Prove or disprove: The product of two functions has antiderivatives if one of the functions is continuous and the other has antiderivatives I ⊆ R is an interval and f , g : I → R .f has antiderivatives (there exists F : I → R with F ′ = f)g is continuous on IProve that the product f ⋅ g has antiderivativesThis is true when f is continuous or bounded, but I could neither prove it in the general case, nor find a counterexample

Question

Prove or disprove: The product of two functions has antiderivatives if one of the functions is continuous and the other has antiderivatives  I  ⊆      R   is an interval and   f  ,  g  :  I  →      R  .f has antiderivatives (there exists   F  :  I  →      R   with       F    ′    =  f)g is continuous on IProve that the product   f  ⋅  g has antiderivativesThis is true when f is continuous or bounded, but I could neither prove it in the general case, nor find a counterexample

mar1nerne · Accepted Answer

Step 1This is not true. For example, take   I  =  [  0  ,  1  ] and                    f        (        x        )        =                  {                                                                      1                                      x                                                  sin                ⁡                                  1                  x                                                            for                 x                ∈                (                0                ,                1                ]                ,                                                                    0                                            for                 x                =                0                ,                                                                      g        (        x        )        =                  {                                                                      x                                sin                ⁡                                  1                  x                                                            for                 x                ∈                (                0                ,                1                ]                ,                                                                    0                                            for                 x                =                0.                                                                  Step 2Then f has an antiderivative, namely                    F        (        x        )        =        x                  x                cos        ⁡                  1          x                −                  3          2                          ∫          0          x                          t                cos        ⁡                  1          t                d        t        ,            and g is continuous, but                    f        (        x        )        g        (        x        )        =                  {                                                                      sin                  2                                ⁡                                  1                  x                                                            for                 x                ∈                (                0                ,                1                ]                ,                                                                    0                                            for                 x                =                0                ,                                                                  does not have an antiderivative, because it has the &quot;wrong value&quot; at   x  =  0.

Prove or disprove: The product of two functions has antiderivatives if one of the functions is conti

Answered question

Answer & Explanation

New Questions in Calculus 1