Riemann Sum Approximations - Calculus 2

1. A student calculates a left Riemann sum for $\int_0^4 h(x) \, dx$ using $n = 8$ subintervals and obtains 32. If the same integral is approximated using a left Riemann sum with $n = 4$ subintervals, and $h(x)$ is linear on each subinterval $[0,1]$, $[1,2]$, $[2,3]$, and $[3,4]$, what can be concluded about the $n = 4$ approximation?
   Answer: The $n = 4$ approximation equals 32 since left sums are exact for piecewise linear functions when intervals align with linear pieces When using left Riemann sums with subintervals that align exactly with the linear pieces of a piecewise linear function, the approximation becomes exact. This is because on each linear piece $[k, k+1]$, the left endpoint value $h(k)$ multiplied by the interval length 1 gives exactly the area under that linear segment. Therefore, both the $n=4$ and $n=8$ approximations equal 32. Choice A incorrectly applies a scaling relationship. Choices C and D incorrectly assume the approximation has error when it's actually exact.
2. Consider the integral $\int_0^2 e^{-x^2} \, dx$. A student uses both the midpoint rule and trapezoidal rule with $n = 4$ subintervals to approximate this integral. Which statement about the relative accuracy of these approximations is most likely correct?
   Answer: The midpoint rule is more accurate because it generally has error that is half that of the trapezoidal rule for smooth functions For smooth functions, the midpoint rule typically has error approximately half that of the trapezoidal rule with the same number of subintervals. This is a general theoretical result: the midpoint rule error is $O(h^2)$ with a smaller constant than the trapezoidal rule. Choice A is wrong because both use the same number of evaluations. Choice B incorrectly analyzes concavity ($e^{-x^2}$ changes concavity). Choice C is wrong because both are second-order methods, but with different error constants.
3. A continuous function $w(t)$ represents the rate of water flow into a tank (in gallons per minute) over a 12-minute period. Using a left Riemann sum with $\Delta t = 3$ minutes, the total volume is approximated as $3[w(0) + w(3) + w(6) + w(9)]$. If this gives 84 gallons, and using $\Delta t = 2$ minutes gives $2[w(0) + w(2) + w(4) + w(6) + w(8) + w(10)] = 78$ gallons, what conclusion is most reasonable?
   Answer: The flow rate $w(t)$ is increasing over the 12-minute period since the finer approximation gives a smaller result For left Riemann sums, if the function is increasing, using smaller intervals (finer approximation) gives a larger sum, approaching the true value from below. If the function is decreasing, smaller intervals give smaller sums. Since the $\Delta t = 2$ approximation (78) is smaller than the $\Delta t = 3$ approximation (84), this suggests $w(t)$ is increasing. Choice A incorrectly assumes the average is exact. Choice B has the trend backwards. Choice D is unnecessarily cautious given the clear pattern.
4. A function $f(x)$ is continuous on $[2, 10]$. Using a right Riemann sum with $n = 4$ subintervals, the approximation yields 84. Using a left Riemann sum with $n = 4$ subintervals, the approximation yields 76. If $f(x)$ is strictly increasing on $[2, 10]$, what is the best estimate for the actual value of $\int_2^{10} f(x) \, dx$?
   Answer: The integral is between 76 and 84, with 80 being a reasonable estimate For a strictly increasing function, the left Riemann sum underestimates the integral (gives 76) and the right Riemann sum overestimates it (gives 84). Therefore, the true value lies between 76 and 84. The average of these bounds, 80, serves as a reasonable estimate. Choice A incorrectly assumes the average is exact. Choice C incorrectly claims both sums overestimate. Choice D incorrectly claims both sums underestimate.
5. Consider approximating $\int_0^3 \sqrt{9-x^2} \, dx$ using a midpoint Riemann sum with $n = 6$ subintervals. Which statement best describes the relationship between this approximation and the exact value?
   Answer: The midpoint sum overestimates because $\sqrt{9-x^2}$ is concave down on $[0,3]$, and midpoint sums overestimate concave down functions The function $f(x) = \sqrt{9-x^2}$ has $f''(x) = -\frac{9}{(9-x^2)^{3/2}} < 0$ on $(0,3)$, so it's concave down. For concave down functions, the midpoint sum overestimates the integral because the midpoint of each chord lies above the curve. Choice B has the correct concavity but wrong direction. Choices C and D incorrectly identify the concavity as upward.

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