Results 1 to 2 of 2

Math Help - cal problem

  1. #1
    Junior Member
    Joined
    Jun 2006
    Posts
    53

    cal problem

    suppose f is continuous on [0,1], f(x) is greater than or equal to 0 and f(1/2) is greater than 0.

    a. prove that the integral of f(x) dx from 0 to 1 is greater than 0.

    b. show, by counterexample, that the assumption of continuity is necessary.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    a) On [0,1], f(x) is non-negative so the integral can be at least 0. But we know that f(1/2) > 0. So because of continuity, there exists an e > 0 for which f(x) > 0 on the interval (1/2-e,1/2+e). Then:

    <br />
\int\limits_0^1 {f\left( x \right)dx}  = \int\limits_0^{\frac{1}{2} - \varepsilon } {f\left( x \right)dx}  + \int\limits_{\frac{1}{2} - \varepsilon }^{\frac{1}{2} + \varepsilon } {f\left( x \right)dx}  + \int\limits_{\frac{1}{2} + \varepsilon }^1 {f\left( x \right)dx} <br />

    Now, the first and third integrals may be 0, but the second one is > 0.

    b) If f isn't continuous, then you can construct a function for which f(x) > 0 for x in (1/2-e,1/2+e) isn't true. For example; keep f(x) at 0 everywhere, except at x = 1/2.

    <br />
\left\{ {\begin{array}{*{20}c}<br />
   {f\left( x \right) = 0} & {x \ne \frac{1}{2}}  \\<br />
   {f\left( x \right) > 0} & {x = \frac{1}{2}}  \\<br />
\end{array}} \right.<br />
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum