Tag: Absolute Values

Compute the integral from 0 to π/2 of |sin x – cos x|

by RoRi

August 3, 2015

Compute the following integral:

$\int_0^{\frac{\pi}{2}} | \sin x - \cos x | \, dx.$

Since $\cos x - \sin x \geq 0$ on the integral $\left[ 0, \frac{\pi}{4} \right]$ and $\cos x - \sin x < 0$ on $\left( \frac{pi}{4}, \frac{\pi}{2} \right]$ we have,

$\begin{align*} \int_0^{\frac{\pi}{2}} | \sin x - \cos x | \, dx &= \int_0^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - \cos x) \, dx \\ &= \sin \frac{\pi}{4} + \cos \frac{\pi}{4} - 1 + \cos \frac{\pi}{4} - \cos \frac{\pi}{2} - \sin \frac{\pi}{2} + \sin \frac{\pi}{4} \\ &= \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} - 1 + \frac{\sqrt{2}}{2} - 0 - 1 + \frac{\sqrt{2}}{2} \\ &= 2 \sqrt{2} - 2. \end{align*}$

Compute the area between the graphs of functions

by RoRi

July 29, 2015

Let

$f(x) = |x|+|x-1|, \qquad g(x) = 0.$

Find the area between the graphs of $f$ and $g$ on the interval $[-1,2]$ .

First, we draw the graph, shading the region $S$ between the two graphs in blue.

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In this problem $f(x) \geq g(x)$ on the entire interval $[-1,2]$ ; however, the function $f(x) = |x|+|x-1|$ is defined piecewise on this interval.

$\begin{align*} \int_{-1}^2 (f(x) - g(x)) \, dx &= \int_{-1}^2 \left(|x| + |x-1|\right) \, dx \\ &= \int_{-1}^2 |x| \, dx + \int_{-1}^2 | x-1| \, dx. \end{align*}$

The functions $|x|$ and $|x-1|$ are defined piecewise on the interval $[-1,2]$ by

$|x| = \begin{cases} -x & \text{if } x \in [-1,0) \\ x & \text{if } x \in [0,2] \end{cases}$

and

$|x-1| = \begin{cases} -(x-1) & \text{if } x \in [-1,1) \\ (x-1) & \text{if } x \in [1,2] \end{cases}.$

So, we break each integral into two pieces and compute:

$\begin{align*} \int_{-1}^2 |x| \, dx + \int_{-1}^2 |x-1| \, dx &= \int_{-1}^0 (-x) \, dx + \int_0^2 x \, dx + \int_{-1}^1 -(x-1) \, dx + \int_1^2 (x-1) \, dx \\ &= \left. \left(-\frac{x^2}{2}\right) \right|_{-1}^0 + \left. \left( \frac{x^2}{2} \right) \right|_0^2 + \left. \left( - \frac{x^2}{2} \right) \right|_{-1}^1 + (x)\biggr \rvert_{-1}^1 + \left. \left( \frac{x^2}{2} \right) \right|_1^2 + (-x) \biggr \rvert_1^2 \\ &= \frac{1}{2} + 2 - \frac{1}{2} + \frac{1}{2} + 1 + 1 + 2 - \frac{1}{2} - 2 + 1 \\ &= 5. \end{align*}$

Compute the area between the graphs of functions

by RoRi

July 29, 2015

Let

$f(x) = 2|x|, \qquad g(x) = 1-3x^3.$

Find the area between the graphs of $f$ and $g$ on the interval $\left[-\frac{\sqrt{3}}{3},\frac{1}{3}\right]$ .

First, we draw the graph, shading the region $S$ between the two graphs in blue.

Rendered by QuickLaTeX.com

In this problem $g(x) \geq f(x)$ on the entire interval $\left[-\frac{\sqrt{3}}{3},\frac{1}{3}\right]$ ; however, the function $f(x) = 2|x|$ is defined piecewise on this interval as

$f(x) = |x| = \begin{cases} -x & \text{if } x \in \left[-\frac{\sqrt{3}}{3},0\right) \\ x & \text{if } x \in \left[0,\frac{1}{3}\right] \end{cases}.$

Thus, we break the integral into intervals $\left[-\frac{\sqrt{3}}{3},0\right]$ and $\left[0,\frac{1}{3}\right]$ , and in both cases we are taking the integral of $g(x) - f(x)$ , but the definition of $f$ will depend on which interval we are in. The computation is as follows:

$\begin{align*} a(S) = \int_{-\frac{\sqrt{3}}{3}}^{\frac{1}{3}} (g(x) - f(x)) \, dx &= \int_{-\frac{\sqrt{3}}{3}}^{\frac{1}{3}} g(x) \, dx - \int_{-\frac{\sqrt{3}}{3}}^{\frac{1}{3}} f(x) \, dx \\ &\phantom{=} \\ &= \int_{-\frac{\sqrt{3}}{3}}^{\frac{1}{3}} (1 - 3x^3) \, dx - \int_{-\frac{\sqrt{3}}{3}}^0 (-2x) \, dx - \int_0^{\frac{1}{3}} 2x \, dx \\ & \phantom{=} \\ &= \left. \left( x - \frac{3}{4}x^4 \right) \right|_{-\frac{\sqrt{3}}{3}}^{\frac{1}{3}} + (x^2) \biggr \rvert_{-\frac{\sqrt{3}}{3}}^0 - (x^2) \biggr \rvert_0^{\frac{1}{3}} \\ & \phantom{=} \\ &= \left( \frac{1}{3} - \frac{1}{108} + \frac{\sqrt{3}}{3} + \frac{1}{12}\right) + \left(-\frac{1}{3}\right) - \left( \frac{1}{9} \right) \\ &= \frac{9\sqrt{3} - 1}{27}. \end{align*}$

Compute the area between the graphs of functions

by RoRi

July 28, 2015

Let

$f(x) = |x-1|, \qquad g(x) = x^2-2x.$

Find the area between the graphs of $f$ and $g$ on the interval $[0,2]$ .

First, we draw the graph, shading the region $S$ between the two graphs in blue.

Rendered by QuickLaTeX.com

In this problem $f(x) \geq g(x)$ on the entire interval $[-1,1]$ ; however, the function $f(x) = |x|$ is defined piecewise on this interval as

$f(x) = |x-1| = \begin{cases} -(x-1) & \text{if } x \in [0,1) \\ x-1 & \text{if } x \in [1,2] \end{cases}.$

Thus, we break the integral into intervals $[0,1]$ and $[1,2]$ , and in both cases we are taking the integral of $f(x) - g(x)$ , but the definition of $f$ will depend on which interval we are in. The computation is as follows:

$\begin{align*} a(S) = \int_0^2 (f(x) - g(x)) \, dx &= \int_0^1 (-(x-1) - x^2 + 2x) \, dx + \int_1^2 (x-1 - x^2 + 2x) \, dx \\ &= \int_0^1 (-x^2 + x + 1) \,dx + \int_1^2 (-x^2 + 3x - 1) \, dx \\ &= \left. \left( -\frac{x^3}{3} + \frac{x^2}{2} + x \right) \right|_0^1 + \left. \left( -\frac{x^3}{3} + \frac{3x^2}{2} - x \right) \right|_1^2 \\ &= \left( -\frac{1}{3} + \frac{1}{2} + 1 \right) + \left( -\frac{8}{3} + 6 - 2 + \frac{1}{3} - \frac{3}{2} + 1\right) \\ &= \frac{7}{6} + \frac{7}{6} \\ &= \frac{7}{3}. \end{align*}$

Compute the area between the graphs of functions

by RoRi

July 28, 2015

Let

$f(x) = |x|, \qquad g(x) = x^2-1.$

Find the area between the graphs of $f$ and $g$ on the interval $[-1,1]$ .

First, we draw the graph, shading the region $S$ between the two graphs in blue.

Rendered by QuickLaTeX.com

In this problem $f(x) \geq g(x)$ on the entire interval $[-1,1]$ ; however, the function $f(x) = |x|$ is defined piecewise on this interval as

$f(x) = |x| = \begin{cases} -x & \text{if } x \in [-1,0) \\ x & \text{if } x \in [0,1] \end{cases}.$

Thus, we break the integral into intervals $[-1,0]$ and $[0,1]$ , and in both cases we are taking the integral of $f(x) - g(x)$ , but the definition of $f$ will depend on which interval we are in. The computation is as follows:

$\begin{align*} a(S) = \int_{-1}^{1} (f(x) - g(x)) \, dx &= \int_{-1}^0 (-x - x^2 + 1) \, dx + \int_0^1 (x - x^2 + 1) \, dx \\ &= \left. \left(-\frac{x^2}{2} - \frac{x^3}{3} + x \right) \right|_{-1}^0 + \left. \left( \frac{x^2}{2} - \frac{x^3}{3} + x \right) \right|_0^1 \\ &= \left( \frac{1}{2} - \frac{1}{3} + 1 \right) + \left( \frac{1}{2} - \frac{1}{3} + 1 \right) \\ &= \frac{7}{3}. \end{align*}$

Find an interval of integration to make some integral equations true

by RoRi

July 25, 2015

Find all real numbers $c$ such that the following equations hold:

$\displaystyle{\int_0^c x(1-x) \, dx = 0}$ .
$\displaystyle{\int_0^c \left| x(1-x) \right| \, dx = 0}$ .

We evaluate the integral:
$\int_0^c x(1-x) \, dx = 0 \ \implies \ \left. \left( \frac{-x^3}{3} + \frac{x^2}{2} \right) \right|_0^c = 0 \implies \ \frac{-c^3}{3} + \frac{c^2}{2} = 0.$

Solving for $c$ we have,

$3c^2 - 2c^3 = 0 \quad \implies \quad c = 0, \text{ or } c = \frac{3}{2}.$
Here, since $\left| x(1-x) \right| \geq 0$ for all $x$ , we can use the comparison theorem (Theorem 1.20 in Apostol) to see that if $\left| x(1-x) \right| > 0$ for any $x$ then,
$\int_0^c \left| x(1-x) \right| \, dx > \int_0^c 0 \, dx = 0.$

So, for the equation to hold we must have

$\left|x(1-x) \right| = 0 \quad \text{for all } 0 \leq x \leq c.$

Since the expression will be nonzero for any $0 < x < 1$ , we must have $c = 0$ .

Compute an integral of |(x-1)(3x-1)|

by RoRi

July 24, 2015

Compute

$\int_0^2 \left|(x-1)(3x-1)\right| \, dx.$

Since we are taking the integral of an absolute value, we want to break the expression in the absolute value into intervals on which it is either always positive or always negative, and evaluate the integrals of the pieces separately. So, examining $|(x-1)(3x-1)|$ we have zeros of $(x-1)(3x-1)$ at $x = 1, \frac{1}{3}$ . If the expression is going to change sign it must do so at these points. So,

$\begin{align*} x < \frac{1}{3} && \implies && (x-1)(3x-1) > 0 && \implies && |(x-1)(3x-1)| &= (x-1)(3x-1) \\ \frac{1}{3} < x < 1 && \implies && (x-1)(3x-1) < 0 && \implies && |(x-1)(3x-1)| &= -(x-1)(3x-1) \\ x > 1 && \implies && (x-1)(3x-1) > 0 && \implies && |(x-1)(3x-1)| &= (x-1)(3x-1). \end{align*}$

So, we break the integral into pieces (which we can do since the integral is additive with respect to the interval of integration by Theorem 1.17 of Apostol),

$\int_0^2 \left| (x-1)(3x-1) \right| \, dx &= \int_0^{1/3} (x-1)(3x-1) \, dx + \int_{1/3}^1 -(x-1)(3x-1) \, dx + \int_1^2 (x-1)(3x-1).$

Now we can use the formula for the integral of a polynomial to calculate each of these separately,

$\begin{align*} \int_0^{1/3} (x-1)(3x-1) \, dx &= \int_0^{1/3} (3x^2 - 4x + 1) \, dx \\ &= \left. \left( 3 \frac{x^3}{3} - 4 \frac{x^2}{2} + x \right) \right|_0^{1/3} \\ &= \left( \frac{1}{27} - \frac{2}{9} + \frac{1}{3} \right) - 0 \\ &= \frac{4}{27} \\ \int_{1/3}^1 -(x-1)(3x-1) \, dx &= -\int_{1/3}^1 (3x^2 - 4x + 1) \, dx \\ &= - \left. \left( 3 \frac{x^3}{3} - 4 \frac{x^2}{2} + x \right) \right|_{1/3}^1 \\ &= - \left( (1-2 + 1) - \left(\frac{1}{27} - \frac{2}{9} + \frac{1}{3} \right) \right) \\ &= \frac{4}{27} \\ \int_1^2 (x-1)(3x-1) \, dx &= \int_1^2 (3x^2 - 4x + 1) \, dx \\ &= \left. \left( 3 \frac{x^3}{3} - 4 \frac{x^2}{2} + x \right) \right|_1^2 \\ &= (8 - 8 + 2) - (1 - 2 + 1) \\ &= 2 \end{align*}$

Putting all of these pieces together, we have,

$\int_0^2 \left| (x-1)(3x-1) \right| \, dx = \frac{4}{27} + \frac{4}{27} + 2 = \frac{62}{27}.$

Compute some values of phi(x) = |x-3| + |x-1|

by RoRi

July 11, 2015

Define $\varphi(x) = |x-3| + |x-1|$ for all $x \in \mathbb{R}$ . Compute:

$\varphi(0) = |0-3| + |0-1| = 3 + 1 = 4$ .
$\varphi(1) = |1-3| + |1-1| = 2 + 0 = 2$ .
$\varphi(2) = |2-3| + |2-1| = 1 + 1 = 2$ .
$\varphi(3) = |3-3| + |3-1| = 0 + 2 = 2$ .
$\varphi(-1) = |-1-3| + |-1-1| = |-4| + |-2| = 4+2 = 6$ .
$\varphi(-2) = |-2-3| + |-2-1| = |-5| + |-3| = 5+3 = 8$ .
Find all $t \in \mathbb{R}$ such that $\varphi(t+2) = \varphi(t)$ .
$\begin{align*} &&\varphi(t+2) &= \varphi(t) \\ \implies && |t+2 - 3| + |t+2-1| &= |t-3|+|t-1| \\ \implies && |t-1| + |t+1| &= |t-3| + |t-1| \\ \implies && |t+1| &= |t-3| \implies && t = 1 \end{align*}$

Some true/false questions on absolute values

by RoRi

July 7, 2015

Decide whether the following are true or false.

$x < 5 \implies |x| < 5$ .
$|x-5| < 2 \implies 3 < x < 7$ .
$|1 + 3x| \leq 1 \implies x \geq -\frac{2}{3}$ .
$|x-1| = |x-2|$ is not satisfied by any $x \in \mathbb{R}$ .
For all $x \in \mathbb{R}_{>0}$ there exists a $y \in \mathbb{R}_{>0}$ such that $|2x+y| = 5$ .

False. Let $x = -6$ , then $x < 5$ , but $|x| = |-6| = 6 > 5$ .
True.

$|x-5| < 2 \ \implies \ -(x-5) < 2 \qquad \text{and} (x-5) < 2$

Hence,

$x >3 \qquad \text{and} x < 7.$

This means $3 < x < 7$ .
True.

$|1+3x| \leq 1 \ \implies \ -\frac{2}{3} \leq x \leq 0 \ \implies \ -\frac{2}{3} \leq x.$
False. This equality is true for $x = \frac{3}{2}$ since

$\left| \frac{3}{2} - 1 \right| = \frac{1}{2} \qquad \text{and} \qquad \left| \frac{3}{2} - 2 \right| = \frac{1}{2}.$
False. Take $x = 3$ , then

$|2x+y| = 5 \ \implies \ |6+y| = 5 \ \implies \ y=-11 \text{ or } y = -1.$

Hence, $y < 0$ in either case and so $y \notin \mathbb{R}_{>0}$ .

Translate inequalities involving absolute values to inequalities without absolute values

by RoRi

July 7, 2015

Write the following inequalities in equivalent forms without the absolute values.

a1 is equivalent to b2:

$|x| < 3 \iff -3 < x < 3.$

a2 is equivalent to b5:

$|x-1| < 3 \iff -3 < x-1 < 3 \iff -2 < x < 4.$

a3 is equivalent to b7:

$|3-2x| < 1 \iff -1 < 3-2x < 1 \iff 1 > 2x -3 > -1 \iff 4 > 2x > 2 \iff 1 < x < 2.$

a4 is equivalent to b10:

$|1+2x| \leq 1 \iff -1 \leq 1+2x \leq 1 \iff -2 \leq 2x \leq 0 \iff -1 \leq x \leq 0.$

a5 is equivalent to b3:

$|x-1| > 2 \iff x-1>2 \text{ or } x-1 < -2. \quad \text{Thus, } x > 3 \text{ or } x< -1.$

a6 is equivalent to b8:

$|x+2| \geq 5 \iff x+2 \geq 5 \text{ or } x+2 \leq -5. \quad \text{Thus, } x\geq 3 \text{ or } x \leq -7.$

a7 is equivalent to b9:

$\begin{align*} |5 - x^{-1}|< 1 &\iff -1 < 5-\frac{1}{x} < 1 \\ &\iff 1 > \frac{1}{x} - 5 > -1 \\ &\iff 6 > \frac{1}{x} > 4 \\ &\iff \frac{1}{6} < x < \frac{1}{4}. \end{align*}$

a8 is equivalent to b4:

$\begin{align*} |x-5| < |x+1| &\iff (x-5)^2 < (x+1)^2 \\ &\iff x^2 - 10x + 25 < x^2 + 2x + 1 \\ &\iff -12x < -24 \iff x > 2. \end{align*}$

a9 is equivalent to b6:

$\begin{align*} |x^2 - 2| \leq 1 &\iff -1 \leq x^2 - 2 \leq 1 \\ &\iff 1 \leq x^2 \leq 3 \\ &\iff -\sqrt{3} \leq x \leq -1 \text{ or } 1 \leq x \leq \sqrt{3}. \end{align*}$

a10 is equivalent to b1: $x < x^2 - 12 < 4x \iff 0 < x^2 - x - 12$ and $x^2 -4x - 12 < 0$ . Thus, we must have $0 < (x-4)(x+3)$ and $(x-6)(x+2) < 0$ . The first of these inequalities requires $(x-4)$ and $(x+3)$ to both be positive or both be negative. Thus, $x < -3$ or $x > 4$ . The second requires $(x-6)$ and $(x+2)$ to have opposite signs. So, $-2 < x < 6$ . Combining these restrictions on $x$ we have $4 < x < 6$ (b1).