Why isn't there a good product formula for antiderivatives?Computing antiderivatives is more challenging than computing derivatives, in part due to the lack of a "product formula''; namely, while (fg)′ can be expressed in terms of f, f′, g, g′, there seems to be no way to express ∫ f g in terms of f , ∫ f , g ∫ g and related quantities. Is there any intuitive reason or heuristic explanation for why no such formula exists? I'm looking for a non-rigorous explanation which can be understood by first-year calculus students.

Question

Why isn&#039;t there a good product formula for antiderivatives?Computing antiderivatives is more challenging than computing derivatives, in part due to the lack of a &quot;product formula&#039;&#039;; namely, while (fg)′ can be expressed in terms of f, f′, g, g′, there seems to be no way to express   ∫  f  g in terms of   f  ,  ∫  f  ,  g  ∫  g and related quantities. Is there any intuitive reason or heuristic explanation for why no such formula exists? I&#039;m looking for a non-rigorous explanation which can be understood by first-year calculus students.

Nollothwnfcm · Accepted Answer

Step 1Differentiating is somewhat like taking the difference of two things (that are close together).Integrating is somewhat like taking the sum of many things (that may be fluctuating).Suppose f and g are functions that we &quot;know&quot; how to differentiate and integrate. That&#039;s somewhat analogous to saying we have expressions for       f    2    −      f    1        g    2    −      g    1        f    1    +  ⋯  +      f    n        g    1    +  ⋯  +      g    n  Step 2Wanting to differentiate or integrate the product fg is somewhat analogous to wanting formulas for       f    2        g    2    −      f    1        g    1   or for       f    1        g    1    +  ⋯  +      f    n        g    n  .With a difference of two things, you can do algebraic &quot;tricks&quot; such as       f    2        g    2    −      f    1        g    1    =      f    2        g    2    −      f    1        g    2    +      f    1        g    2    −      f    1        g    1  .It&#039;s not so easy to find an algebraic &quot;trick&quot; for expressing       f    1        g    1    +  ⋯  +      f    n        g    n   in terms of       f    1    +  ⋯  +      f    n   and       g    1    +  ⋯  +      g    n  .

3c4ar1bzki1u · Accepted Answer

Step 1Maybe use the notion of &quot;elementary function&quot;. There are elementary functions A, B such that both   ∫  A and   ∫  B are elementary, but   ∫  (  A  B  ) is not elementary. So there can be no simple &quot;product rule&quot; for integration, since any simple combinations of elementary functions is again elementary.Step 2Example:   A  (  x  )  =  cos  ⁡  x,   B  (  x  )  =      1    x  . Then   ∫  cos  ⁡  x    d  x  =  sin  ⁡  x  +  C and   ∫      1    x      d  x  =  ln  ⁡  x  +  C are elementary, but   ∫            cos      ⁡      x        x      d  x is not.

Why isn't there a good product formula for antiderivatives? Computing antiderivatives is more chall

Answered question

Answer & Explanation

New Questions in Calculus 1