Science:Math Exam Resources/Courses/MATH101/April 2009/Question 01 (f)/Solution 2

(The following solution takes advantage of symmetry about $\theta =\pi /4$.)

The average value of $\left|\sin \theta -\cos \theta \right|$ over the interval $0\leq \theta \leq \pi /2$ is
${\begin{aligned}f_{ave}={\frac {1}{\pi /2}}\int _{0}^{\pi /2}\left|\sin \theta -\cos \theta \right|d\theta .\end{aligned}}$

If we draw a picture of the integrand, we can see that it is symmetric about $\theta =\pi /4$. Motivated by this symmetry, we make a change of variables $\theta =\pi /4+x$.

${\begin{aligned}f_{ave}={\frac {2}{\pi }}\int _{-\pi /4}^{\pi /4}\left|\sin(\pi /4+x)-\cos(\pi /4+x)\right|dx.\end{aligned}}$

Now, this integrand is symmetric about the origin since

${\begin{aligned}\left|\sin(\pi /4+x)-\cos(\pi /4+x)\right|&=\left|\cos(\pi /2-(\pi /4+x))-\sin(\pi /2-(\pi /4+x))\right|\\&=\left|\cos(\pi /4-x)-\sin(\pi /4-x)\right|\\&=\left|\sin(\pi /4-x)-\cos(\pi /4-x)\right|.\end{aligned}}$

Hence,

${\begin{aligned}f_{ave}={\frac {2}{\pi }}\cdot 2\int _{0}^{\pi /4}\left|\sin(\pi /4+x)-\cos(\pi /4+x)\right|dx.\end{aligned}}$

Since $\sin(\pi /4+x)-\cos(\pi /4+x)$ is positive for $0\leq x\leq \pi /4$,

${\begin{aligned}f_{ave}&={\frac {4}{\pi }}\int _{0}^{\pi /4}\left(\sin(\pi /4+x)-\cos(\pi /4+x)\right)dx\\&={\frac {4}{\pi }}\left[-\cos(\pi /4+x)-\sin(\pi /4+x)\right]_{0}^{\pi /4}\\&={\frac {4}{\pi }}\left(-\cos(\pi /2)-\sin(\pi /2)+\cos(\pi /4)+\sin(\pi /4)\right)\\&={\frac {4}{\pi }}\left(0-1+{\frac {2}{\sqrt {2}}}\right)\\&={\frac {4}{\pi }}\left({\sqrt {2}}-1\right)\end{aligned}}$