Explanation Given information: The integral function is ∫ 0 9 1 x − 1 3 d x . Calculation: Show the integral function as follows ∫ 0 9 1 x − 1 3 d x (1) Show the definition of an Improper Integral of Type 2 : (a) If f is continuous on [ a , b ] and is discontinuous at b ,then ∫ a b f ( x ) d x = lim t → b − ∫ a t f ( x ) d x Provided this limit exists (as a finite number). (b) If f is continuous on [ a , b ] and is discontinuous at a , then ∫ a b f ( x ) d x = lim t → − a + ∫ t b f ( x ) d x Provided this limit exists (as a finite number). The improper integral ∫ a b f ( x ) d x is called convergent if the corresponding limit exists. The improper integral ∫ a b f ( x ) d x is called divergent if the corresponding limit does not exist. (c) If f has a discontinuity at c, where a < C < b , and both ∫ a c f ( x ) d x and ∫ c b f ( x ) d x are convergent, then ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x Apply Part (c) of the definition of improper integral type 2 in Equation (1). ∫ 0 9 1 x − 1 3 d x = ∫ 0 1 1 x − 1 3 d x + ∫ 1 9 1 x − 1 3 d x = I 1 + I 2 (2) Calculate the value of the integral function ∫ 0 1 1 x − 1 3 d x part of Equation (2). I 1 = ∫ 0 1 1 x − 1 3 d x (3) Consider the function f ( x ) = 1 x − 1 3 The function has vertical asymptote x = 1 . Therefore, the infinite discontinuity occurs at right endpoint of [ 0 , 1 ] . Apply Part (a) of the definition of improper integral type 2 in Equation (3). I 1 = ∫ 0 1 1 x − 1 3 d x = lim t → 1 − ∫ 0 t 1 x − 1 3 d x = lim t → 1 − ∫ 0 t 1 ( x − 1 ) 1 3 d x = lim t → 1 − ∫ 0 t ( x − 1 ) − 1 3 d x (4) Consider the value of the function u = ( x − 1 ) (5) Differentiate Equation (5) with respect to x . d u d x = d d x ( x − 1 ) d u d x = 1 d u = d x Modify Equation (4). Substitute u for ( x − 1 ) and d u for d x in Equation (4). I 1 = ∫ 0 1 1 x − 1 3 d x = lim t → 1 − ∫ 0 t ( x − 1 ) − 1 3 d x = lim t → 1 − ∫ 0 t u − 1 3 d u = lim t → 1 − [ u − 1 3 + 1 − 1 3 + 1 ] 0 t I 1 = lim t → 1 − [ 3 2 u 2 3 ] 0 t (6) Substitute ( x − 1 ) for u in Equation (6). I 1 = lim t → 1 − [ 3 2 ( x − 1 ) 2 3 ] 0 t = 3 2 lim t → 1 − [ ( x − 1 ) 2 3 ] 0 t = 3 2 lim t → 1 − [ ( t − 1 ) 2 3 − ( 0 − 1 ) 2 3 ] = 3 2 lim t → 1 − [ ( t − 1 ) 2 3 − 1 ] I 1 = 3 2 [ ( 1 − 1 ) 2 3 − 1 ] = 3 2 [ 0 − 1 ] = − 3 2 (7) Calculate the value of the integral function ∫ 1 9 1 x − 1 3 d x part of Equation (2)

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ISBN: 9780176743826

Author: Stewart

Publisher: VST

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Chapter 7.8, Problem 33E

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To find: whether theintegralfunction is convergent or divergent.

To evaluate: The integral function if it is convergent.

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Chapter 7.8, Problem 32E

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whether theintegralfunction is convergent or divergent. To evaluate: The integral function if it is convergent.

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To find: whether theintegralfunction is convergent or divergent. To evaluate: The integral function if it is convergent.

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Chapter 7 Solutions

To find: whether theintegralfunction is convergent or divergent.

To evaluate: The integral function if it is convergent.