For f ( x ) = ∫ x x + 1 sin ⁡ ( e t ) d t Prove that : e x | f ( x ) | ≤ 2

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For       f    (    x    )    =          ∫      x              x        +        1              sin    ⁡    (          e      t        )    d    t  Prove that :             e      x              |      f      (      x      )      |        ≤    2

zupa1z · Accepted Answer

f    (    x    )    =          ∫      x              x        +        1              sin    ⁡    (          e      t        )    d    t        ⟹                      e        x            f      (      x      )      =              ∫        x                  x          +          1                            e        x            sin      ⁡      (              e        t            )      d      t      ≤              ∫        x                  x          +          1                            e        t            sin      ⁡      (              e        t            )      d      t      Now substitute   z  =      e    t   which gives            e      x        f    (    x    )    ≤          ∫                        e          x                                      e                      x            +            1                                sin    ⁡    (    z    )    d    z    =    cos    ⁡    (          e              x              )    −    cos    ⁡    (          e              x        +        1              )        ⟹              e      x              |        f    (    x    )          |        ≤          |        cos    ⁡    (          e              x              )          |        +          |        cos    ⁡    (          e              x        +        1              )          |        ≤    2

For f(x)=int_x^(x+1) sin(e^t)dt. Prove that: e^x|f(x)|<=2

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