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Compute the integral from 0 to x of f(t) for the given functions

Find a formula to compute

    \[ F(x) = \int_0^x f(t) \, dt \]

for all x \in \mathbb{R} for the following function f(t).

  1. \displaystyle{f(t) = (t + |t|)^2}.
  2. The function,

        \[ f(t) = \begin{cases} 1-t^2 & \text{if } |t| \leq 1, \\ 1 - |t| & \text{if } |t| > 1. \end{cases} \]

  3. f(t) = e^{-|t|}.
  4. f(t) = the maximum of 1 and t^2.

  1. We know from this exercise (Section 5.5, Exercise #13) that

        \[ F(x) = \int_0^x (t + |t|)^2 \, dt = \frac{2}{3} x^2 (x + |x|). \]

  2. If \abs{x} \leq 1, then f(t) = 1-t^2 over the whole integral, and so

        \[ F(x) = \int_0^x f(t) \, dt = \int_0^x (1-t^2) \, dt = \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^x  = x - \frac{x^3}{3}. \]

    Then, if x > 1 we have

        \begin{align*}   F(x) &= \int_0^x f(t) \, dt \\[9pt]  &= \int_0^1 (1-t^2) \, dt + \int_1^x 1-t \, dt \\[9pt]  &= \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^1 + \left( t - \frac{t^2}{2} \right) \Bigr \rvert_1^x \\[9pt]  &= \frac{2}{3} + x - \frac{x^2}{2} - \frac{1}{2} \\[9pt]  &= x - \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6}. \end{align*}

    (Since x > 1 we have \frac{|x|}{x} = 1 so this equation works. This is the form Apostol wrote these answers as in the back of the book, so I’m getting our answers to match his. I wouldn’t have written them this way otherwise.)

    Finally, if x < -1 we have

        \begin{align*}  F(x) &= \int_0^x f(t) \, dt \\[9pt]  &= \int_0^{-1} (1-t^2) \, dt + \int_{-1}^x (1+t) \, dt \\[9pt]  &= \left( t - \frac{t^3}{3} \right) \Bigr \rvert_0^{-1} + \left( t + \frac{t^2}{2} \right) \Bigr \rvert_{-1}^x \\[9pt]  &= -\frac{2}{3} + x + \frac{x^2}{2} + \frac{1}{2} \\[9pt]  &= x + \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6}. \end{align*}

    Since the formulas for F(x) are the same for x < -1 and x > 1 are the same we have

        \[ F(x) = x + \frac{1}{2} x|x| + \frac{|x|}{x} \frac{1}{6} \]

    for |x| > 1.

  3. We consider two cases. If x \geq 0 then

        \[ F(x) = \int_0^x e^{-|t|} \, dt = \int_0^x e^{-t} \, dt = -e^{-t} \Bigr \rvert_0^x = 1-e^{-x}. \]

    If x < 0 then

        \[ F(x) = \int_0^x e^{-|t|} \, dt = \int_0^x e^t \, dt = e^x - 1. \]

  4. Since the maximum of 1 and t^2 is equal to 1 if |t| \leq 1 and is equal to t^2 if |t| > 1 we consider three cases (|x| \leq 1, x > 1 and x < -1).

    For |x| \leq 1 we have

        \[ F(x) = \int_0^x f(t) \, dt = \int_0^x dt = x. \]

    For x > 1 we have

        \[ F(x) = \int_0^1 dt + \int_1^x t^2 \, dt = 1 + \frac{t^3}{3}\Bigr \rvert_1^x = \frac{x^3}{3} + \frac{2}{3} = \frac{x^3}{3} + \frac{|x|}{x} \frac{2}{3} \]

    For x < -1 we have

        \[ F(x) = \int_0^{-1} dt + \int_{-1}^x t^2 \, dt = -1 + \frac{t^3}{3} \Bigr \rvert_{-1}^x = +\frac{x^3}{3} + \frac{|x|}{x} \frac{2}{3}. \]

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