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CED topics

• 6.1 Exploring Accumulations of Change
• 6.2 Approximating Areas with Riemann Sums
• 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation

Individual problems also preview:

• 6.7 The Fundamental Theorem of Calculus and Definite Integrals

The free-response question also touches:

• 2.1 Defining Average and Instantaneous Rates of Change at a Point
• 5.1 Using the Mean Value Theorem

Big idea and practices

Big idea: accumulation. Rate × time is an amount; summed, the total change; graphed, area. CHA-4 (integrals as accumulation), LIM-5 (Riemann sums approximate integrals); the continuous raise previews FUN-6.

Students leave able to:

• estimate from a table: left, right, midpoint, trapezoid, unequal widths
• call over or under from increasing/decreasing and concavity
• interpret a rate's integral in context, with units
• convert among sigma, Riemann sum, and integral notation

Practices: 1.E, 1.F (approximation versus actual); 2.B, 2.C (sigma to integral); 3.C, 3.D (Mean Value Theorem, FRQ (d)), 3.F (FRQ (b)); 4.B, 4.C (units, notation).

45-minute plan

1. 0–5 minLaunch: hooks aloud; pairs read "How long, how far?" together and pick in it (or circle on the printout). Hands up on the first.
2. 5–13 minChallenges 1–3 in pairs. After Challenge 2, draw the weekly staircase.
3. 13–21 minThe explorable and its four “Try these” prompts.
4. 21–28 minConcept and rule. Units first: ${\tfrac{\text{yd}}{\text{s}} \cdot \text{s} = \text{yd}}$.
5. 28–36 minWorked example, Challenges 4 and 5, the reveal.
6. 36–40 minFluency 1 and 2.
7. 40–45 minExit ticket, reflection.

Homework: Fluency 3–6, mixed review, FRQ (score it next class).

Launch questions

• $20 a week: is $1,500 more or less than a year away? Check without a calculator?
• A radar gun gives speed at every instant. How do you get distance?
• 10.8 yards per second held for 4.5 seconds: how far? Why does a real sprinter fall short?
• Yards per second times seconds? Dollars per week times weeks?

Misconceptions and how to respond

• “Area is just a picture.” Units dropped. Label axes with units; multiply units first (FRQ part (b)).
• Equal widths from an unequal table. 0.5 s throughout gives 23.8. Write the widths first.
• Forgetting to halve. Watch for 62.5 on Challenge 4 and 84 on Fluency 4. The trapezoid sum averages the left and right sums.
• Left and right mixed up. Ask “which time gives this bar's height?” and shade the bar.
• Over or under as a property of the sum. Tie it to the function: increasing/decreasing decides left and right; concavity, trapezoid and midpoint. The hatched (over) and dotted (under) regions show it.
• Whole weeks versus continuous time. On Challenge 2 the common miss is 47: after week 47 you have $1,480.50. On the follow-up, 47.178 weeks is partway through week 48.
• 125 weeks. $20 a week for 125 weeks is $2,500. Multiply back.

Support and extension

Support

• One bar first: at a constant $20 a week, savings are one rectangle's area; the raise makes a staircase.
• Explorable at ${n = 1}$, 2, and 4 before the formulas: what do the bars miss?
• Fill in the first table's widths for them; not the next.
• Pairs: one computes, one checks units and over or under.

Extension

• Prove the trapezoid sum averages the left and right sums.
• Why does a midpoint sum overestimate a concave-down function? (Its rectangle equals the trapezoid under the midpoint tangent, which lies above the curve.)
• How many subintervals put each sum within 0.1 yd of the exact area, 39.99116 yd? (Answer: about 240 for left and right, 11 for midpoint, and 16 for trapezoid. Why are the last two faster?)
• Challenge 2's pay is a left sum of ${r(t) = 20 + 0.5t}$, width 1 week. Show the continuous model leads by exactly 0.25n dollars after ${n}$ weeks (a triangle of area 0.25 a week).
• Graphing calculator: when does the sprint model reach 40 yards?

Without devices

• Print the lesson page as the handout: no buttons or hints, room to write. Students circle their predictions.
• Board: the sprint curve from the table, four bars 1.125 s wide. Model sums at ${n = 4}$: left 32.56, right 44.67, midpoint 40.66, and trapezoid 38.62 yd, against an area of 39.99 yd.
• Check work with the answer key below.
• Collect the exit ticket on paper.

Answer key

Every problem in the order the lesson gives them, with the answer the page accepts and my full write-up. Answer boxes check only the final number, so ask students for their work.

1. Challenge 1

   No calculator · CED 6.1

   You save $20 a week toward a $1,500 necklace. After how many weeks can you buy it?

   Answer: 75

   1. After ${n}$ weeks: ${20n}$ dollars.
   2. Solve ${20n = 1{,}500}$: ${n = 1{,}500 \div 20 = 75}$.
   3. 75 weeks: exactly $1,500, about 1.44 years (52 weeks/year).

   Common wrong answers

   • 125: 125 weeks at $20 a week is $2,500. Divide $1,500 by $20.
   • 1.44: That's years; the question asks for weeks.

2. Challenge 2

   No calculator · CED 6.1

   Now a raise every week. You get $20 in week 1, $20.50 in week 2, $21 in week 3, and so on. After which week do you first have at least $1,500?

   Answer: 48

   1. Week ${k}$ pays ${20 + 0.5(k - 1)}$. Total after ${n}$ weeks: $\displaystyle S(n) = 20n + 0.5 \cdot \frac{n(n - 1)}{2}$$S(n)$$=20n+0.5\cdot \frac{n(n-1)}{2}$$\displaystyle = 20n + 0.25n(n - 1)$$=20n+0.25n(n-1)$
   2. After 40 weeks: ${800 + 390 = 1{,}190}$; after 50: ${1{,}000 + 612.5 = 1{,}612.5}$. The answer is in between.
   3. After week 47: $\displaystyle S(47) = 20(47) + 0.25(47)(46)$$S(47)$$=20(47)+0.25(47)(46)$$\displaystyle = 940 + 540.5$$=940+540.5$$\displaystyle = 1{,}480.5$$=1,480.5$ $1,480.50: not enough.
   4. After week 48: $\displaystyle S(48) = 20(48) + 0.25(48)(47)$$S(48)$$=20(48)+0.25(48)(47)$$\displaystyle = 960 + 564$$=960+564$$\displaystyle = 1{,}524$$=1,524$ $1,524, so week 48 is the first with at least $1,500.

   Common wrong answers

   • 47: After week 47 you have $1,480.50: still $19.50 short.
   • 75: 75 weeks is without raises; raises get you there sooner.

3. Challenge 3

   Calculator · CED 6.1, 6.7

   Follow-up: a continuous raise, ${r(t) = 20 + 0.5t}$ dollars per week at ${t}$ weeks. The area under ${r}$ from ${0}$ to ${T}$, a trapezoid, is ${20T + 0.25T^2}$ dollars. Solve ${20T + 0.25T^2 = 1{,}500}$ for ${T}$.

   Answer: 47.178 (or 47.177)

   1. Multiply both sides by 4: ${T^2 + 80T - 6{,}000 = 0}$.
   2. Quadratic formula. Under the root: ${80^2 + 4 \times 6{,}000 = 30{,}400}$, which is ${1{,}600 \times 19}$, and ${\sqrt{1{,}600}}$ is 40: $\displaystyle T = \frac{-80 \pm \sqrt{30{,}400}}{2}$$T$$=\frac{-80\pm \sqrt{30,400}}{2}$$\displaystyle = -40 \pm 20\sqrt{19}$$=-40\pm 20\sqrt{19}$simplify the root
   3. Time can't be negative: $\displaystyle T = -40 + 20\sqrt{19} \approx \boxed{47.178}$$T$$=-40+20\sqrt{19}\approx \boxed{47.178}$ 47.178 weeks, partway through week 48, as in the whole-week answer.

   Common wrong answers

   • 48: 48 is the whole-week answer. Continuously, you pass $1,500 partway through a week: solve the equation.
   • -127.178: Negative time: take the other root.

4. Worked example

   No calculator · CED 6.2

   A cyclist's speed ${v(t)}$ is in the table. Estimate the distance from ${t = 0}$ to ${t = 2}$: left Riemann sum, the table's three subintervals. $$\begin{array}{c|cccc} t \text{ (h)} & 0 & 0.5 & 1 & 2 \\ \hline v(t) \text{ (mi/h)} & 12 & 16 & 15 & 18 \end{array}$$

   Worked on the page: 29

   1. Subintervals ${[0, 0.5]}$, ${[0.5, 1]}$, ${[1, 2]}$: widths 0.5 h, 0.5 h, 1 h. Unequal is fine.
   2. Left sum: starting speeds 12, 16, 15 mi/h.
   3. Each bar: width (time) times height (speed) is a distance. Add the bars: $\displaystyle \int_0^2 v(t)\,dt \approx \underbrace{\orange{0.5}\rule[-0.25em]{0em}{1em}}_{\text{width}} \underbrace{\orange{(12)}}_{\text{height}} + \orange{0.5(16)} + \orange{1(15)}$${\int }_0^{2}v(t)dt$$\approx \underset{\text{width}}{\underbrace{0.5}}\underset{\text{height}}{\underbrace{(12)}}+0.5(16)+1(15)$$\displaystyle = 6 + 8 + 15$$=6+8+15$
   4. Left sum: about 29 miles.

5. Challenge 4

   No calculator · CED 6.2

   Same ride, trapezoid sum. First subinterval done: ${\frac{0.5}{2}(12 + 16) = 7}$ miles. Finish the other two and add. $$\begin{array}{c|cccc} t \text{ (h)} & 0 & 0.5 & 1 & 2 \\ \hline v(t) \text{ (mi/h)} & 12 & 16 & 15 & 18 \end{array}$$

   Answer: 31.25

   1. ${[0, 0.5]}$: ${\frac{0.5}{2}(12 + 16) = 7}$ (given).
   2. ${[0.5, 1]}$: ${\frac{0.5}{2}(16 + 15) = 7.75}$.
   3. ${[1, 2]}$: ${\frac{1}{2}(15 + 18) = 16.5}$.
   4. Total, 31.25 miles: ${7 + 7.75 + 16.5 = \boxed{31.25}}$. Check: left sum 29, right sum ${0.5(16) + 0.5(15) + 1(18)}$, which is 33.5; their average is 31.25.

   Common wrong answers

   • 29: That's the left sum. Use both ends of each subinterval.
   • 33.5: That's the right sum. Use both ends of each subinterval.
   • 62.5: Average each pair: divide by 2.

6. Challenge 5

   No calculator · CED 6.1, 6.2

   A sprinter's velocity ${v(t)}$ is differentiable, increasing, and concave down, with values in the table. Approximate the distance run in the first 4.5 seconds with a trapezoid sum on the table's six subintervals. Enter the estimate in yards. Then decide, with a reason, whether it is an over- or underestimate; the write-up answers that part. $$\begin{array}{c|ccccccc} t \text{ (s)} & 0 & 0.5 & 1 & 1.5 & 2 & 3 & 4.5 \\ \hline v(t) \text{ (yd/s)} & 0 & 5.0 & 7.7 & 9.1 & 9.9 & 10.5 & 10.8 \end{array}$$

   Answer: 39.55

   1. Each trapezoid is ${\frac{\Delta t}{2}(v_{\text{start}} + v_{\text{end}})}$. Widths: 0.5 s (first four), 1 s, 1.5 s.
   2. First four, 0.5 s each: $\displaystyle \int_0^2 v(t)\,dt \approx \frac{0.5}{2}(0 + 5.0) + \frac{0.5}{2}(5.0 + 7.7) + \frac{0.5}{2}(7.7 + 9.1) + \frac{0.5}{2}(9.1 + 9.9)$${\int }_0^{2}v(t)dt$$\approx \frac{0.5}{2}(0+5.0)+\frac{0.5}{2}(5.0+7.7)+\frac{0.5}{2}(7.7+9.1)+\frac{0.5}{2}(9.1+9.9)$$\displaystyle = \frac{0.5}{2}(53.5)$$=\frac{0.5}{2}(53.5)$factor out ${\tfrac{0.5}{2}}$ and add$\displaystyle = 13.375$$=13.375$
   3. Next, 1 s: $\displaystyle \int_2^3 v(t)\,dt \approx \frac{1}{2}(9.9 + 10.5)$${\int }_2^{3}v(t)dt$$\approx \frac{1}{2}(9.9+10.5)$$\displaystyle = 0.5(20.4)$$=0.5(20.4)$$\displaystyle = 10.2$$=10.2$
   4. Last, 1.5 s: $\displaystyle \int_3^{4.5} v(t)\,dt \approx \frac{1.5}{2}(10.5 + 10.8)$${\int }_3^{4.5}v(t)dt$$\approx \frac{1.5}{2}(10.5+10.8)$$\displaystyle = 0.75(21.3)$$=0.75(21.3)$$\displaystyle = 15.975$$=15.975$
   5. Add: $\displaystyle \int_0^{4.5} v(t)\,dt \approx 13.375 + 10.2 + 15.975$${\int }_0^{4.5}v(t)dt$$\approx 13.375+10.2+15.975$$\displaystyle = \boxed{39.55}$$=\boxed{39.55}$ Since ${v}$ is concave down, each slanted top lies below the curve, so 39.55 yd is an underestimate.

   Common wrong answers

   • 36.55: That's the left sum. Average both ends of each subinterval.
   • 42.55: That's the right sum. Average both ends of each subinterval.
   • 23.8: Not all widths are 0.5 s: t = 2 to 3 is 1 s, and t = 3 to 4.5 is 1.5 s.

7. Fluency 1

   No calculator · CED 6.2

   Approximate the area under ${f(x) = x^2}$ from ${x = 0}$ to ${x = 2}$: left Riemann sum, 4 equal subintervals.

   Answer: 1.75

   1. ${\Delta x = \frac{2 - 0}{4} = 0.5}$; left endpoints ${0}$, ${0.5}$, ${1}$, ${1.5}$.
   2. ${f}$ there: ${0}$, ${0.25}$, ${1}$, ${2.25}$.
   3. Heights: ${0 + 0.25 + 1 + 2.25 = 3.5}$, so ${L_4 = 0.5(3.5) = \boxed{1.75}}$ square units.

   Common wrong answers

   • 3.75: That's the right sum. Left uses x = 0, 0.5, 1, 1.5.
   • 3.5: Multiply by the width, 0.5.

8. Fluency 2

   No calculator · CED 6.2, 6.3

   Approximate ${\int_1^3 (3x + 1)\,dx}$: right Riemann sum, 4 equal subintervals.

   Answer: 15.5

   1. ${\Delta x = \frac{3 - 1}{4} = 0.5}$; right endpoints ${1.5}$, ${2}$, ${2.5}$, ${3}$.
   2. ${3x + 1}$ there: ${5.5}$, ${7}$, ${8.5}$, ${10}$.
   3. Heights: ${5.5 + 7 + 8.5 + 10 = 31}$, so ${R_4 = 0.5(31) = \boxed{15.5}}$ square units.

   Common wrong answers

   • 12.5: That's the left sum. Right uses x = 1.5, 2, 2.5, 3.
   • 14: 14 is the exact integral; the question asks for the right sum.

9. Fluency 3

   No calculator · CED 6.2

   Approximate the area under ${g(x) = x^2 + 1}$ from ${x = 0}$ to ${x = 4}$: midpoint sum, 2 equal subintervals.

   Answer: 24

   1. ${\Delta x = 2}$; subintervals ${[0, 2]}$ and ${[2, 4]}$, midpoints ${1}$ and ${3}$.
   2. ${g(1) = 2}$ and ${g(3) = 10}$.
   3. Midpoint sum: ${M_2 = 2(2 + 10) = \boxed{24}}$ square units.

   Common wrong answers

   • 12: Those are the left endpoints. Use the midpoints, 1 and 3.
   • 44: Those are the right endpoints. Use the midpoints, 1 and 3.

10. Fluency 4

    No calculator · CED 6.1, 6.2

    Water flows into a tank at ${R(t)}$ liters per minute. Estimate the inflow from ${t = 0}$ to ${t = 6}$: trapezoid sum, the table's three subintervals. $$\begin{array}{c|cccc} t \text{ (min)} & 0 & 2 & 5 & 6 \\ \hline R(t) \text{ (L/min)} & 4 & 7 & 9 & 5 \end{array}$$

    Answer: 42

    1. Subintervals ${[0, 2]}$, ${[2, 5]}$, ${[5, 6]}$: widths 2, 3, 1 minutes.
    2. Add the three trapezoids: $\displaystyle \int_0^6 R(t)\,dt \approx \frac{2}{2}(4 + 7) + \frac{3}{2}(7 + 9) + \frac{1}{2}(9 + 5)$${\int }_0^{6}R(t)dt$$\approx \frac{2}{2}(4+7)+\frac{3}{2}(7+9)+\frac{1}{2}(9+5)$$\displaystyle = 11 + 24 + 7$$=11+24+7$
    3. About 42 liters (liters per minute times minutes).

    Common wrong answers

    • 84: Average each pair: divide by 2.
    • 41: The widths are 2, 3, and 1 minutes, not all 2.

11. Fluency 5

    No calculator · CED 6.3

    Evaluate the Riemann sum ${\displaystyle\sum_{k=1}^{4} \left(1 + \frac{k}{2}\right)^2 \cdot \frac{1}{2}}$.

    Answer: 10.75

    1. For ${k = 1, 2, 3, 4}$: ${1 + \frac{k}{2}}$ is ${1.5}$, ${2}$, ${2.5}$, ${3}$.
    2. Squares: ${2.25}$, ${4}$, ${6.25}$, ${9}$; sum ${21.5}$.
    3. Times ${\frac{1}{2}}$: 10.75, a right Riemann sum for ${\int_1^3 x^2\,dx}$ with 4 subintervals.

    Common wrong answers

    • 21.5: Keep the factor 1/2: the width.
    • 6.75: k runs from 1 to 4, so the first term uses 1 + 1/2 = 1.5, not 1.

12. Fluency 6

    No calculator · CED 6.1, 6.2

    Fans enter a stadium at ${F(t)}$ fans per minute, ${t}$ minutes after the gates open. Estimate how many fans enter in the first 40 minutes: midpoint sum, 2 equal subintervals. $$\begin{array}{c|ccccc} t \text{ (min)} & 0 & 10 & 20 & 30 & 40 \\ \hline F(t) \text{ (fans/min)} & 20 & 60 & 90 & 70 & 30 \end{array}$$

    Answer: 2600

    1. Subintervals ${[0, 20]}$ and ${[20, 40]}$, 20 minutes each.
    2. Midpoints ${t = 10}$ and ${t = 30}$: ${F = 60}$ and ${F = 70}$ fans per minute.
    3. Midpoint sum: ${20(60 + 70) = \boxed{2{,}600}}$, so about 2,600 fans.

    Common wrong answers

    • 1300: Each subinterval is 20 minutes wide, not 10.

13. Review 1

    From Is the Ball Ever Stopped?, Fluency 2 · No calculator · CED 2.1, 2.3

    A runner's distance from the start, ${D(t)}$, in meters, is measured every 10 seconds. Use the values at ${t = 10}$ and ${t = 30}$ to estimate ${D'(20)}$, the velocity at ${t = 20}$. $$\begin{array}{c|cccc} t \text{ (s)} & 0 & 10 & 20 & 30 \\ \hline D(t) \text{ (m)} & 0 & 62 & 128 & 188 \end{array}$$

    Answer: 6.3

    1. ${[10, 30]}$ is centered at ${t = 20}$, so its average rate of change estimates ${D'(20)}$.
    2. From the table: $\displaystyle \frac{D(30) - D(10)}{30 - 10} = \frac{188 - 62}{20}$$\frac{D(30)-D(10)}{30-10}$$=\frac{188-62}{20}$$\displaystyle = \frac{126}{20}$$=\frac{126}{20}$
    3. ${\dfrac{126}{20} = \boxed{6.3}}$, so ${D'(20)}$ is about 6.3 meters per second.

    Common wrong answers

    • 6.4: That's the average from t = 0 to t = 20. Use t = 10 and t = 30.
    • 6.6: That uses t = 10 and t = 20. Use t = 10 and t = 30.
    • 6: That uses t = 20 and t = 30. Use t = 10 and t = 30.
    • 126: Divide by the time between the readings, 30 − 10 = 20 seconds.

14. Review 2

    From Best Price, Fluency 5 · No calculator · CED 5.5

    Find the absolute maximum value of ${f(x) = 12x - x^3}$ on ${[0, 3]}$.

    Answer: 16

    1. ${f'(x) = 12 - 3x^2}$, which is 0 at ${x = \pm 2}$. Only ${x = 2}$ is in ${[0, 3]}$.
    2. Candidates: ${f(0) = 0}$, ${f(2) = 24 - 8 = 16}$, and ${f(3) = 36 - 27 = 9}$.
    3. The absolute maximum value is 16, at ${x = 2}$.

    Common wrong answers

    • 2: The maximum is at x = 2. The question asks for f(2).
    • 9: That's f(3), an endpoint. Check the critical point too.

15. Exit ticket

    No calculator · CED 6.1, 6.2

    A summer job pays ${r(t)}$ dollars per week, ${t}$ weeks into the summer. Estimate earnings in the first 6 weeks: left Riemann sum, the table's three subintervals. $$\begin{array}{c|cccc} t \text{ (weeks)} & 0 & 2 & 5 & 6 \\ \hline r(t) \text{ (dollars/week)} & 40 & 50 & 60 & 45 \end{array}$$

    Answer: $290.00

    1. Widths 2, 3, 1 weeks; starting rates 40, 50, 60 dollars per week.
    2. Add width times rate: $\displaystyle \int_0^6 r(t)\,dt \approx 2(40) + 3(50) + 1(60)$${\int }_0^{6}r(t)dt$$\approx 2(40)+3(50)+1(60)$$\displaystyle = 80 + 150 + 60$$=80+150+60$
    3. About $290 (dollars per week times weeks).

    Common wrong answers

    • $325.00: That's the right sum. Use the rate at the start of each subinterval.
    • $307.50: That's the trapezoid sum. Use only the starting rates.

Free response: scoring guideline

1. Part (a) · CED 6.2, 6.3

   Use a trapezoidal sum with the six subintervals indicated by the table to approximate ${\int_0^{4.5} v(t)\,dt}$. Show the computations that lead to your answer, and indicate units of measure.

   Answer: 39.55

   Scoring

   2 points. 1 point: a trapezoidal sum over all six subintervals, unequal widths: $\displaystyle \int_0^{4.5} v(t)\,dt \approx \frac{0.5}{2}(0 + 5.0) + \frac{0.5}{2}(5.0 + 7.7) + \frac{0.5}{2}(7.7 + 9.1) + \frac{0.5}{2}(9.1 + 9.9) + \frac{1}{2}(9.9 + 10.5) + \frac{1.5}{2}(10.5 + 10.8)$${\int }_0^{4.5}v(t)dt$$\approx \frac{0.5}{2}(0+5.0)+\frac{0.5}{2}(5.0+7.7)+\frac{0.5}{2}(7.7+9.1)+\frac{0.5}{2}(9.1+9.9)+\frac{1}{2}(9.9+10.5)+\frac{1.5}{2}(10.5+10.8)$ 1 point: 39.55 yards, with units.

2. Part (b) · CED 6.1, 6.3

   Using correct units, explain the meaning of ${\int_0^{4.5} v(t)\,dt}$ in the context of the problem.

   Scoring

   1 point: ${\int_0^{4.5} v(t)\,dt}$ is the distance, in yards, the sprinter runs from ${t = 0}$ to ${t = 4.5}$ seconds. Needs the interval and units. Accept "displacement" or "change in position": since ${v(t) > 0}$ is given, displacement equals distance.

3. Part (c) · CED 2.1

   Using the data in the table, find the average acceleration of the sprinter over the interval ${2 \le t \le 3}$. Show the computations that lead to your answer, and indicate units of measure.

   Answer: 0.6

   Scoring

   1 point: $\displaystyle \frac{v(3) - v(2)}{3 - 2} = \frac{10.5 - 9.9}{1}$$\frac{v(3)-v(2)}{3-2}$$=\frac{10.5-9.9}{1}$$\displaystyle = 0.6$$=0.6$ 0.6 yards per second per second.

4. Part (d) · CED 5.1

   Must there be a time ${c}$, with ${2 < c < 3}$, at which the sprinter's acceleration ${v'(c)}$ is ${0.6}$ yards per second per second? Justify your answer.

   Scoring

   2 points. 1 point: ${v}$ is differentiable, so continuous on ${[2, 3]}$ and differentiable on ${(2, 3)}$. 1 point: yes, by the Mean Value Theorem some ${c}$ with ${2 < c < 3}$ has ${v'(c) = \dfrac{v(3) - v(2)}{3 - 2} = 0.6}$, part (c)'s average acceleration.