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Okay so I can't make heads or tails of this supposed solution given by my lecturer. The result we have to prove is obvious; clearly a function that is positive over an interval has a positive definite integral over that interval. I just don't understand the thinking behind the solution at all. He says 'by continuity' which I associate with the epsilon-delta argument, but he only uses epsilon. Also he sets c = f(x0) making c a point on the y-axis, but everything else considers c as a point on the x-axis.

Maybe I'm just being thick here, but I think this proof is messy and unclear at best, and I wouldn't be surprised to hear that it's completely wrong. Whether or not that's the case could someone please talk me through the correct thinking here?

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Refnom95
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2 Answers2

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The given solution does more harm than help... here is a corrected version.

Let $c:= f(x_0)>0$ (where $x_0$ is defined in the question). By continuity, there exists an $\delta>0$ such that $a \le x_0-\delta < x_0+\delta \le b$ and such that $f(x) \ge c/2$ for $x \in [x_0-\delta,x_0+\delta]$. Since $f(x) \ge 0$ on $[a,b]$, $$\int_a^b f(x) \mathop{dx} \ge \int_{x_0-\delta}^{x_0+\delta} f(x) \mathop{dx} \ge \frac{c}{2} \cdot 2 \delta > 0.$$

angryavian
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I think what he may mean is:

Let $f(x_0) = c>0$ for some $x_0 \in (a,b)$. Fix $\varepsilon = c/2$. By continuity, there exists a $\delta > 0$ such that $a \leq x_0 - \delta < x_0 + \delta \leq b$ and such that $x \in [x_0-\delta,x_0+\delta]$ implies that $f(x) \geq c/2$. Since $f(x) \geq 0$ on $[a,b]$,

$\int_{a}^{b} f(x)dx \geq \int_{x_0-\delta}^{x_0+\delta} f(x)dx \geq (c/2)\delta > 0$.

What this is more or less saying is, we can draw a little rectangle somewhere under the curve which is easy to calculate the integral of. We know that the integral of our curve has a greater value than the integral of this little rectangle, which has a greater value than zero.

mathochist
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