I’ve found that $a= \pi$ , $b= -2$, $C=0$ $d= \pi /2 $
Now I have been told to find $a_0$ without using integration ?
and the answer is $a_0 = \frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} (\pi -2t ) dt = \frac{1}{2} \cdot \frac{\pi}{2} \cdot \pi \cdot \frac{1}{\pi} = \frac{\pi}{4} $
What are they doing here ? I don’t get how it’s done
$a_0$ equals $1/\pi$ times the area under the graph over the generic interval. The area is that of the right triangle with sides $\pi$ and $\pi/2.$ This gives
$a_0=\frac{1}{\pi}\times\frac{1}{2}\times\pi\times\frac{\pi}{2}=\frac{\pi}{4}.$