definite integral negative variable Man, it's been so long since I did this. I am trying to do this:
NB: limits are $-\pi$ and 0, but I can't get the minus in the limits. If anybody knows how do to that please let me know, the $\pi$ symbol jumps into the integrand when I type minus before it...
$$
\frac{1}{\pi}\int^0_{-\pi} -t dt
$$
I move the minus out of the integral:
$$
-\frac{1}{\pi}\int^0_{-\pi} t dt
$$
I do the integration:
$$
-\frac{1}{\pi} \cdot \left[\frac{t^2}{2}\right]^0_{-\pi}
$$
I insert the limits, the final limit is zero, so the second term is removed:
$$
-\frac{1}{\pi} \cdot \frac{(-\pi)^2}{2}
$$
$$
-\frac{1}{\pi} \cdot \frac{\pi^2}{2}
$$
$$
-\frac{\pi}{2}
$$
But this is incorrect. It should be:
$$
+\frac{\pi}{2}
$$
So my question is: Do I have to change the signing when I do definite integrals on the left side of the y axis?

Note that you evaluate $$-\frac{1}{\pi}\left(\dfrac{t^2}{2}\right)\Big|_{-\pi}^0$$
Which is $$-\frac 1{\pi}\left[\left(\dfrac{0^2}{2}\right) - \left(\dfrac{(-\pi)^2}{2}\right)\right] = -\frac 1{\pi}\left[0 - \left(\dfrac{(-\pi)^2}{2}\right)\right] -\frac 1{\pi}\cdot -\frac{\pi^2}{2} = \frac{\pi}{2}$$
