) = e y , k ( z ) = e 3 z Thus, f ( x , y , z ) = g ( x ) h ( y ) k ( z ) Now to verify ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x = ( ∫ 1 4 g ( x ) d x ) ( ∫ 0 3 h ( y ) d y ) ( ∫ 2 6 k ( z ) d z ) First evaluate left-hand side ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x , first evaluate inner most integral treating x and y as constant, Thus, ∫ 2 6 x 2 e y + 3 z d z substitute y + 3 z = u ⇒ 3 d z = d u And limit changes as z = 2 ; u = y + 6 z = 6 ; u = y + 18 Thus, ∫ 2 6 x 2 e y + 3 z d z = ∫ y + 6 y + 18 x 2 3 e u d u = x 2 3 [ e u ] y + 6 y + 18 = x 2 3 [ e y + 18 − e y + 6 ] Now integrate above with respect to y, treating x as constant ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y = ∫ 0 3 x 2 3 [ e y + 18 − e y + 6 ] d y = x 2 3 [ e y + 18 − e y + 6 ] 0 3 = x 2 3 <

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ISBN: 9781319379421

Author: Rogawski

Publisher: MAC HIGHER

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Chapter 15.3, Problem 28E

To determine

Show that the integral of ∫14∫03∫26x2ey+3zdzdydx can be expressed as a product of g(x)h(y)k(z) and verify:

∫14∫03∫26x2ey+3zdzdydx=(∫14g(x)dx)(∫03h(y)dy)(∫26k(z)dz)

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Chapter 15.3, Problem 27E

Chapter 15.3, Problem 29E

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Chapter 15 Solutions

CALCULUS 4E (HC) W/ ACHIEVE ACCESS

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Show that the integral of ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x can be expressed as a product of g ( x ) h ( y ) k ( z ) and verify: ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x = ( ∫ 1 4 g ( x ) d x ) ( ∫ 0 3 h ( y ) d y ) ( ∫ 2 6 k ( z ) d z )

Solution Summary: The author explains that the integral of displaystyle 'underset' can be expressed as a product of g(x)h(y)k(z).

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Show that the integral of ∫14∫03∫26x2ey+3zdzdydx can be expressed as a product of g(x)h(y)k(z) and verify:

∫14∫03∫26x2ey+3zdzdydx=(∫14g(x)dx)(∫03h(y)dy)(∫26k(z)dz)

Want to see the full answer?

Chapter 15 Solutions

Show that the integral of ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x can be expressed as a product of g ( x ) h ( y ) k ( z ) and verify: ∫ 1 4 ∫ 0 3 ∫ 2 6 x 2 e y + 3 z d z d y d x = ( ∫ 1 4 g ( x ) d x ) ( ∫ 0 3 h ( y ) d y ) ( ∫ 2 6 k ( z ) d z )

Solution Summary: The author explains that the integral of displaystyle 'underset' can be expressed as a product of g(x)h(y)k(z).

Videos

Show that the integral of ∫14∫03∫26x2ey+3zdzdydx can be expressed as a product of g(x)h(y)k(z) and verify: ∫14∫03∫26x2ey+3zdzdydx=(∫14g(x)dx)(∫03h(y)dy)(∫26k(z)dz)

Want to see the full answer?

Chapter 15 Solutions

Show that the integral of ∫14∫03∫26x2ey+3zdzdydx can be expressed as a product of g(x)h(y)k(z) and verify:

∫14∫03∫26x2ey+3zdzdydx=(∫14g(x)dx)(∫03h(y)dy)(∫26k(z)dz)