- #1
mahmoud2011
- 88
- 0
[itex]\int^{\pi}_{0} f(x) dx[/itex] where ,
f(x) = sin x if [itex]0 \leq x < \frac{\pi}{2}[/itex]
and f(x) = cos(x) if [itex]\frac{\pi}{2} \leq x \leq \pi[/itex]
The problem is that the version I am using of Fundamental theorem is if f is continuous on some closed interval , I wrote the integral as
[itex]\int^{\pi / 2}_{0} f(x) dx + \int^{\pi}_{\pi /2} f(x) dx[/itex]
but I have in the first integral f still is not continuous on [itex][0,\pi/2][/itex]
I tried to open some references and reached another version for the theorem there f is integrable on f , and g' =f , but I couldn't do any thing
Thanks
f(x) = sin x if [itex]0 \leq x < \frac{\pi}{2}[/itex]
and f(x) = cos(x) if [itex]\frac{\pi}{2} \leq x \leq \pi[/itex]
The problem is that the version I am using of Fundamental theorem is if f is continuous on some closed interval , I wrote the integral as
[itex]\int^{\pi / 2}_{0} f(x) dx + \int^{\pi}_{\pi /2} f(x) dx[/itex]
but I have in the first integral f still is not continuous on [itex][0,\pi/2][/itex]
I tried to open some references and reached another version for the theorem there f is integrable on f , and g' =f , but I couldn't do any thing
Thanks