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Calculate some values for some even and odd functions with given properties

Let f be an odd function, integrable everywhere, with

    \[ f(5) = 7, \qquad f(0) = 0. \]

Let g be an even function, integrable everywhere, with

    \[ g(x) = f(x+5), \qquad f(x) = \int_0^x g(t) \, dt \qquad \text{for all } x. \]

Prove the following:

  1. f(x-5) = - g(x) for all x;
  2. \displaystyle{\int_0^5 f(t) \, dt = 7;
  3. \displaystyle{\int_0^x f(t) \, dt = g(0) - g(x).

  1. Proof. We compute using the given properties,

        \begin{align*}  f(x+5) = g(x) && \implies && g(-x) &= f(-x + 5) \\  && \implies && -g(x) &= f(x-5) & (g \text{ is odd}, f \text{ is even}). \ \blacksquare \end{align*}

  2. Again, we compute using the given properties of f and g,

        \begin{align*}  \int_0^5 f(t) \, dt &= \int_0^5 g(t-5) \, dt \\  &= \int_{-5}^0 g(t) \, dt & (\text{Translation})\\  &= \int_5^0 g(-t) \, dt \\  &= - \int_0^5 g(-t) \, dt \\  &= \int_0^5 g(t) \, dt & (g \text{ is odd}) \\  &= f(5) = 7. \qquad \blacksquare \end{align*}

  3. Finally,

        \begin{align*}  \int_0^x f(t) \, dt &= \int_0^x g(t-5) \, dt \\  &= \int_{-5}^{x-5} g(t) \, dt \\  &= \int_{-5}^0 g(t) \, dt  + \int_0^{x-5} g(t) \, dt\\   &= f(5) + f(x-5) \\  &= g(0) - g(x). \qquad \blacksquare \end{align*}

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