AP Calculus AB: Riemann Sums and Definite Integral Notation (Drill 1)

Question 1. The table below shows selected values of a continuous function \( f \).

x	0	2	4	6
f(x)	1	5	3	9

Using a right Riemann sum with 3 subintervals of equal width, which of the following approximates \( \int_0^6 f(x)\,dx \)?

• A) \( 18 \)
• B) \( 34 \) ✓
• C) \( 26 \)
• D) \( 17 \)

Explanation: Choice B is correct. The three subintervals are \( [0,2] \), \( [2,4] \), and \( [4,6] \), each with width \( \Delta x = 2 \). A right Riemann sum uses the right endpoint of each subinterval: \( f(2)+f(4)+f(6) = 5+3+9 = 17 \). Multiplying by the width: \( 2 \times 17 = 34 \). Choice A is incorrect because \( 18 \) is the left Riemann sum, \( 2(f(0)+f(2)+f(4)) = 2(1+5+3) = 18 \), the student used left endpoints instead of right endpoints. Choice C is incorrect because \( 26 \) results from applying the trapezoidal rule, \( \frac{2}{2}[1+2(5)+2(3)+9] = 26 \), a different approximation method. Choice D is incorrect because \( 17 \) is the correct sum of the right-endpoint values but without multiplying by the subinterval width \( \Delta x = 2 \), a student who computes the sum correctly but forgets the width factor makes this error.

Question 2. Which of the following expressions is equivalent to \( \int_1^4 x^2\,dx \)?

• A) \( \displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \left(\dfrac{3k}{n}\right)^2 \cdot \dfrac{3}{n} \)
• B) \( \displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \left(1+\dfrac{3k}{n}\right)^2 \cdot \dfrac{3}{n} \) ✓
• C) \( \displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \left(1+\dfrac{3k}{n}\right)^2 \cdot \dfrac{1}{n} \)
• D) \( \displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \left(1+\dfrac{k}{n}\right)^2 \cdot \dfrac{1}{n} \)

Explanation: Choice B is correct. On \( [1,4] \) with \( n \) equal subintervals, \( \Delta x = \frac{3}{n} \) and the right endpoint of the \( k \)th subinterval is \( x_k = 1 + k \cdot \frac{3}{n} \). The right Riemann sum is therefore \( \sum_{k=1}^{n}\left(1+\frac{3k}{n}\right)^2 \cdot \frac{3}{n} \), and its limit as \( n \to \infty \) equals \( \int_1^4 x^2\,dx \). Choice A is incorrect because the sample points \( \frac{3k}{n} \) omit the left endpoint 1, placing the partition on \( [0,3] \) instead of \( [1,4] \), a student who forgets to add the starting value makes this error. Choice C is incorrect because while the sample points \( 1+\frac{3k}{n} \) are correct, \( \Delta x = \frac{1}{n} \) corresponds to a total width of 1, not 3, the student identified the correct sample points but used the wrong subinterval width. Choice D is incorrect because both the sample points \( 1+\frac{k}{n} \) and the width \( \frac{1}{n} \) correspond to a unit-length partition of \( [1,2] \), not \( [1,4] \).

Question 3. A continuous function \( g \) is positive and strictly decreasing on \( [2,8] \). Which of the following must be true about Riemann sum approximations of \( \int_2^8 g(x)\,dx \)?

• A) The left Riemann sum is an underestimate and the right Riemann sum is an overestimate.
• B) The left Riemann sum is an overestimate and the right Riemann sum is an underestimate. ✓
• C) Whether the left Riemann sum is an overestimate or underestimate depends on the concavity of \( g \).
• D) Both the left Riemann sum and the right Riemann sum are overestimates.

Explanation: Choice B is correct. When a function is strictly decreasing, the left endpoint of each subinterval yields the maximum value of \( g \) on that subinterval, so each left-sum rectangle is taller than the actual area, making the left Riemann sum an overestimate. Conversely, the right endpoint yields the minimum value on each subinterval, so the right Riemann sum underestimates the integral. Choice A is incorrect because it reverses the relationship; this would be true if \( g \) were increasing, not decreasing. A student who only memorizes the rule for increasing functions will select this choice. Choice C is incorrect because for monotone functions, over/underestimate is determined entirely by whether the function is increasing or decreasing; concavity affects the accuracy of the trapezoidal rule, not the direction of left/right Riemann sum error. Choice D is incorrect because both sums cannot simultaneously overestimate a positive function, for a decreasing function, the right sum is always smaller than the true integral.

Question 4. The table below shows selected values of a continuous function \( h \).

x	0	3	6	9
h(x)	2	8	4	6

Using the trapezoidal rule with 3 subintervals of equal width, which of the following approximates \( \int_0^9 h(x)\,dx \)?

• A) \( 42 \)
• B) \( 30 \)
• C) \( 54 \)
• D) \( 48 \) ✓

Explanation: Choice D is correct. The trapezoidal rule with \( \Delta x = 3 \) gives \( \frac{3}{2}[h(0)+2h(3)+2h(6)+h(9)] = \frac{3}{2}[2+16+8+6] = \frac{3}{2}(32) = 48 \). Choice A is incorrect because \( 42 \) is the left Riemann sum, \( 3[h(0)+h(3)+h(6)] = 3(14) = 42 \), the student selected the wrong approximation method. Choice B is incorrect because \( 30 \) results from omitting the factor of 2 on the interior values: \( \frac{3}{2}[2+8+4+6] = \frac{3}{2}(20) = 30 \), the trapezoidal rule requires doubling all values except the first and last endpoints. Choice C is incorrect because \( 54 \) is the right Riemann sum, \( 3[h(3)+h(6)+h(9)] = 3(18) = 54 \), again the wrong method.

Question 5. A particle moves along a straight line with velocity \( v(t) \) feet per second, where \( v(t) \) may be positive or negative. A student writes the expression \( \displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} v\left(\dfrac{5k}{n}\right) \cdot \dfrac{5}{n} \) to model a physical quantity over \( [0,5] \). Which of the following best describes what this expression represents?

• A) The average velocity of the particle over \( [0,5] \)
• B) The total distance traveled by the particle over \( [0,5] \)
• C) The displacement of the particle over \( [0,5] \) ✓
• D) The speed of the particle at \( t = 5 \)

Explanation: Choice C is correct. The expression \( \lim_{n \to \infty} \sum_{k=1}^{n} v\left(\frac{5k}{n}\right) \cdot \frac{5}{n} \) is a right Riemann sum for \( \int_0^5 v(t)\,dt \). By the net change interpretation of the definite integral, \( \int_0^5 v(t)\,dt \) gives the net displacement, the signed change in position, of the particle from \( t=0 \) to \( t=5 \). Because \( v(t) \) may be negative, the integral accumulates signed area, yielding displacement rather than total distance. Choice A is incorrect because average velocity requires dividing the integral by the length of the interval: \( \frac{1}{5}\int_0^5 v(t)\,dt \), the expression as written does not include this division. Choice B is incorrect because total distance requires integrating the absolute value of velocity, \( \int_0^5 |v(t)|\,dt \), using \( v(t) \) without absolute value gives displacement, which differs from total distance whenever the particle changes direction. This is one of the most frequently tested misconceptions in the integration unit. Choice D is incorrect because \( v(5) \) would give the instantaneous velocity at a single moment; a limit of a Riemann sum represents an accumulated quantity over an interval, not an instantaneous value.