Riemann integral

**mechaniac** · November 28th 2011, 06:35 AM

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

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

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

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

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

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

* if $x\leq \pi /2$

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

* if $x> \pi /2$

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

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

Thanks!

**Plato** · November 28th 2011, 07:01 AM

Originally Posted by mechaniac

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

$I'(x)=f(x)$ according to fundamental theorem of calculus
then $I(x)=F(x) + C$
* if $x\leq \pi /2$
$I(x)=\int_{0}^{x}f(t)dt = 1-cos(x)$
* if $x> \pi /2$
$\int_{0}^{\pi/2}sin(x)dx + \int_{\pi/2+\delta }^{x}sin(x)dx=1+x-\frac{\pi}{2},\delta \to \0$

Yes that is correct. I would write as:
$I(x) = \left\{ {\begin{array}{rl} {1 - \cos (x),} & {x \leqslant \frac{\pi }{2}} \\ {x + 1 - \frac{\pi }{2}} & {x > \frac{\pi }{2}} \\ \end{array} } \right.$

**mechaniac** · November 28th 2011, 07:29 AM

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

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