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Primitive of a composite function

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0

$begingroup$

I'm reading Zorich, Mathematical Analysis I, and I found a not clear step in the paragraph on Primitives.
The particular sentence is shown below (adapted).

From the definition of primitive of a function on an interval, and from the properties of differentiation, the following relation holds:
$$int (fcirc phi)(t)phi '(t)mathrm{d}t=(Fcirc phi)(t)+c$$
where $F$ is the primitive of $f$ on an interval $Isubsetmathbb{R}$, and $phi:I_tsubsetmathbb{R} to I_xsubset I$ is a function $C^{(1)}(I_t)$.

I can not understand why it is necessary that $phi '(t)$ has to be continuous in $I_t$, what would be wrong if it were not? I can not see the problem in that case.

Thanks.

real-analysis indefinite-integrals function-and-relation-composition

share|cite|improve this question

asked Mar 9 at 16:37

NamelessNameless

788

$endgroup$

add a comment | 

0

$begingroup$

I'm reading Zorich, Mathematical Analysis I, and I found a not clear step in the paragraph on Primitives.
The particular sentence is shown below (adapted).

From the definition of primitive of a function on an interval, and from the properties of differentiation, the following relation holds:
$$int (fcirc phi)(t)phi '(t)mathrm{d}t=(Fcirc phi)(t)+c$$
where $F$ is the primitive of $f$ on an interval $Isubsetmathbb{R}$, and $phi:I_tsubsetmathbb{R} to I_xsubset I$ is a function $C^{(1)}(I_t)$.

I can not understand why it is necessary that $phi '(t)$ has to be continuous in $I_t$, what would be wrong if it were not? I can not see the problem in that case.

Thanks.

real-analysis indefinite-integrals function-and-relation-composition

share|cite|improve this question

asked Mar 9 at 16:37

NamelessNameless

788

$endgroup$

add a comment | 

$begingroup$

I'm reading Zorich, Mathematical Analysis I, and I found a not clear step in the paragraph on Primitives.
The particular sentence is shown below (adapted).

From the definition of primitive of a function on an interval, and from the properties of differentiation, the following relation holds:
$$int (fcirc phi)(t)phi '(t)mathrm{d}t=(Fcirc phi)(t)+c$$
where $F$ is the primitive of $f$ on an interval $Isubsetmathbb{R}$, and $phi:I_tsubsetmathbb{R} to I_xsubset I$ is a function $C^{(1)}(I_t)$.

I can not understand why it is necessary that $phi '(t)$ has to be continuous in $I_t$, what would be wrong if it were not? I can not see the problem in that case.

Thanks.

real-analysis indefinite-integrals function-and-relation-composition

share|cite|improve this question

asked Mar 9 at 16:37

NamelessNameless

788

$endgroup$

share|cite|improve this question

asked Mar 9 at 16:37

NamelessNameless

788

share|cite|improve this question

asked Mar 9 at 16:37

NamelessNameless

788

asked Mar 9 at 16:37

NamelessNameless

788

asked Mar 9 at 16:37

NamelessNameless

788

1 Answer
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0

$begingroup$

Assuming the notation $int g(t),dt = G(t), tin I,$ means simply that $G'(t)=g(t),tin I,$ you are correct. You might have problems with

$$int_a^b (fcirc phi)(t)phi'(t),dt = (Fcirc phi)(b)-(Fcirc phi)(a)),$$ however. Now that I think about it, you could have problems with that even with Zorich's formulation.

share|cite|improve this answer

answered Mar 9 at 17:17

zhw.zhw.

74.1k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the feedback.
  $endgroup$
  – Nameless
  Mar 9 at 21:45
  

  
  
  
  

add a comment | 

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0

$begingroup$

Assuming the notation $int g(t),dt = G(t), tin I,$ means simply that $G'(t)=g(t),tin I,$ you are correct. You might have problems with

$$int_a^b (fcirc phi)(t)phi'(t),dt = (Fcirc phi)(b)-(Fcirc phi)(a)),$$ however. Now that I think about it, you could have problems with that even with Zorich's formulation.

share|cite|improve this answer

answered Mar 9 at 17:17

zhw.zhw.

74.1k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the feedback.
  $endgroup$
  – Nameless
  Mar 9 at 21:45
  

  
  
  
  

add a comment | 

0

$begingroup$

Assuming the notation $int g(t),dt = G(t), tin I,$ means simply that $G'(t)=g(t),tin I,$ you are correct. You might have problems with

$$int_a^b (fcirc phi)(t)phi'(t),dt = (Fcirc phi)(b)-(Fcirc phi)(a)),$$ however. Now that I think about it, you could have problems with that even with Zorich's formulation.

share|cite|improve this answer

answered Mar 9 at 17:17

zhw.zhw.

74.1k43175

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the feedback.
  $endgroup$
  – Nameless
  Mar 9 at 21:45
  

  
  
  
  

add a comment | 

$begingroup$

Assuming the notation $int g(t),dt = G(t), tin I,$ means simply that $G'(t)=g(t),tin I,$ you are correct. You might have problems with

$$int_a^b (fcirc phi)(t)phi'(t),dt = (Fcirc phi)(b)-(Fcirc phi)(a)),$$ however. Now that I think about it, you could have problems with that even with Zorich's formulation.

share|cite|improve this answer

answered Mar 9 at 17:17

zhw.zhw.

74.1k43175

$endgroup$

share|cite|improve this answer

answered Mar 9 at 17:17

zhw.zhw.

74.1k43175

answered Mar 9 at 17:17

zhw.zhw.

74.1k43175

answered Mar 9 at 17:17

zhw.zhw.

74.1k43175