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 $f(t)$ .

$\displaystyle{f(t) = (t + |t|)^2}$ .
The function,
$f(t) = \begin{cases} 1-t^2 & \text{if } |t| \leq 1, \\ 1 - |t| & \text{if } |t| > 1. \end{cases}$
$f(t) = e^{-|t|}$ .
$f(t) =$ the maximum of 1 and $t^2$ .

We know from this exercise (Section 5.5, Exercise #13) that
$F(x) = \int_0^x (t + |t|)^2 \, dt = \frac{2}{3} x^2 (x + |x|).$
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 $|x| > 1$ .
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

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

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

One comment

August 14, 2019 at 11:58 am

Artem says:

I think (b) is wrong. You cannot simply convert $x^2$ into $x|x|$ . The second term for the negative case (< -1) should be $+\frac{x^2}{2}$ . The answer in Apostol is wrong, which is easy to check if you plot it.

Stumbling Robot

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

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