Finding a certain antiderivativeThe problem says:If f ( x ) = b 2 x 2 + b 3 x 3 + . . . + b m x m find a function g with g ′ = f..This problem is stated in the differentiation part of the book, integration comes later.I tried starting with a simple example and trying to find a pattern from there:If f ( x ) = b 2 x 2 Then we can try: g ( x ) = − b 2 x g ′ ( x ) = − b 2 ( − 1 x 2 ) = b 2 x 2 = f ( x )Then if f ( x ) = b 2 x 2 + b 3 x 3 .I tried: g ( x ) = − b 2 x − b 3 x 3 / 2 g ′ ( x ) = − b 2 ( − 1 x 2 ) − b 3 ( − 3 2 x 1 / 2 x 3 ) Which is not what I wanted. The denominator in the second summand is ok, but the numerator is not what I need.But I'm stuck on what I need to look for, any hints for deriving the correct pattern would be appreciated.

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Finding a certain antiderivativeThe problem says:If   f  (  x  )  =            b      2              x      2        +            b      3              x      3        +  .  .  .  +            b      m              x      m       find a function g with       g    ′    =  f..This problem is stated in the differentiation part of the book, integration comes later.I tried starting with a simple example and trying to find a pattern from there:If   f  (  x  )  =            b      2              x      2      Then we can try:   g  (  x  )  =  −            b      2        x        g    ′    (  x  )  =  −      b    2        (                  −        1                    x        2              )    =            b      2              x      2        =  f  (  x  )Then if   f  (  x  )  =            b      2              x      2        +            b      3              x      3      .I tried:   g  (  x  )  =  −            b      2        x    −            b      3              x              3                  /                2                  g    ′    (  x  )  =  −      b    2        (                  −        1                    x        2              )    −      b    3        (                  −                  3          2                          x                      1                          /                        2                                      x        3              )  Which is not what I wanted. The denominator in the second summand is ok, but the numerator is not what I need.But I&#039;m stuck on what I need to look for, any hints for deriving the correct pattern would be appreciated.

Ann Mathis · Accepted Answer

Step 1Solve the problem for   p  (  x  )  =      b          x      k        =  b      x          −      k      .Then apply the trick to each summand and sum the functions you get. Since the derivative of a sum is the sum of the derivatives, you&#039;re done.Now, if we want to find a function q(x) such that       q    ′    =  p, we may first try with a monomial,   q  (  x  )  =  a      x    h  . Then we know that       q    ′    (  x  )  =  h  a      x          h      −      1       so we can take   h  a  =  b and   h  −  1  =  −  k. This means   h  =  −  k  +  1 and  a  =      b          −      k      +      1       which is possible provided   k  ≠  1.Step 2Putting together the pieces, you can take   g  (  x  )  =            b      2              (      −      2      +      1      )      x        +            b      3              (      −      3      +      1      )              x        2              +  ⋯  +            b      m              (      −      m      +      1      )              x                  m          −          1                      =  −            b      2        x    −            b      3              2              x        2              −  ⋯  −            b      m              (      m      −      1      )              x                  m          −          1

narratz5dz · Accepted Answer

Step 1      f    (    x    )    =                  b        2                    x        2              +                  b        3                    x        3              +    .    .    .    +                  b        m                    x        m            Say   g  (  x  )  =      a    1    (  x  )  +      a    2    (  x  )  +  .  .  .  +      a          m      −      1        (  x  ), such that       a    i    ′    (  x  ) gives the       i          t      h       term of f(x); that is,            a      i      ′        (    x    )    =                  b                  i          +          1                            x                  i          +          1                    The first term of f(x) has an       x    2   in the denominator. You know that when we differentiate 1/x, we get       x    2   in the denominator. So       a    1    (  x  )  =  a      /    x  ,      a    1    ′    (  x  )  =  −  a      /        x    2    =      b    2        /        x    2      ⟹    a  =  −      b    2  .Step 2The second term of f(x) has an       x    3   in the denominator. This is one degree higher than       x    2  , so it makes sense to chose       a    2    (  x  ) having       x    2   in the denominator, which is one degree higher than x.      a    2    (  x  )  =  c      /        x    2    ,      a    2    ′    (  x  )  =  −  2  c      /        x    3    =      b    3        /        x    3      ⟹    c  =  −      b    3        /    2Similarly chose       a    3    (  x  ) having       x    3   in the denominator, to obtain             a      3        (    x    )    =    −                  b        4                    3                  x          3                    .You might be able to see the pattern:             a      i        (    x    )    =    −                  b                  i          +          1                            i                  x          i                    .  ∴      g    (    x    )    =    −                  b        2            x        −                  b        3                    2                  x          2                      −                  b        4                    3                  x          3                      .    .    .    −                  b        m                    (        m        −        1        )                  x                      m            −            1                                +    c  We put an arbitrary constant c because it vanishes on differentiation.

Finding a certain antiderivative The problem says: If f <mo stretchy="false">( x <mo

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