Integration by substitution, ∫ 1 a f ( x s ) d x = ∫ 1 a s f ( x ) 1 s x 1 − 1 / s d x

Question

Integration by substitution,       ∫    1    a    f  (      x    s    )  d  x  =      ∫    1                  a        s              f  (  x  )            1              s                  x                      1            −            1                          /                        s                                d  x

odmeravan5c · Accepted Answer

Note the following: In your case   ϕ  (  t  )  =      t    s  . So write:      ∫    1    a    f  (      x    s    )         d    x  =      ∫    1    a    f  (  ϕ  (  t  )  )         d    t  =      ∫    1    a    f  (  ϕ  (  t  )  )                    ϕ        ′            (      t      )                      ϕ        ′            (      t      )               d    t  =      ∫    1    a    g  (  ϕ  (  t  )  )      ϕ    ′    (  t  )         d    tWe had set   g  (  ϕ  (  t  )  )  :=            f      (      ϕ      (      t      )      )                      ϕ        ′            (      t      )      . This seems a little weird at first, but we will just accept it as of now. NOW use substitution   u  =      t    s        ∫    1    a    g  (  ϕ  (  t  )  )      ϕ    ′    (  t  )         d    t  =      ∫    1                  a        s              g  (  u  )         d    uThe difficulty is to write             f      (      ϕ      (      t      )      )                      ϕ        ′            (      t      )       in terms of u. Take a look:            f      (      ϕ      (      t      )      )                      ϕ        ′            (      t      )        =            f      (              t        s            )              s              t                  s          −          1                      =            f      (      u      )              s              t                  s          −          1                    Since   u  =      t    s  , we get   t  =      u    s  . So:            f      (      u      )              s              t                  s          −          1                      =            f      (      u      )              s              u                  1          −                      1            s                              This means:      ∫    1    a    f  (      x    s    )         d    x  =      ∫    1                  a        s                        f      (      u      )              s              u                  1          −                      1            s                                       d    u

Integration by substitution, <msubsup> ∫ 1 a </msubsup> f (

Answered question

Answer & Explanation

New Questions in Calculus 1