If f ( x ) is even, then what can we say about: ∫ − 2 2 f ( x ) d xIf f ( x ) is odd, then what can we say about ∫ − 2 2 f ( x ) d xAre they both zero? For the first one if its even wouldn't this be the same as ∫ a a f ( x ) d x = 0Now if its odd f ( − x ) = − f ( x ). Would FTOC make this zero as well?

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If   f  (  x  ) is even, then what can we say about:      ∫          −      2              2        f  (  x  )  d  xIf   f  (  x  ) is odd, then what can we say about      ∫          −      2              2        f  (  x  )  d  xAre they both zero? For the first one if its even wouldn&#039;t this be the same as      ∫          a              a        f  (  x  )  d  x  =  0Now if its odd   f  (  −  x  )  =  −  f  (  x  ). Would FTOC make this zero as well?

iceniessyoy · Accepted Answer

If   f  (  x  ) is even then   f  (  −  x  )  =  f  (  x  ). So      ∫          −      2        2    f  (  x  )        d    x  =      ∫          −      2        0    f  (  x  )        d    x  +      ∫    0    2    f  (  x  )        d    x  =      ∫    0    2    f  (  −  x  )        d    x  +      ∫    0    2    f  (  x  )        d    xBut then   f  (  −  x  )  =  f  (  x  )  f  (  −  x  )  =  f  (  x  ) so that simplifies to   2      ∫    0    2    f.Similarly, if   f is odd - that is:   f  (  −  x  )  =  −  f  (  x  ) we get      ∫          −      2        2    f  (  x  )        d    x  =      ∫          −      2        0    f  (  x  )        d    x  +      ∫    0    2    f  (  x  )        d    x  =      ∫    0    2    f  (  −  x  )        d    x  +      ∫    0    2    f  (  x  )        d    x  =  0

Emanuel Keith · Accepted Answer

Not exactly:      {                                                      ∫                              −                a                            a                        f            (            x            )                                      d                                    x            =            2                          ∫                              0                            a                        f            (            x            )                                      d                                    x                                    if                     f                     is even,                                                                ∫                              −                a                            a                        f            (            x            )                                      d                                    x            =            0                                    if                     f                     is odd.                        To see it, make the substitution     t  =  −  x,         d      x  =  −      d      t:      ∫          −      a        0    f  (  x  )        d      x  =  −      ∫          a        0    f  (  −  t  )        d      t  =      ∫          0        a    f  (  −  t  )        d      t  =      {                                                      ∫                              0                            a                        f            (            −            t            )                                      d                                    t                                    (          f                    even          )          ,                                                  −                          ∫                              0                            a                        f            (            −            t            )                                      d                                    t                                    (          f                    odd          )          ,                        then use Chasles relation.

If f ( x ) is even, then what can we say about: <msubsup> ∫<!-- ∫ --

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