For time, $t$, in hours, $0 \leq t \leq 1$, a bug is crawling at a velocity, $v$, in meters/hour given by \[ v=\frac{5}{5+t} \] Use $\Delta t=0.2$ 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 $=$

Step 1 ：Divide the interval [0,1] into five equal parts: [0,0.2], [0.2,0.4], [0.4,0.6], [0.6,0.8], [0.8,1].

Step 2 ：Evaluate the function at t=0, t=0.2, t=0.4, t=0.6, t=0.8 using the formula \(v(t) = \frac{5}{5+t}\).

Step 3 ：\(v(0) = \frac{5}{5+0} = 1\)

Step 4 ：\(v(0.2) = \frac{5}{5+0.2} = 0.833\)

Step 5 ：\(v(0.4) = \frac{5}{5+0.4} = 0.714\)

Step 6 ：\(v(0.6) = \frac{5}{5+0.6} = 0.625\)

Step 7 ：\(v(0.8) = \frac{5}{5+0.8} = 0.556\)

Step 8 ：Calculate the left-hand Riemann sum using the formula \(\Delta t \times (v(0) + v(0.2) + v(0.4) + v(0.6) + v(0.8))\).

Step 9 ：Substitute the values into the formula: \(0.2 \times (1 + 0.833 + 0.714 + 0.625 + 0.556) = 0.7456\) meters.

Step 10 ：\(\boxed{0.7456}\) meters is the underestimate for the distance the bug crawls during this hour.