Calculus Quesiton

Question

Your velocity is given by v(t)=2t^2+2 in m/sec, with t in seconds. Estimate the distance, s, traveled between t=0 and t=8. Use the average of the left and right sums with 4 subdivisions. 

Answer: s=

Answer 1

Good point about the velocity being positive so that distance = displacement!

In case you need to show the two sums separately instead of the shortcut in the first answer:

Four subdivisions means 8/4 = 2 units in t per division. Make a table t = {0,2,4,6,8} and v = {2,10,34,74,130}. From the concept distance = rate*time, s(left) is the sum of velocities 1 - 4 times the interval of 2 seconds, s(left) = (2+10+34+74)*2 = 240 m; s(right) = (10+34+74+130)*2 = 496 m. The average of these is indeed 368 m. (The exact value from integral calculus is 357 1/3 m)

Answer 2

Two aspects that make this question significantly easier: The velocity function is always positive, so total distance traveled = displacement, which means we can integrate velocity since it is the same as the speed function. The second is that we don't have to find both the left- and right-hand sums and average them; the trapezoidal sum will = this average (as it always does). And the trapezoidal sum has a nice formula, worth remembering:

Sum_trap = (b - a) / (2n) · [f(a) + 2f(x₁) + ... + 2f(x_n-1) + f(b)]

= 8/8 · [2 + 20 + 68 + 148 + 130] = 368 m