Practice Problems

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Course: MA005: Calculus I
Book: Practice Problems
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Date: Friday, July 19, 2024, 3:24 AM

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Work through the odd-numbered problems 1-61. Once you have completed the problem set, check your answers.

Table of contents

Practice Problems

In problems 1 – 5 , rewrite the sigma notation as a summation and perform the indicated addition.

1. \sum_{\mathrm{k}=2}^{4} \mathrm{k}^{2}

3. \sum_{n=1}^{3}(1+n)^{2}

5. \sum_{j=0}^{5} \cos (\pi \mathrm{j})


In problems 7 – 11, rewrite each summation using the sigma notation. Do not evaluate the sums.

7. 3+4+5+\ldots+93+94

9. 9+16+25+36+\ldots+144

11. 1 \cdot 2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+\ldots+7 \cdot 2^{7}


In problems 13 – 15 , use the values of \mathrm{a}_{\mathrm{K}} and \mathrm{b}_{\mathrm{K}} in Table 3 and verify that the value in part (a) does equal the value in part (b).

\mathrm{k} \mathrm{a}_{\mathrm{K}} \mathrm{b}_{\mathrm{K}}
1 1 2
2 2 2
3 3 2

                                       Table 3

13. (a)  \sum_{k=1}^{3}\left(a_{k}+b_{k}\right)          (b) \sum_{\mathrm{k}=1}^{3} \mathrm{a}_{\mathrm{k}}+\sum_{\mathrm{k}=1}^{3} \mathrm{~b}_{\mathrm{k}}

15. (a)  \sum_{k=1}^{3} 5 \mathrm{a}_{\mathrm{k}}                   (b)  5 \cdot \sum_{k=1}^{3} \mathrm{a}_{\mathrm{k}}


For problems 15 – 17 , use the values of \mathrm{a}_{\mathrm{K}} and \mathrm{b}_{\mathrm{K}} in Table 3 and verify that the value in part (a) does not equal the value in part (b).

17.  (a)  \sum_{\mathrm{k}=1}^{3}\left(\mathrm{a}_{\mathrm{k}}^{2}\right)            (b)  \left(\sum_{\mathrm{k}=1}^{3} \mathrm{a}_{\mathrm{k}}\right)^{2}


For problems 19 – 29, f(x)=x^{2}, g(x)=3 x, and \mathrm{h}(\mathrm{x})=2 / \mathrm{x}. Evaluate each sum.

19. \sum_{\mathrm{k}=0}^{3} \mathrm{f}(\mathrm{k})

21. \sum_{j=0}^{3} 2 \mathrm{f}(\mathrm{j})

23. \sum_{\mathrm{m}=1}^{3} \mathrm{~g}(\mathrm{~m})

25. \sum_{j=1}^{3} g^{2}(j)

27. \sum_{\mathrm{k}=2}^{4} \mathrm{~h}(\mathrm{k}))

29. \sum_{n=1}^{3} f(n) \cdot h(n)


For problems 31 – 35 , write out each summation and simplify the result. These are examples of "telescoping sums"

31. \sum_{k=1}^{7}\left((k)^{2}-(k-1)^{2}\right)

33. \sum_{\mathrm{k}=1}^{5}\left(\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+1}\right)

35. \sum_{k=0}^{8}(\sqrt{k+1}-\sqrt{k})


For problems 37– 43 , (i) list the subintervals determined by the partition P, (ii) find the values of \Delta \mathrm{x}_{\mathrm{i}}, (iii) find the mesh of P, and (iv) calculate \sum_{\mathrm{i}=1}^{\mathrm{n}} \Delta \mathrm{x}_{\mathrm{i}}.

37. \mathrm{P}=\{2,3,4.5,6,7\}

39. \mathrm{P}=\{-3,-1,0,1.5,2\}


41. P is given in Fig. 19.


43. For \Delta \mathrm{x}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}-\mathrm{x}_{\mathrm{i}-1}, verify that \sum_{\mathrm{i}=1}^{\mathrm{n}} \Delta \mathrm{x}_{\mathrm{i}} = length of the interval [a, b].


For problems 45 – 47 , (i) sketch the graph of f on the given interval, (ii) draw vertical lines at each point of the given partition, (iii) evaluate each f(ci) and sketch the corresponding rectangle, and (iv) calculate and add together the areas of the rectangles.

45. f(x)=4-x^{2}, P=\{0,1,1.5,2\}           (a)  \mathrm{c}_{1}=0, \mathrm{c}_{2}=1, and  \mathrm{c}_{3}=2 .             (b)  c_{1}=1, c_{2}=1.5  and \mathrm{c}_{3}=1.5.

47. \mathrm{f}(\mathrm{x})=\sin (\mathrm{x}), \mathrm{P}=\{0, \pi / 4, \pi / 2, \pi\}        (a)  c_{1}=0, c_{2}=\pi / 4, and c_{3}=\pi / 2.   (b)  \mathrm{c}_{1}=\pi / 4, \mathrm{c}_{2}=\pi
        / 2, and  \mathrm{c}_{3}=\pi.


For problems 49 – 51 , sketch the function and find the smallest possible value and the largest possible value for a Riemann sum of the given function and partition.

49. f(x)=1+x^{2}               (a)  \mathrm{P}=\{1,2,4,5\}           (b)  \mathrm{P}=\{1,2,3,4,5\}           (c)  \mathrm{P}=\{1,1.5,2,3,4,5\}

51. f(x)=\sin (x)                (a)  \mathrm{P}=\{0, \pi / 2, \pi\}          (b)  \mathrm{P}=\{0, \pi / 4, \pi / 2, \pi\}      (c)  \mathrm{P}=\{0, \pi / 4,3 \pi / 4, \pi\}


Upper and Lower Sum Problems

53. Suppose \mathrm{LS}_{\mathrm{P}}=7.362 and  USP = 7.402 for a positive function f and a partition P of the interval [ 1, 5]. We can be certain that every Riemann sum for the partition P is within what distance of the exact value of the area between the graph of f and the interval [ 1,5] ? (b) What if \mathrm{LS}_{\mathrm{P}}=7.372 and \mathrm{US}_{\mathrm{P}}=7.390?

55. If we divide the interval [ 2, 4] into 50 equally wide subintervals and calculate a Riemann sum for f(x)=1+x^{3} by "randomly" selecting a point \mathrm{c}_{\mathrm{i}} in each subinterval (any point \mathrm{c}_{\mathrm{i}} in the subinterval), then we can be certain that the Riemann sum is within what distance of the exact value of the area between f and the interval [ 2, 4] ?


Summing Powers of Consecutive Integers

Explicit formulas for some commonly encountered summations are known and are useful for explicitly evaluating some Riemann sums and their limits. The formulas below are included here for your reference. They will not be used or needed in the following sections. The summation formula for the first n positive integers is relatively well–known, has several easy but clever proofs, and even has an interesting story.

1+2+3+\ldots+(n-1)+\mathrm{n}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}


Proof: Let S represent the sum 1+2+3+\ldots+(n-2)+(n-1)+n. Rearranging the summands, we also know \mathrm{S}=\mathrm{n}+(\mathrm{n}-1)+(\mathrm{n}-2)+\ldots+3+2+1. Adding these 2 representations of S together,

\begin{array}{rllllll} \mathrm{S} & =1 & +2 & +3 & +\ldots & +(\mathrm{n}-2) & +(\mathrm{n}-1) & +\mathrm{n} \\ +\mathrm{S} & =\mathrm{n} & +(\mathrm{n}-1) & +(\mathrm{n}-2) & +\ldots & +3 & +2 &
    +1 \\ \hline \end{array}
2 \mathrm{~S}=(\mathrm{n}+1)+(\mathrm{n}+1)+(\mathrm{n}+1) \quad+\ldots+(\mathrm{n}+1)+(\mathrm{n}+1)+(\mathrm{n}+1)=\mathrm{n} \cdot(\mathrm{n}+1)

so S=\frac{n(n+1)}{2}, the desired formula.


It is said that Gauss discovered this formula for himself at the age of 5. His teacher, planning on keeping the class busy for a while, asked the students to add the integers from 1 to 100. Gauss thought a few minutes, wrote his answer on his slate, and turned it in. According to the story, he then sat smugly while his classmates struggled with the problem.

57. Find the sum of the first 100 positive integers in 2 ways: (1) using Gauss' formula, and (2) using Gauss' method.

59. Find the sum of the integers from 10 to 20.


Formulas for other integer powers of the first n integers have been discovered:

\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k} \quad=\frac{1}{2} \mathrm{n}^{2}+\frac{1}{2} \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)}{2} \quad \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{2}=\frac{1}{3} \mathrm{n}^{3}+\frac{1}{2} \mathrm{n}^{2}+\frac{2}{12}
    \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+)}{6}

\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{3}=\frac{1}{4} \mathrm{n}^{4}+\frac{1}{2} \mathrm{n}^{3}+\frac{3}{12} \mathrm{n}^{2}+0 \cdot \mathrm{n}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}

\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{4}=\frac{1}{5} \mathrm{n}^{5}+\frac{1}{2} \mathrm{n}^{4}+\frac{4}{12} \mathrm{n}^{3}+0 \cdot \mathrm{n}^{2}-\frac{1}{30} \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\left(3 \mathrm{n}^{2}+3 \mathrm{n}-1\right)}{30}


In problem 61, use the properties of summation and the formulas for powers to evaluate each sum

61. \sum_{\mathrm{k}=1}^{10} \mathrm{k} \cdot\left(\mathrm{k}^{2}+1\right)


Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-5.2-Sigma-Notation-and-Riemann-Sums.pdf
Creative Commons License This work is licensed under a Creative Commons Attribution 3.0 License.

Answers

1. 2^{2}+3^{2}+4^{2}=29

3. (1+1)^{2}+(1+2)^{2}+(1+3)^{2}=29

5. \cos (0)+\cos (\pi)+\cos (2 \pi)+\cos (3 \pi)+\cos (4 \pi)+\cos (5 \pi)=1+(-1)+1+(-1)+1+(-1)=0


7. \sum_{\mathrm{k}=3}^{94} \mathrm{k}

9. \sum_{k=3}^{12} k^{2}

11. \sum_{k=1}^{7} k \cdot 2^{k}


13. (a)  (1+2)+(2+2)+(3+2)=3+4+5=12           (b) (1+2+3)+(2+2+2)=12

15. (a)  5 \cdot 1+5 \cdot 2+5 \cdot 3=5+10+15=30                        (b)  5 \cdot(1+2+3)=5 \cdot 6=30


17. (a)  1^{2}+2^{2}+3^{2}=1+4+9=14                                     (b)  (1+2+3)^{2}=6^{2}=36


19. f(0)+f(1)+f(2)+f(3)=0^{2}+1^{2}+2^{2}+3^{2}=14

21. 2 \cdot f(0)+2 \cdot f(1)+2 \cdot f(2)+2 \cdot f(3)=2 \cdot 0+2 \cdot 1+2 \cdot 4+2 \cdot 9=28

23. g(1)+g(2)+g(3)=3+6+9=18

25. g^{2}(1)+g^{2}(2)+g^{2}(3)=3^{2}+6^{2}+9^{2}=126

27. \mathrm{h}(2)+\mathrm{h}(3)+\mathrm{h}(4)=\frac{2}{2}+\frac{2}{3}+\frac{2}{4}=\frac{13}{6}

29. \mathrm{f}(1) \mathrm{h}(1)+\mathrm{f}(2) \mathrm{h}(2)+\mathrm{f}(3) \mathrm{h}(3)=(1)(2)+(4)(1)+(9)(2 / 3)=12


31. \left(1^{2}-0^{2}\right)+\left(2^{2}-1^{2}\right)+\left(3^{2}-2^{2}\right)+\left(4^{2}-3^{2}\right)+\ldots+\left(7^{2}-6^{2}\right)=7^{2}-0^{2}=49

33. \left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)=1-\frac{1}{6}=\frac{5}{6}

35. (\sqrt{1}-\sqrt{0})+(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\ldots+(\sqrt{9}-\sqrt{8})=\sqrt{9}-\sqrt{0}=3


37. (i)  [2,3],[3,4.5],[4.5,6],[6,7]                   (ii)  1,1.5,1.5,1              (iii)  \text {mesh }=1.5             (iv)  1+1.5+1.5+1=5

39. (i)  [-3,-1],[-1,0],[0,1.5],[1.5,2]         (ii)  2,1,1.5,0.5             (iii)  \text {mesh }=2                 (iv)  2+1+1.5+0.5=5


41. (i)  [3,3.8],[3.8,4.5],[4.5,5.2],[5.2,7]      (ii)  0.8,0.7,0.7,1.8       (iii)  \text { mesh }=1.8             (iv)  0.8+0.7+0.7+1.8=4


43. \begin{aligned} &\Delta x_{1}+\Delta x_{2}+\Delta x_{3}+\ldots+\Delta x_{n}=\left(x_{1}-x_{0}\right)+\left(x_{2}-x_{1}\right)+\left(x_{3}-x_{2}\right)+\ldots+\left(x_{n}-\right. \\ &\left.x_{n-1}\right)=x_{n}-x_{0} \end{aligned}


45.
(a)  f(0)(1)+f(1)(0.5)+f(2)(0.5)=(4)(1)+(3)(0.5)+(0)(0.5)=5.5
(b)  \mathrm{f}(1)(1)+\mathrm{f}(1.5)(0.5)+\mathrm{f}(1.5)(0.5)=(3)(1)+(1.75)(0.5)+(1.75)(0.5)=4.75

47. a. (i) and (ii) See the graph
(iii)  \mathrm{f}(0)=0, \mathrm{f}(\pi / 4) \approx 0.707, \mathrm{f}(\pi / 2)=1
(iv)  (\pi / 4)(0)+(\pi / 4)(0.707)+(\pi / 2)(1) \approx 2.13


49.

(a)  (2)(1)+(5)(2)+(17)(1) \leq \mathrm{RS} \leq(5)(1)+(17)(2)+(26)(1) \text { so } 29 \leq \mathrm{RS} \leq 65
(b)  (2)(1)+(5)(1)+(10)(1)+(17)(1) \leq \mathrm{RS} \leq(5)(1)+(10)(1)+(17)(1)+(26)(1) \text { so } 34 \leq
    \mathrm{RS} \leq 58
(c)  (2)(0.5)+(3.25)(0.5)+(5)(1)+(10)(1)+(17)(1) \leq \mathrm{RS} \leq(3.25)(0.5)+(5)(0.5)+(10)(1)+(17)(1)+(26)(1) so 34.625 \leq \mathrm{RS} \leq 57.125


51.
(a)  (0)(\pi / 2)+(0)(\pi / 2) \leq \mathrm{RS} \leq(1)(\pi / 2)+(1)(\pi / 2) \text { so } 0 \leq \mathrm{RS} \leq \pi
(b)  (0)(\pi / 4)+(0.707)(\pi / 4)+(0)(\pi / 2) \leq \mathrm{RS} \leq(0.707)(\pi / 4)+(1)(\pi / 4)+(1)(\pi
    / 2) \text { so } 0.56 \leq \mathrm{RS} \leq 2.91
(c)  (0)(\pi / 4)+(0.707)(\pi /)+(0.707)(\pi / 4)+(0)(\pi / 4) \leq \mathrm{RS} \leq(0.707)(\pi / 4)+(1)(\pi / 4)+(1)(\pi / 4)+(0.707)(\pi / 4) so 1.11 \leq \mathrm{RS} \leq
    2.68


53. (a)  17.402-7.362 \mid=0.04                      (b)  |7.390-7.372|=0.018


55. \text { I error } \mathrm{I} \leq\left(\text { base) }(\text { height })=\frac{4-2}{50}(65-9)=\frac{2}{50}(56)=\frac{112}{50}=2.24\right.


57. (a)  \frac{100(101)}{2}=5050                                       (b)  \begin{aligned}
&1+2+3+\ldots+100 \\
&100+99+98+\ldots+1 \\ \hline &101+101+101+\ldots+101=100(101)=10100 . \quad \frac{1}{2}(10100)=5050
\end{aligned}     


59. 10+11+12+\ldots+20=(1+2+3+\ldots+20)-(1+2+3+\ldots+9)=\frac{20(21)}{2}-\frac{9(10)}{2}=210-45=165


61. \sum_{k=1}^{10}\left(k^{3}+k\right)=\sum_{k=1}^{10} \mathrm{k}^{3}+\sum_{k=1}^{10} \mathrm{k}=\left(\frac{10(11)}{2}\right)^{2}+\frac{10(11)}{2}=(55)^{2}+55=3080