Question

for time, t, in hours, 0 ≤ t ≤ 1, a bug is crawling at a velocity, v, in meters/hour given by v = \frac{3}{3 + t}. use δt = 0.2 to estimate the distance that the bug crawls during this hour. use left - and right - hand riemann sums to find an overestimate and an underestimate. then average the two to get a new estimate. underestimate = overestimate = average = (for each, include units)

Explanation:

Step1: Find the partition points

Given \(a = 0\), \(b=1\), \(\Delta t=0.2\). The number of sub - intervals \(n=\frac{b - a}{\Delta t}=\frac{1 - 0}{0.2}=5\).
The partition points are \(t_0 = 0\), \(t_1=0.2\), \(t_2 = 0.4\), \(t_3=0.6\), \(t_4 = 0.8\), \(t_5=1\).

Step2: Calculate the left - hand Riemann sum (under - estimate)

The left - hand Riemann sum formula is \(L=\sum_{i = 0}^{n - 1}v(t_i)\Delta t\).
\(v(t)=\frac{3}{3 + t}\)
\(v(t_0)=\frac{3}{3+0}=1\), \(v(t_1)=\frac{3}{3 + 0.2}=\frac{3}{3.2}=\frac{15}{16}\), \(v(t_2)=\frac{3}{3+0.4}=\frac{3}{3.4}=\frac{15}{17}\), \(v(t_3)=\frac{3}{3+0.6}=\frac{3}{3.6}=\frac{5}{6}\), \(v(t_4)=\frac{3}{3+0.8}=\frac{3}{3.8}=\frac{15}{19}\)
\(L=(1+\frac{15}{16}+\frac{15}{17}+\frac{5}{6}+\frac{15}{19})\times0.2\)

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Step3: Calculate the right - hand Riemann sum (over - estimate)

The right - hand Riemann sum formula is \(R=\sum_{i = 1}^{n}v(t_i)\Delta t\)
\(v(t_1)=\frac{15}{16}\), \(v(t_2)=\frac{15}{17}\), \(v(t_3)=\frac{5}{6}\), \(v(t_4)=\frac{15}{19}\), \(v(t_5)=\frac{3}{3 + 1}=\frac{3}{4}\)
\(R=(\frac{15}{16}+\frac{15}{17}+\frac{5}{6}+\frac{15}{19}+\frac{3}{4})\times0.2\)

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Step4: Calculate the average

The average of \(L\) and \(R\) is \(\frac{L + R}{2}\)
\(\frac{0.89+0.84}{2}=\frac{1.73}{2}=0.865\approx0.87\text{ meters}\)

Answer:

underestimate = \(0.89\) meters, overestimate = \(0.84\) meters, average = \(0.87\) meters