5.1 Approximating areas  (Page 9/17)

Add the numbers to get 8.1-in. net increase.

309,389,957

$L_8=3+2+1+2+3+4+5+4=24$

$L_8=3+5+7+6+8+6+5+4=44$

$L_{10}\approx 1.7604,L_{30}\approx 1.7625,L_{50}\approx 1.76265$

$R_1=\mathrm{-1},L_1=1,R_{10}=\mathrm{-0.1},L_{10}=0.1,L_{100}=0.01,$ and $R_{100}=\mathrm{-0.1}.$ By symmetry of the graph, the exact area is zero.

$R_1=0,L_1=0,R_{10}=2.4499,L_{10}=2.4499,R_{100}=2.1365,L_{100}=2.1365$

If $\left[c,d\right]$ is a subinterval of $\left[a,b\right]$ under one of the left-endpoint sum rectangles, then the area of the rectangle contributing to the left-endpoint estimate is $f\left(c\right)\left(d-c\right).$ But, $f\left(c\right)\le f\left(x\right)$ for $c\le x\le d,$ so the area under the graph of f between c and d is $f\left(c\right)\left(d-c\right)$ plus the area below the graph of f but above the horizontal line segment at height $f\left(c\right),$ which is positive. As this is true for each left-endpoint sum interval, it follows that the left Riemann sum is less than or equal to the area below the graph of f on $\left[a,b\right].$

$L_N=\frac{b-a}{N}{\displaystyle \sum _{i=1}^{N}f\left(a+\left(b-a\right)\frac{i-1}{N}\right)}=\frac{b-a}{N}{\displaystyle \sum _{i=0}^{N-1}f\left(a+\left(b-a\right)\frac{i}{N}\right)}$ and $R_N=\frac{b-a}{N}{\displaystyle \sum _{i=1}^{N}f\left(a+\left(b-a\right)\frac{i}{N}\right)}.$ The left sum has a term corresponding to $i=0$ and the right sum has a term corresponding to $i=N.$ In $R_N-L_N,$ any term corresponding to $i=1,2\text{,…,}N-1$ occurs once with a plus sign and once with a minus sign, so each such term cancels and one is left with $R_N-L_N=\frac{b-a}{N}\left(f\left(a+\left(b-a\right)\right)\frac{N}{N}\right)-\left(f\left(a\right)+\left(b-a\right)\frac{0}{N}\right)=\frac{b-a}{N}\left(f\left(b\right)-f\left(a\right)\right).$

Graph 1: a. $L\left(A\right)=0,B\left(A\right)=20;$ b. $U\left(A\right)=20.$ Graph 2: a. $L\left(A\right)=9;$ b. $B\left(A\right)=11,U\left(A\right)=20.$ Graph 3: a. $L\left(A\right)=11.0;$ b. $B\left(A\right)=4.5,U\left(A\right)=15.5.$

Let A be the area of the unit circle. The circle encloses n congruent triangles each of area $\frac{\text{sin}\left(\frac{2\pi }{n}\right)}{2},$ so $\frac{n}{2}\text{sin}\left(\frac{2\pi }{n}\right)\le A.$ Similarly, the circle is contained inside n congruent triangles each of area $\frac{BH}{2}=\frac{1}{2}\left(\text{cos}\left(\frac{\pi }{n}\right)+\text{sin}\left(\frac{\pi }{n}\right)\text{tan}\left(\frac{\pi }{n}\right)\right)\text{sin}\left(\frac{2\pi }{n}\right),$ so $A\le \frac{n}{2}\text{sin}\left(\frac{2\pi }{n}\right)\left(\text{cos}\left(\frac{\pi }{n}\right)\right)+\text{sin}\left(\frac{\pi }{n}\right)\text{tan}\left(\frac{\pi }{n}\right).$ As $n\to \infty ,\frac{n}{2}\text{sin}\left(\frac{2\pi }{n}\right)=\frac{\pi \text{sin}\left(\frac{2\pi }{n}\right)}{\left(\frac{2\pi }{n}\right)}\to \pi ,$ so we conclude $\pi \le A.$ Also, as $n\to \infty ,\text{cos}\left(\frac{\pi }{n}\right)+\text{sin}\left(\frac{\pi }{n}\right)\text{tan}\left(\frac{\pi }{n}\right)\to 1,$ so we also have $A\le \pi .$ By the squeeze theorem for limits, we conclude that $A=\pi .$