# 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?

• @scott thanks, curly brackets. Got it! – DrOnline Apr 28 '13 at 23:34
• No problem DrOnline! – Scott H. Apr 28 '13 at 23:35

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