# Prove: Let $f$ be continuous on $[a,b]$, then suppose $f(x) \geq 0$ for all $x \in [a,b]$

Prove: Let $f$ be continuous on $[a,b]$, then suppose $f(x) \geq 0$ for all $x \in [a,b]$. if there exists a point $c \in [a,b]$, such that $f(c) > 0. then $$\int_{a}^{b} f(x) dx>0. This is what i have so far, since f(x) is continuous and f(c)>0 \exists [t,s] such that f(x) > f(x)/2, x \in [t,s] ... and I have no idea how to continuous to prove that, anyone help me? Thanks • Did you mean to say that f(x)>\dfrac{f(c)}{2} for x in an interval [t,s] containing c? Commented Mar 8, 2016 at 3:43 • @JohnWaylandBales yes～ Commented Mar 8, 2016 at 3:44 • So then a rectangle with base the interval [t,s] and height \dfrac{f(c)}{2} would lie between the x-axis and the graph of f. Would that not imply that \int_a^bf(x)\,dx\ge (s-t)\cdot\dfrac{f(c)}{2}? Commented Mar 8, 2016 at 3:48 • Let g(x) = \int_a^x f(x) dx Then g'(x) \ge 0 etc... Would that work? Commented Mar 8, 2016 at 7:30 • I know it will \geq while I just need > and necessary not = Commented Mar 8, 2016 at 18:40 ## 1 Answer First prove that f(x) \geqslant 0 implies \displaystyle \int_a^bf(x) \, dx \geqslant 0. This follows because for any partition Q and lower Darboux sum L(Q,f)$$0 \leqslant L(Q,f) \leqslant \sup_{P} L(P,f) = \int_a^bf(x) \, dx.$$As you observed, if there is at least one point c \in [a,b] where f is continuous and f(c) > 0, then by continuity there exists a subinterval [\alpha,\beta] with c \in (\alpha, \beta) and where f(x) > f(c)/2 > 0 for all x \in [\alpha,\beta]. The infimum of f on [\alpha,\beta] must, therefore, be strictly greater than 0. Hence,$$\int_a^b f(x) \, dx \geqslant \int_\alpha^\beta f(x) \, dx\geqslant \inf_{x \in [\alpha,\beta]}f(x)(\beta - \alpha) > 0.$$Here we are using \displaystyle f(x) \geqslant g(x) \implies \int_a^b f(x) \, dx \geqslant \int_a^b g(x) \, dx which follows from the first part of the proof using f(x) - g(x) \geqslant 0 as the integrand. In particular,$$f(x) \geqslant f(x) \chi_{[\alpha,\beta]} \implies \int_a^b f(x) \, dx \geqslant \int_\alpha^\beta f(x) \, dx, $$and on [\alpha,\beta],$$f(x) \geqslant \inf_{x \in [\alpha,\beta]}f(x) \implies \int_\alpha^\beta f(x) \, dx\geqslant \inf_{x \in [\alpha,\beta]}f(x)(\beta - \alpha) .$\$

• while I need to prove is > 0 rather than >&= 0 , it's still work? Commented Mar 8, 2016 at 3:43
• Yes it does -- will modify above.
– RRL
Commented Mar 8, 2016 at 3:45
• I see ....Thank you so much Commented Mar 8, 2016 at 3:46
• @Helen: You're welcome.
– RRL
Commented Mar 8, 2016 at 3:47