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Thread: Riemann integral

  1. #1
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    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!
    Last edited by mechaniac; Nov 28th 2011 at 07:31 AM.
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  2. #2
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    Re: Riemann integral

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

    $\displaystyle I'(x)=f(x)$ 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:
    $\displaystyle I(x) = \left\{ {\begin{array}{rl} {1 - \cos (x),} & {x \leqslant \frac{\pi }{2}} \\ {x + 1 - \frac{\pi }{2}} & {x > \frac{\pi }{2}} \\ \end{array} } \right.$
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  3. #3
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    Re: Riemann integral

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