cal problem

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:

$ \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} $

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.

$ \left\{ {\begin{array}{*{20}c} {f\left( x \right) = 0} & {x \ne \frac{1}{2}} \\ {f\left( x \right) > 0} & {x = \frac{1}{2}} \\ \end{array}} \right. $