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Prove formulas for the integral from -b to b of even and odd functions

We define an even function to be a function f such that whenever x is in the domain of f so is -x, and for all x in the domain, we have

    \[ f(-x) = f(x). \]

We define an odd function to be a function f such that whenever x is in the domain of f so is -x, and for all x in the domain, we have

    \[ f(-x) = -f(x). \]

Then, for f integrable on [0,b] prove the following.

  1. If f is an even function then

        \[ \int_{-b}^b f(x) \, dx = 2 \int_0^b f(x) \, dx. \]

  2. If f is an odd function then

        \[ \int_{-b}^b f(x) \, dx = 0. \]


  1. Proof. First, since the integral is additive with respect to the interval of integration (Theorem 1.17 of Apostol), we have

        \[ \int_{-b}^b f(x) \, dx = \int_{-b}^0 f(x) \, dx + \int_0^b f(x) \, dx . \]

    Then, for the first integral we use the expansion/contraction of the interval of integration with k=-1 to get

        \[ \int_{-b}^0 f(x) \, dx = - \int_b^0 f(-x) \, dx. \]

    Since f(x) is an even function by assumption, we have f(-x) = f(x) for all x \in [0,b]. Since -\int_b^0 = \int_0^b we then have,

        \[ -\int_b^0 f(-x) \, dx = \int_0^b f(x) \, dx. \]

    So, putting this all together we have,

        \[ \int_{-b}^b f(x) \, dx = \int_0^b f(x) \, dx + \int_0^b f(x) \, dx = 2 \int_0^b f(x) \, dx. \qquad \blacksquare \]

  2. Proof. Similar to part (a) we have,

        \[ \int_{-b}^b f(x) \, dx = \int_{-b}^0 f(x) \, dx + \int_0^b f(x) \, dx = \int_0^b f(-x) \, dx + \int_0^b f(x) \, dx. \]

    But in this case, since f is an odd function, we have f(-x) = -f(x) for all x \in [0,b]. Thus,

        \[ \int_0^b f(-x) \, dx = - \int_0^b f(x) \, dx \quad \implies \quad \int_{-b}^b f(x) \, dx = 0. \qquad \blacksquare \]

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