Evaluating a Definite Integral in a Real-World Context
A car's velocity (v(t)) in metres per second over a 10-second interval is measured at discrete times: (t=0,2,4,6,8,10) with velocities 0, 5, 12, 20, 25, 30 respectively. The car's acceleration is not constant. A student uses the trapezoidal rule with these data points to estimate the distance travelled. Another student uses Simpson's rule with the same points. The teacher asks you to compare the two estimates and discuss which is more reliable given the shape of the velocity-time graph, which appears to be concave up for the first half and concave down for the second half.
1. Calculate the distance travelled using the trapezoidal rule with the given data. Show all steps.
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The trapezoidal rule with (n=5) subintervals of width (h=2) is: [\text{Distance} \approx \frac{h}{2}\left[v(0)+2v(2)+2v(4)+2v(6)+2v(8)+v(10)\right]] [= \frac{2}{2}\left[0 + 2(5) + 2(12) + 2(20) + 2(25) + 30\right]] [= 1 \times [0 + 10 + 24 + 40 + 50 + 30] = 154 \text{ m}.] So the trapezoidal estimate is 154 m.
2. Calculate the distance travelled using Simpson's rule with the given data. Compare the two estimates and discuss which is likely more accurate, considering the concavity of the velocity-time graph.
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Simpson's rule with (n=5) subintervals (odd number) cannot be applied directly because Simpson's rule requires an even number of subintervals. With 6 data points, there are 5 subintervals, which is odd. To apply Simpson's rule, we need an even number of subintervals. We can use Simpson's rule on the first 4 subintervals (0 to 8) and the trapezoidal rule on the last subinterval (8 to 10), or use Simpson's 3/8 rule? But standard Simpson's rule requires even number. Alternatively, we could use Simpson's rule on the first 4 subintervals (0 to 8) and then trapezoidal for the last, but the question says 'using Simpson's rule with the same points'-this is a common trap: Simpson's rule cannot be applied directly with 5 subintervals. So we must note that. However, some might use Simpson's rule on the first 4 and trapezoidal on the last, or use a composite method. Let's assume we use Simpson's rule on the first 4 subintervals (0 to 8) and trapezoidal on the last (8 to 10). Then: Simpson's on [0,8] with (h=2): (\frac{h}{3}[v(0)+4v(2)+2v(4)+4v(6)+v(8)] = \frac{2}{3}[0+4(5)+2(12)+4(20)+25] = \frac{2}{3}[0+20+24+80+25] = \frac{2}{3}(149) = 99.333...) m. Trapezoidal on [8,10]: (\frac{2}{2}[v(8)+v(10)] = 1(25+30)=55) m. Total = 154.333... m. But this is not pure Simpson's. Alternatively, we could use Simpson's rule on all 6 points by treating it as 3 subintervals? No, Simpson's rule requires pairs. So the correct response is to point out that Simpson's rule cannot be applied with 5 subintervals; it requires an even number. Therefore, the student's claim to use Simpson's rule with the same points is invalid. The trapezoidal estimate is 154 m. If we approximate using Simpson's on the first 4 and trapezoidal on last, we get about 154.33 m, very close. Given the concavity changes (concave up then concave down), the trapezoidal rule may overestimate or underestimate depending on the interval. For concave up, trapezoidal overestimates; for concave down, it underestimates. Since the graph is concave up first half and concave down second half, errors may partially cancel. Simpson's rule is exact for cubics and often more accurate, but here it cannot be applied directly. So the trapezoidal estimate is reasonable, but we should note the limitation.