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Math Help - Integral proof

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    Integral proof

    Suppose that g: [2,8] --> R is continuous in [2,8] and g(x) greater than or equal to 0 for all x in [2,8].
    Prove that if the integral from 2 to 8 of g dx =0, then g(x) =0 for all x in [2,8]
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    Quote Originally Posted by Coda202 View Post
    Suppose that g: [2,8] --> R is continuous in [2,8] and g(x) greater than or equal to 0 for all x in [2,8].
    Prove that if the integral from 2 to 8 of g dx =0, then g(x) =0 for all x in [2,8]
    This is a special case of a more general theorem.

    Theorem: Let f:[a,b]\to \mathbb{R} be a continous function with f\geq 0 and \smallint_a^b f =0 then f=0.

    Proof: Assume by contradiction that there is a point x_0 \in (a,b) so that f(x_0) = 0. The reason why we can ignore the endpoints is because if it was the case that f(x) = 0 on (a,b) and f(a),f(b)\not = 0 then that would make f non-continous at either a or b. Now if f(x_0)\not = 0 then there is a \delta > 0 so that on [x_0 - \delta,x_0+\delta]\subseteq [a,b] we have f\not = 0 i.e. f > 0. Since f is continous on this closed subinterval it means it attains a mimimum m>0. Now define the function g:[a,b]\to \mathbb{R} in the following way: if x\in [x_0-\delta,x_0+\delta] then g(x) = m otherwise g(x) = 0. Note g is integrable. It follows that \smallint_a^b g \leq \smallint_a^b f \implies 2m\delta \leq \smallint_a^b f.
    But 2m\delta > 0 would would mean \smallint_a^b f > 0 a contradiction.
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