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Left Riemann sum estimate of distance traveled

A car travels along a straight road. Its velocity v(t)v(t), in meters per second, is measured every 2 seconds and recorded in the table below.

tt (s)0246810
v(t)v(t) (m/s)5812151718

Using the data in the table, estimate the total distance traveled by the car over the interval t=0t = 0 to t=10t = 10 seconds by computing the left-endpoint Riemann sum with n=5n = 5 subintervals. What is your estimate for the distance traveled, in meters?

Hint

Each subinterval has width Δt=2\Delta t = 2. The left sum uses the velocity value at the start of each subinterval: L5=Δt[v(0)+v(2)+v(4)+v(6)+v(8)]L_5 = \Delta t \left[v(0) + v(2) + v(4) + v(6) + v(8)\right]. Note the value at t=10t=10 is not used.

Step-by-step solution

The interval [0,10][0,10] is divided into n=5n=5 subintervals of width Δt=(10−0)/5=2\Delta t = (10-0)/5 = 2 seconds. For a left-endpoint sum, we use the velocity at the left endpoint of each subinterval, meaning we use v(0),v(2),v(4),v(6),v(8)v(0), v(2), v(4), v(6), v(8) and omit v(10)v(10).

Summing these left-endpoint velocities gives v(0)+v(2)+v(4)+v(6)+v(8)=5+8+12+15+17=57v(0)+v(2)+v(4)+v(6)+v(8) = 5+8+12+15+17 = 57. Multiplying by Δt=2\Delta t = 2 gives the estimated distance L5=2(57)=114 meters.L_5 = 2(57) = 114 \text{ meters}.

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