Integral by substitution of an absolute value of a derivative

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I found the deformation below in a book. Could you explain how is it possible?
beginequation
int_alpha^beta left|fracdbfedsright|ds = int_alpha^beta left|fracdbfedtright|fracdtdsds=int_a^b left|fracdbfedtright|dt
endequation
where $aleq tleq b$ corresponds to $alpha leq s leq beta$.



I think it does not make a big difference, but in that book, $left|fracdbfedsright|$ was written as $(fracdbfeds cdot fracdbfeds)^frac12$ instead.




integration absolute-value substitution




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  • What's $bf e$.
    – Nosrati
    Jul 15 at 12:13














up vote
1
down vote

favorite












I found the deformation below in a book. Could you explain how is it possible?
beginequation
int_alpha^beta left|fracdbfedsright|ds = int_alpha^beta left|fracdbfedtright|fracdtdsds=int_a^b left|fracdbfedtright|dt
endequation
where $aleq tleq b$ corresponds to $alpha leq s leq beta$.



I think it does not make a big difference, but in that book, $left|fracdbfedsright|$ was written as $(fracdbfeds cdot fracdbfeds)^frac12$ instead.




integration absolute-value substitution




share|cite|improve this question





















  • What's $bf e$.
    – Nosrati
    Jul 15 at 12:13












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I found the deformation below in a book. Could you explain how is it possible?
beginequation
int_alpha^beta left|fracdbfedsright|ds = int_alpha^beta left|fracdbfedtright|fracdtdsds=int_a^b left|fracdbfedtright|dt
endequation
where $aleq tleq b$ corresponds to $alpha leq s leq beta$.



I think it does not make a big difference, but in that book, $left|fracdbfedsright|$ was written as $(fracdbfeds cdot fracdbfeds)^frac12$ instead.




integration absolute-value substitution




share|cite|improve this question













I found the deformation below in a book. Could you explain how is it possible?
beginequation
int_alpha^beta left|fracdbfedsright|ds = int_alpha^beta left|fracdbfedtright|fracdtdsds=int_a^b left|fracdbfedtright|dt
endequation
where $aleq tleq b$ corresponds to $alpha leq s leq beta$.



I think it does not make a big difference, but in that book, $left|fracdbfedsright|$ was written as $(fracdbfeds cdot fracdbfeds)^frac12$ instead.





integration absolute-value substitution





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share|cite|improve this question




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edited Jul 15 at 10:44
























asked Jul 15 at 10:37









oremata

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  • What's $bf e$.
    – Nosrati
    Jul 15 at 12:13
















  • What's $bf e$.
    – Nosrati
    Jul 15 at 12:13















What's $bf e$.
– Nosrati
Jul 15 at 12:13




What's $bf e$.
– Nosrati
Jul 15 at 12:13










1 Answer
1






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The first transformation is a result of using the chain rule. Suppose that $t$ is a function of $s$, then,
$$
fracdtextbfeds = fracdtextbfedtfracdtds tag1
$$



Taking the norm of (1) yields,
$$
beginalign
left| fracdtextbfeds right| =& left( fracdtextbfeds cdot fracdtextbfeds right)^frac12 \
=& left[fracdtextbfedt cdot fracdtextbfedtleft(fracdtds right)^2 right]^frac12 \
=& left| fracdtextbfedt right|fracdtds
endalign
$$



The final part is a result of using Integration by Substitution, which gives that,
$$
int_t(alpha)^t(beta)left| fracdtextbfedt right| dt = int_alpha^beta left| fracdtextbfedt right|fracdtds ds
$$






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    The first transformation is a result of using the chain rule. Suppose that $t$ is a function of $s$, then,
    $$
    fracdtextbfeds = fracdtextbfedtfracdtds tag1
    $$



    Taking the norm of (1) yields,
    $$
    beginalign
    left| fracdtextbfeds right| =& left( fracdtextbfeds cdot fracdtextbfeds right)^frac12 \
    =& left[fracdtextbfedt cdot fracdtextbfedtleft(fracdtds right)^2 right]^frac12 \
    =& left| fracdtextbfedt right|fracdtds
    endalign
    $$



    The final part is a result of using Integration by Substitution, which gives that,
    $$
    int_t(alpha)^t(beta)left| fracdtextbfedt right| dt = int_alpha^beta left| fracdtextbfedt right|fracdtds ds
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      The first transformation is a result of using the chain rule. Suppose that $t$ is a function of $s$, then,
      $$
      fracdtextbfeds = fracdtextbfedtfracdtds tag1
      $$



      Taking the norm of (1) yields,
      $$
      beginalign
      left| fracdtextbfeds right| =& left( fracdtextbfeds cdot fracdtextbfeds right)^frac12 \
      =& left[fracdtextbfedt cdot fracdtextbfedtleft(fracdtds right)^2 right]^frac12 \
      =& left| fracdtextbfedt right|fracdtds
      endalign
      $$



      The final part is a result of using Integration by Substitution, which gives that,
      $$
      int_t(alpha)^t(beta)left| fracdtextbfedt right| dt = int_alpha^beta left| fracdtextbfedt right|fracdtds ds
      $$






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        The first transformation is a result of using the chain rule. Suppose that $t$ is a function of $s$, then,
        $$
        fracdtextbfeds = fracdtextbfedtfracdtds tag1
        $$



        Taking the norm of (1) yields,
        $$
        beginalign
        left| fracdtextbfeds right| =& left( fracdtextbfeds cdot fracdtextbfeds right)^frac12 \
        =& left[fracdtextbfedt cdot fracdtextbfedtleft(fracdtds right)^2 right]^frac12 \
        =& left| fracdtextbfedt right|fracdtds
        endalign
        $$



        The final part is a result of using Integration by Substitution, which gives that,
        $$
        int_t(alpha)^t(beta)left| fracdtextbfedt right| dt = int_alpha^beta left| fracdtextbfedt right|fracdtds ds
        $$






        share|cite|improve this answer













        The first transformation is a result of using the chain rule. Suppose that $t$ is a function of $s$, then,
        $$
        fracdtextbfeds = fracdtextbfedtfracdtds tag1
        $$



        Taking the norm of (1) yields,
        $$
        beginalign
        left| fracdtextbfeds right| =& left( fracdtextbfeds cdot fracdtextbfeds right)^frac12 \
        =& left[fracdtextbfedt cdot fracdtextbfedtleft(fracdtds right)^2 right]^frac12 \
        =& left| fracdtextbfedt right|fracdtds
        endalign
        $$



        The final part is a result of using Integration by Substitution, which gives that,
        $$
        int_t(alpha)^t(beta)left| fracdtextbfedt right| dt = int_alpha^beta left| fracdtextbfedt right|fracdtds ds
        $$







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 15 at 12:26









        Daniel Beale

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