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Let $A$ be the exact area over $[a,b]$ under $y=f(x)$.

If $f(x) \geq 0$ (positive), and increasing, then $\forall x \in [a,b]$, Left Riemann Sum $\leq$ A $\leq$ Right Riemann Sum.

How do I prove this? I don't know where to start

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  • $\begingroup$ Draw a picture? $\endgroup$ – carmichael561 Jan 8 '17 at 16:44
  • $\begingroup$ I need a solid math proof photo wont work $\endgroup$ – K Split X Jan 8 '17 at 16:44
  • $\begingroup$ Thats why im confused right now $\endgroup$ – K Split X Jan 8 '17 at 16:45
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    $\begingroup$ Of course, but the picture should make it clear what's going on. $\endgroup$ – carmichael561 Jan 8 '17 at 16:45
  • $\begingroup$ Picture should help you develop a rigorous proof. $\endgroup$ – edm Jan 8 '17 at 16:48
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Well, for a single interval and nondecreasing $f$:

$$a\le x\le b\implies f(a)\le f(x) \le f(b) \implies \int_a^b f(a)\,dx\le \int_a^bf(x)\,dx \le \int_a^bf(b)\,dx$$

$$\implies (b-a)f(a) \le \int_a^bf(x)\,dx \le (b-a)f(b)$$

The general case comes from adding this inequality up across intervals.

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  • $\begingroup$ What is the purpose of multiplying both sides by $(b-a)$? $\endgroup$ – K Split X Jan 8 '17 at 22:53
  • $\begingroup$ @KSplitX $\int_a^b c \,dx=(b-a)c$ for constants $c$ - it's what you get by integrating the values $f(a),f(b)$ over the given interval. $\endgroup$ – πr8 Jan 8 '17 at 23:30

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