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Suppose that $\int_a^b{f(x) dx} \le \int_a^b{g(x)dx}$ . Is it true that $f(x)\le g(x)$ for all $x \in [a,b]$? Explain.
Is it true that there exists $c \in [a,b]$ such that $f(c)\le g(c)$?

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What is the $x$ in $f(x) \le g(x)$? You need a quantifier. –  user66345 Apr 8 '13 at 3:43
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@Ray Yang: It’s utterly inappropriate to tag something homework unless the OP explicitly acknowledges it to be such. –  Brian M. Scott Apr 8 '13 at 3:47
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@user66345: It appears that the quantifier got attached to the wrong thing: I’d just about bet that ‘for all $x\in[a,b]$’ belongs with $f(x)\le g(x)$, since it makes no sense where it is. –  Brian M. Scott Apr 8 '13 at 3:48
    
Ah. My apologies about the homework tag then. –  Ray Yang Apr 8 '13 at 3:49

3 Answers 3

$$0=\int\limits_0^1 0\,dx\le\int\limits_0^1\left(-\frac{4}{3}x+1\right)dx=\frac{1}{3}$$

But it isn't true that

$$0\le -\frac{4}{3}x+1\;\;\forall\,x\in[0,1]$$

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HINT for the first question: Take $g$ to be the function that is identically $0$; surely you can find a nice function $f$ such that $f(a)>0$ but $\int_a^bf\le 0$.

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Since the first part has been answered, for the second part:

Suppose not, then for all $x \in [a,b], f(x) > g(x)$. What does this say about the integrals?

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