Exercise 6.2

Question

Exercise 2: 
    Suppose $f\ge 0$, $f$ is continuous on $[a,b]$, and $\int_a^bf(x)\,dx=0$. Prove that $f(x)=0$ for all $x\in[a,b]$.

ghostofgarborg · Accepted Answer

(	extit{Matt ``Frito'' Lundy}) \
    Suppose there exists an $x \in [a,b]$ such that $f(x) > 0$.
    $f$ is continuous at on $[a,b]$ means there exists a $\delta$ such that
    $|t - x| < \delta$ and $a \leq t \leq b$ implies $f(t) > 0$.
    Let $P = \{a, x - \delta, x+ \delta, b\}$,
    then $L(P,f) > 0$, and for any partition $P$
    
    $$\int_a^b f(x) \ dx \geq L(P,x) > 0,$$
    a contradiction.

Amit Awasthi · Answer

\begin{proof} Suppose, to get a contradiction, that $f(x_0) > 0$ for some $x_0 \in [a,b]$. By the definition of continuity, given $0 < \epsilon < f(x_0)$, there is an interval $(x_0-\delta,x_0+\delta)$ such that $f > 0$ on this interval. Then, by Theorem $6.12(c)$, \begin{equation*} \int_a^b f(x) dx = \left( \begin{aligned} &\int_a^{x_0-\delta} f(x) dx\ &+ \int_{x_0-\delta}^{x_0+\delta} f(x) dx\ &\quad + \int_{x_0+\delta}^b f(x) dx \end{aligned} ight) > 0 \end{equation*} since $\int_{x_0-\delta}^{x_0+\delta} f(x) dx \ge (f(x_0) - \epsilon)\delta > 0$ by our choice of $\epsilon$ and Definition 6.1, which is a contradiction. \end{proof} \begin{remark} Exercise 6.1 above shows that $\int_a^b f(x) d\alpha$ can be 0 even if $f(x) e 0$ for some $x \in [a,b]$, but only when $f$ is discontinuous, since the continuity of $f$ would make $\int_a^b f(x) d\alpha e 0$ by Exercise 6.2. \end{remark}

Exercise 6.2

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Exercise 6.2

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