Could someone help me solve this problem, I'm not sure where I'm going wrong $$ \int_{0}^{2\pi} |x-\pi| $$
We break the integral up since $[0,\pi] < 0$ and $[\pi-2\pi] > 0$. I then have
$$ \int_0^\pi |x-\pi| + \int_\pi^{2\pi}|x-\pi| $$
Integrating this I get $$ |1-\pi x||^{\pi}_0+|1-\pi x ||^{2 \pi}_{\pi} = \pi^2+3\pi^2=4\pi^2 $$
However, the correct answer is $\pi^2$. Could someone explain where I went wrong? Your help is appreciated.
See where $x-\pi$ is positive or negative. For example if $x\in (0,\pi)$, then $x-\pi<0$. Therefore $$|x-\pi|=\pi -x$$ integrating this gives $\pi x- x^2/2$
When $x \in [0,\pi]$, $x - \pi \le 0$, so $|x - \pi| = \pi - x$.
When $x \in [\pi, 2\pi]$, $x - \pi \ge 0$, so $|x - \pi| = x - \pi$. Now:
$$\int_0^{2\pi}|x - \pi|dx = \int_0^{\pi}|x - \pi|dx + \int_{\pi}^{2\pi}|x - \pi|dx = \int_0^{\pi}(\pi - x)dx + \int_{\pi}^{2\pi}(x - \pi)dx = \frac{\pi^2}{2} + (2\pi^2 - 2\pi^2 - \frac{\pi^2}{2} + \pi^2) = \pi^2 $$
Between $0$ and $\pi$ we have that $|x-\pi| = (\pi-x)$, while between $\pi$ and $2\pi$ we have that $|x-\pi| = (x-\pi)$. This allows us to remove the absolute signs and get normal polynomials.
