∫(from -π to π) |π – |x||dx is equal to:

∫? ² |x-1|-x|dx
Let f (x)=|x-1|-x|
= {|1-2x|, x≤1; 1, x≥1}
A = 1/2+1=3/2
∫? ¹/² (1-2x)dx+∫? /? ¹ (2x-1)+∫? ²1dx
= [x-x²]? ¹/²+ [x]? ² = 3/21dx
= [x-x²]? ¹/²+ [x]? ² = 3/2

 $x^4-2x^3+2x-1={\left(x-1\right)}^2\left(x^2-1\right)$

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

= 2