# How to prove Left Riemann Sum is underestimate and Right Riemann sum is overestimate?

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

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

Well, for a single interval and nondecreasing $f$:
$$a\le x\le b\implies f(a)\le f(x) \le f(b) \implies \int_a^b f(a)\,dx\le \int_a^bf(x)\,dx \le \int_a^bf(b)\,dx$$
$$\implies (b-a)f(a) \le \int_a^bf(x)\,dx \le (b-a)f(b)$$
• What is the purpose of multiplying both sides by $(b-a)$? – K Split X Jan 8 '17 at 22:53
• @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. – πr8 Jan 8 '17 at 23:30