Explanation Given information: The integral is ∫ 0 1 ∫ 1 3 ∫ − 1 2 ( 2 x + 3 y − z ) d x d y d z . Formula used: Integral of power function is ∫ a x n d x = a [ x n + 1 n + 1 ] + C Calculation: Consider the function, ∫ 0 1 ∫ 1 3 ∫ − 1 2 ( 2 x + 3 y − z ) d x d y d z = ∫ 0 1 ( ∫ 1 3 ( ∫ − 1 2 ( 2 x + 3 y − z ) d x ) d y ) d z First evaluate the inner integral, Then, ∫ − 1 2 ( 2 x + 3 y − z ) d x = 2 [ x 1 + 1 1 + 1 ] − 1 2 + 3 y [ x 0 + 1 0 + 1 ] − 1 2 − z [ x 0 + 1 0 + 1 ] − 1 2 = [ x 2 ] − 1 2 + 3 y [ x ] − 1 2 − z [ x ] − 1 2 Now evaluate the limits, ∫ − 1 2 ( 2 x + 3 y − z ) d x = [ x 2 ] − 1 2 + 3 y [ x ] − 1 2 − z [ x ] − 1 2 = [ ( 2 ) 2 − ( − 1 ) 2 ] + 3 y [ ( 2 ) − ( − 1 ) ] − z [ ( 2 ) − ( − 1 ) ] = [ 4 − 1 ] + 3 y [ 2 + 1 ] − z [ 2 + 1 ] = 3 + 9 y − 3 z Now inner integral is ∫ − 1 2 ( 2 x + 3 y − z ) d x = 3 + 9 y − 3 z

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Chapter 6.6, Problem 23E

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To calculate: The value of given integral function ∫01∫13∫−12(2x+3y−z)dxdydz.

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Chapter 6.6, Problem 22E

Chapter 6.6, Problem 24E

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The value of given integral function ∫ 0 1 ∫ 1 3 ∫ − 1 2 ( 2 x + 3 y − z ) d x d y d z .

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To calculate: The value of given integral function ∫01∫13∫−12(2x+3y−z)dxdydz.

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