1. ## Riemann integral

$\displaystyle f(x)=\begin{cases}& \sin(x) , x\leq \pi /2 \\ & \1 , x> \pi /2\end{cases}$

Determine the riemann integral for every $\displaystyle x\in R$

$\displaystyle I(x)=\int_{0}^{x}f(t)dt$

I dont know if i do this right, but this is how i tried:

$\displaystyle I'(x)=f(x)$ according to fundamental theorem of calculus

then $\displaystyle I(x)=F(x) + C$

* if $\displaystyle x\leq \pi /2$

$\displaystyle I(x)=\int_{0}^{x}sin(x)dx = 1-cos(x)$

* if $\displaystyle x> \pi /2$

$\displaystyle \int_{0}^{\pi/2}sin(x)dx + \int_{\pi/2+\delta }^{x}1dx=1+x-\frac{\pi}{2},\delta \to \0$

so did i do totally wrong or does this look good? :P

Thanks!

2. ## Re: Riemann integral

Originally Posted by mechaniac
$\displaystyle f(x)=\begin{cases}& \sin(x) , x\leq \pi /2 \\ & \1 , x> \pi /2\end{cases}$
$\displaystyle I(x)=\int_{0}^{x}f(t)dt$

$\displaystyle I'(x)=f(x)$ according to fundamental theorem of calculus
then $\displaystyle I(x)=F(x) + C$
* if $\displaystyle x\leq \pi /2$
$\displaystyle I(x)=\int_{0}^{x}f(t)dt = 1-cos(x)$
* if $\displaystyle x> \pi /2$
$\displaystyle \int_{0}^{\pi/2}sin(x)dx + \int_{\pi/2+\delta }^{x}sin(x)dx=1+x-\frac{\pi}{2},\delta \to \0$
Yes that is correct. I would write as:
$\displaystyle I(x) = \left\{ {\begin{array}{rl} {1 - \cos (x),} & {x \leqslant \frac{\pi }{2}} \\ {x + 1 - \frac{\pi }{2}} & {x > \frac{\pi }{2}} \\ \end{array} } \right.$

3. ## Re: Riemann integral

nice! i saw now that i did some typos in my original post, corrected now.