Chapter 3, Answers

Chapter 3

3.1.4

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#3

1. (a) \(R(f,3)=22\);(b) \(R(f,4)=26\); (c) \(R(f,3)=20\); (d) \(R(f,2)=8\).
2. \(R(f,4)=10\).
3. \(R(f,4)=\frac{3\pi}{4}\).
4. \(R(f,4)=\frac{1}{6} \left(8+\sqrt{3}\right) \pi\).
5. \(R(f,4)=\frac{1}{18} \left(9+4 \sqrt{3}\right) \pi\).
6. \(R(f,4)= \left(1+e+e^2+e^3\right)/e\).
7. \(R(f,5)= (5+ \ln 720)e = \left(5 + 4\ln 2 + 2\ln 3+ \ln 5\right)e \).
8. \(R(f,4)=\frac{1}{4}\left(1+\sqrt[4]{e}+\sqrt[4]{e^2}+\sqrt[4]{e^3}\right) e^{3/4}\).


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9. \(R(f,7)=\left(\frac{7}{2} +7\ln 2+2\ln 3+\ln 5+\ln 7)\right)\sqrt{e}\).
10. \(R(f,2)= \frac{\pi}{12}\left(12 +\sqrt{2}\right)\).
11. \(R(f,3)= \frac{3\pi}{2}\).
12. See 3.1.2.
13. A midpoint rectangular approximation satisfies the following two properties:
    1. All rectangles have the same width \( \Delta x=\frac{b-a}{n}\).
    2. The height of each rectangle is determined by the \(y\)-coordinate of the point on the graph of \(f\) whose \(x\)-coordinate is the midpoint of the interval that lies at the base of the rectangle.
14. (a) \(M(f,3)=24\); (b) \(M(f,4)=22\); (c) \(M(f,3)=14\); (d) \(M(f,2)=10\).


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15. A left-endpoint rectangular approximation satisfies the following two properties:
    1. All rectangles have the same width \( \Delta x=\frac{b-a}{n}\).
    2. The height of each rectangle is determined by the \(y\)-coordinate of the point on the graph of \(f\) whose \(x\)-coordinate is the left-endpoint of the interval that lies at the base of the rectangle.
16. (a) \(L(f,3)=26\); (b) \(L(f,4)=20\); (c) \(L(f,3)=22\); (d) \(L(f,2)=2\).
17. (b) (i)\(M(f,4)=6\pi\); (ii) \(L(f,4)=6\pi\).
18. (b) (i) \(M(f,3)=38\); (ii) \(L(f,3)=32\).
19. (b) (i) \(R(f,3)= \frac{\pi}{12}\left(12 +\sqrt{2}\right)\); (ii) \(R(f,3)= \frac{\pi}{12}\left(12 -\sqrt{2}\right)\).


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3.2.3

1. (a) 34; (b) 40; (c) \(\frac{3}{4}\); (d) \(\frac{34}{15}\).
2. (a) \(\sum _{k=1}^6 (-1)^{k-1}\frac{1}{k}\);
   (b) \(\sum _{k=1}^6 \frac{k}{3^{k-1}}\);
   (c) \(\sum _{i=1}^4 f\left(x_i\right)\Delta x\);
   (d) \(\sum _{n=1}^4 (-1)^{n-1}\frac{2n-1}{2n}\);
3. \(\sum _{i=1}^1 2^i=2,\sum _{i=1}^2 2^i =2+4=6,\sum _{i=1}^3 2^i =2+4+8=14,\sum _{i=1}^4 2^i =2+4+8+16=30\).

   For a general formula, note the pattern: \(s_1=2=2^2-2\), \(s_2=4=2^3-2\), \(s_3=14=2^4-2\), and \(s_4=30=2^5-2\). Thus, we can conjecture that \(s_k=\sum_{i=1}^k 2^i=2^{k+1}-2\).

4. \( \sum _{i=0}^n \frac{1}{2^i}=2-\frac{1}{2^n}\).
5. (b) \( \sum _{i=1}^n i + \sum _{i=0}^{n-1}i = n^2\). Use the summation formulas to prove the validity of this conjecture.
6. \( \sum _{i=1}^n c a_i = ca_1+ca_2+\cdots ca_n= c(a_1+a_2+\cdots a_n)=c\sum _{i=1}^n a_i\).
7. Hint: write explicitly the summations to see why they are not valid. Then write a brief paragraph explaining your line of reasoning. Or provide numerical examples that show that they are not valid.
8. (a) \(\sum _{k=1}^1 \frac{1}{k(k+1)}=\frac{1}{1\cdot 2}\), \(\sum _{k=1}^2 \frac{1}{k(k+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}=\frac{3+1}{1\cdot
   2\cdot 3}=\frac{2}{1\cdot 3}\),

   \(\sum _{k=1}^3 \frac{1}{k(k+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}=\frac{2}{1\cdot 3}+\frac{1}{3\cdot 4}=\frac{8+1}{1\cdot
   3\cdot 4}=\frac{3}{1\cdot 4}\),

   \(\sum _{k=1}^4 \frac{1}{k(k+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5}=\frac{3}{1\cdot 4}+\frac{1}{4\cdot 5}=\frac{15+1}{1\cdot
   4\cdot 5}=\frac{4}{1\cdot 5}\).

   (b) Conjecture: \(\sum _{k=1}^n \frac{1}{k(k+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\text{…}+\frac{1}{(n-1)\cdot n}=\frac{n}{1\cdot (n+1)}\).

   (c) \(A=1\) and \(B=-1\). Thus, \(\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}\), and \(\sum _{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}=\frac{n}{n+1}\).

9. \(\frac{73}{2}\).
10. 620.
11. 252.
12. \(\frac{1}{4} n^2(n+1)^2+\frac{1}{3}n(n+1)(2n+1)-3n\).
13. \(\frac{27}{4} n^2(n+1)^2+\frac{9}{2}n(n+1)(2n+1)+\frac{9}{2}n(n+1) + n\).
14. \(n + 2(n+1) + \frac{2}{3}\frac{(n+1)(2n+1)}{n}\).
15. Make sure you learn how to prove these summation formulas.

17. (a) \(17+ 18 + \cdots + 24+ 25 = 64 + 125\) or \(189 =189\);
    (c) \(82+83+\cdots + 99+100 = 81 + 100\); \( \sum_{9^2+1}^{10^2} i = 9^3+10^3\) (\(10\)-th row);
    (d) \( \sum_{(n-1)^2+1}^{n^2} i = (n-1)^3+n^3\) (\(n\)-th row);
    (e) use the summation formulas.
18. \(\frac{1}{30} n (n+1) (2 n+1) \left(3 n^2+3 n-1\right)=\frac{n^5}{5}+\frac{n^4}{2}+\frac{n^3}{3}-\frac{n}{30}\). (Hint: for the RHS, find a common
    denominator, factor \(n\), and then use long division by \((n+1)(2n+1) = 2 n^2+3 n+1\).)

3.3.5

1. (b) \(R(f,3)=15\);
   (c) and (d) Step1. \(\Delta x= \frac{3-0}{n}=\frac{3}{n}\), \(x_i=0+i \Delta x=\frac{3i}{n}\)

Step 2. \(f\left(x_i\right)=2+\frac{6 i}{n}\), \(A_i=\) area of the \(i\)-th rectangular region \(=f\left(x_i\right)\Delta x=\left(2+\frac{6 i}{n}\right)\frac{3}{n}\).

Step 3. \(\sum _{i=1}^n f\left(x_i\right)\Delta x
=\sum _{i=1}^n \left(2+\frac{6 i}{n}\right)\frac{3}{n}
=\frac{3}{n}\left(2\sum _{i=1}^n 1+\frac{6}{n}\sum_{i=1}^n i\right)
=\frac{3}{n}\left(2n +\frac{6}{n}\frac{n(n+1)}{2}\right)\).

Step 4. \(\sum _{i=1}^n f\left(x_i\right)\Delta x=3\left(2 +3\left(1+\frac{1}{n}\right)\right)\overset{n\to \infty }{\to }3(2+3(1+0))=15\).

(e) \(A=2\times 3+\frac{1}{2}6\times 3=15\).

2. (a) ; (b) \(R(f,4)=12\);

(c) and (d) Step1. \(\Delta x= \frac{2-(-2)}{n}=\frac{4}{n}\), \(x_i=-2+i \Delta x=-2+\frac{4i}{n}\)

Step 2. \(f\left(x_i\right)=8-\frac{8 i}{n}\), \(A_i=\) area of the \(i\)-th rectangular region \(=f\left(x_i\right)\Delta x=\left(8-\frac{8 i}{n}\right)\frac{4}{n}\).

Step 3. \(\sum _{i=1}^n f\left(x_i\right)\Delta x=\sum _{i=1}^n \left(8-\frac{8 i}{n}\right)\frac{4}{n}=\frac{4}{n}\left(8\sum _{i=1}^n 1-\frac{8}{n}\sum
_{i=1}^n i\right)=\frac{4}{n}\left(8n -\frac{8}{n}\frac{n(n+1)}{2}\right)\).

Step 4. \(\sum _{i=1}^n f\left(x_i\right)\Delta x=4\left(8 -4\left(1+\frac{1}{n}\right)\right)\overset{n\to \infty }{\to }4(8-4(1+0))=16\).

(e) \(A=\frac{1}{2}4\times 8=16\).



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3. (a) ; (b) \(R(f,4)=10\);
   (c) and (d) Step1. \(\Delta x= \frac{4}{n}\), \(x_i=0+i \Delta x=\frac{4i}{n}\)
   Step 2. \(f\left(x_i\right)=-\frac{16 i^2}{n^2}+\frac{16 i}{n}\),

   Step 2.
   \(A_i=\) area of the \(i\)-th rectangular region \(=f\left(x_i\right)\Delta x=\left(-\frac{16
   i^2}{n^2}+\frac{16 i}{n}\right)\frac{4}{n}\).
   Step 3. \(\sum _{i=1}^n f\left(x_i\right)\Delta x=\sum _{i=1}^n \left(-\frac{16 i^2}{n^2}+\frac{16 i}{n}\right)\frac{4}{n}=\frac{4}{n}\left(-\frac{16
   }{n^2}\sum _{i=1}^n i^2+\frac{16 }{n}\sum _{i=1}^n i\right)=\)

   Step 3. \(\sum _{i=1}^n f\left(x_i\right)\Delta x\)
   \(=\frac{4}{n}\left(-\frac{16 }{n^2}\frac{n(n+1)(2n+1)}{6} +\frac{16}{n}\frac{n(n+1)}{2}\right)\).

   Step 4. \(\sum _{i=1}^n f\left(x_i\right)\Delta x=4\left(-\frac{8 }{3}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right) +8\left(1+\frac{1}{n}\right)\right)\overset{n\to
   \infty }{\to }4\left(-\frac{16 }{3}+8\right)=\frac{32 }{3}\).

4. (b) \(R(f,3)=12\);

   (c) Step1. \(\Delta x= \frac{3}{n}\), \(x_i=-1+i \Delta x=-1+\frac{3i}{n}\)

   Step 2. \(f\left(x_i\right)=\frac{27 i^3}{n^3}-\frac{27 i^2}{n^2}+\frac{9 i}{n}\),

   Step 2.
   \(A_i=\) area of the \(i\)-th rectangular region \(=f\left(x_i\right)\Delta x=\left(\frac{27 i^3}{n^3}-\frac{27 i^2}{n^2}+\frac{9 i}{n}\right)\frac{3}{n}\).

   Step 3.

   \(\sum _{i=1}^n f\left(x_i\right)\Delta x
   =\sum _{i=1}^n \left(\frac{27 i^3}{n^3}-\frac{27 i^2}{n^2}+\frac{9 i}{n}\right)\frac{3}{n}
   =\frac{3}{n}\left(\frac{27}{n^3}\sum _{i=1}^n i^3-\frac{27 }{n^2}\sum _{i=1}^n i^2+\frac{9 }{n}\sum _{i=1}^n i\right) \)

   \(\sum _{i=1}^n f\left(x_i\right)\Delta x\)
   \(=\frac{3}{n}(\frac{27 }{n^3}\left(\frac{n(n+1)}{2}\right)^2-\frac{27
   }{n^2}\frac{n(n+1)(2n+1)}{6} +\frac{9}{n}\frac{n(n+1)}{2})\).

   Step 4.
   \(\sum _{i=1}^n f\left(x_i\right)\Delta x=3(\frac{27 }{4}\left(1+\frac{1}{n}\right)^2-\frac{9 }{2}\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)
   +\frac{9 }{2}\left(1+\frac{1}{n}\right))\overset{n\to \infty }{\to }3\left(\frac{27 }{4}-\frac{18}{2}+\frac{9 }{2}\right)=\frac{27 }{4}\).

5. (a) The graph of \(f\) is the upper semicircle with center at the origin and radius 4;
   (b) \(A=\frac{1}{2}\pi 4^2=8\pi\).
6. (a)
   \(\left(\frac{s}{2}\right)^2+h^2=s^2\Longleftrightarrow h^2=\frac{3}{4}s^2\Longleftrightarrow h=\frac{\sqrt{3}}{2}s\).
   Thus, \(A=\frac{1}{2}s\left(\frac{\sqrt{3}}{2}s\right)=\frac{\sqrt{3}}{4}s^2\).

   (b) Step1.
   \(\Delta x= \frac{s/2}{n}=\frac{s}{2n}\), \(x_i=0+i \Delta x=\frac{s i}{2n}\)

   Step 2. \(f\left(x_i\right)=-\sqrt{3}\frac{s i}{2n}+\frac{\sqrt{3}}{2}s\),
   Step 2.
   \(A_i=\) area of the \(i\)-th rectangular region \(=f\left(x_i\right)\Delta
   x=\left(-\sqrt{3}\frac{s i}{2n}+\frac{\sqrt{3}}{2}s\right)\frac{s}{2n}\).

   Step 3. \(\sum _{i=1}^n f\left(x_i\right)\Delta x=\sum _{i=1}^n \left(-\sqrt{3}\frac{s i}{2n}+\frac{\sqrt{3}}{2}s\right)\frac{s}{2n}=\frac{s}{2n}\left(-\sqrt{3}\frac{s
   }{2n}\sum _{i=1}^n i+\frac{\sqrt{3}}{2}s\sum _{i=1}^n 1\right)=\)

   Step 3. \(\sum _{i=1}^n f\left(x_i\right)\Delta x\)
   \(=\frac{s}{2n}\left(-\sqrt{3}\frac{s }{2n}\frac{n(n+1)}{2}+\frac{\sqrt{3}}{2}s n\right)\).

   Step 4. \(\sum _{i=1}^n f\left(x_i\right)\Delta x=\frac{s}{2}(-\sqrt{3}\frac{s }{4}\left(1+\frac{1}{n}\right)+\frac{\sqrt{3}}{2}s)\overset{n\to
   \infty }{\to }\frac{s}{2}\left(-\sqrt{3}\frac{s }{4}+\frac{\sqrt{3}}{2}s\right)=\frac{\sqrt{3}}{8}s^2\).

   Multiplying by 2, we obtain \(A=\frac{\sqrt{3}}{4}s^2\).

3.4.3

2. \(\frac{8}{3}\).
3. \(0\).
4. \(\frac{21}{2}=-3+\frac{27}{2}\).
5. \(4=6-2\).
6. \(\frac{1}{6}\left(9 +3\sqrt{3}+2\pi \right)=-\frac{1}{2}+\frac{1}{6}\left(12 +3\sqrt{3}+2\pi\right )\).
7. \(24 = 29-5\).


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8. (a) True. First, by Theorem 2 Part 3, both \(rf\) and \(sg\) are integrable. Then, by Part 1, the sum is integrable. Thus
   \(\int_a^b(rf + sg)(x)\,dx
   \underset{\text{Part 1}}{=} \int_a^brf(x)\,dx + \int_a^bsg(x)\,dx
   = r\int_a^bf(x)\,dx +s\int_a^bg(x)\,dx\).

   (b) False. For example, \(\int_0^21\,dx=2\), while
   \(2= \int_0^21\cdot 1\,dx\neq \left ( \int_0^21\,dx\right )\left ( \int_0^21\,dx\right )=2\cdot 2 = 4\).

3.4.5

1. \(F(x)=x^2+3x\).
2. \(F(x)=\left \{\begin{array}{ll}
   x^2+7x+10 & -5 \leq x \leq -2\\
   3(x+2) & -2 < x \leq 2
   \end{array}\right . \).
3. \(F(x)=\left \{
   \renewcommand{\arraystretch}{1}
   \begin{array}{ll}
   x^2 – x – 6 & -2 \leq x \leq 2\\
   -x^2+7x-14 & 2 < x \leq \frac{7}{2}
   \end{array}\right . \).
4. \(F(x)= \left \{
   \renewcommand{\arraystretch}{1}
   \begin{array}{ll}
   \frac{1}{2} x^2 +2 x & -4 \leq x \leq 0\\
   \frac{1}{2}x\sqrt{4-x^2}+ 2\left (\frac{\pi}{2}-\arccos x\right ) & 0 < x \leq 2
   \end{array}\right . \).

3.4.7

2. (c) Here is a new integration formula
   \begin{equation}\label{ch03-04-eq09}
   \int_a^b\sqrt[n]{x}\,dx=\frac{n}{n+1}\left (b^{(n+1)/n}-a^{(n+1)/n}\right ),\;\;0\leq a < b,\;\;n\in\mathbb{N}, \; n>1.
   \end{equation}

3.5.2

1. This follows from \(4+1 \leq 4+x^4\leq 4+2^4\) for \(x\in[1,2]\).
2. \(y= e^{-x^2} \) is decreasing for \(x\geq 0\).
3. \(1\leq x \leq 2\Longrightarrow 9 \geq 1+x^3 \geq 1+x^2\geq 2 \Longrightarrow \frac{x}{3}\leq \frac{x}{\sqrt{1+x^3}} \leq \frac{x}{\sqrt{1+x^2}} \leq \frac{x}{\sqrt 2}\).
4. \(\arcsin x\) is increasing.
5. \(1\leq x \leq 2
   \underset{?}{\Longrightarrow} -4\leq x^2-2x-3 \leq -3
   \underset{?}{\Longrightarrow}
   -\frac{x}{3} \leq \frac{x}{x^2-2x-3} \leq -\frac{x}{4} \).
6. \(\sin x\) is increasing, thus \(\sin \sqrt x\) is increasing, (a) \(m=\frac{1}{2}\), \(M= \frac{\sqrt 3}{2}\); (b) \(\frac{1}{24}\pi^2\leq I \leq \frac{\sqrt 3}{24}\pi^2\).
7. \(e^{x^2}\) is increasing. (a) \(1\leq x \leq 2\Longrightarrow e\leq e^{x^2} \leq e^4\); (b) \(e\leq I \leq e^4\).
8. \(\ln x\) is increasing. (a) \( \sqrt{e}\leq x \leq e^2
   \underset{?}{\Longrightarrow} \frac{1}{2} \leq \ln x \leq 2\); (b) \(\frac{1}{2} \left (e^2-\sqrt e\right )\leq I \leq 2 \left (e^2-\sqrt e\right )\).

3.6.2

1. \(\frac{1}{18}\left (9+3\sqrt 3\right ) \).
2. \(\frac{3\pi}{4}\).
3. \(336\ln 2\).
4. \(e^2(6 + 3\ln 2 + \ln 3 \).
5. \( 42\).
6. \(-24 \).
7. \(-6 \).
8. \(-\frac{5}{2}= -\frac{7}{2}+1 \).
9. \(\frac{1}{6}\left (3\sqrt{3}+2\pi\right ) \).
10. (a) \(0<f(x)<\frac{\sqrt 3}{2}\); (b) \(0\leq I \leq \frac{\sqrt 3\pi^2}{9}\).
11. (a) \(\frac{\pi}{6}\leq f(x)\leq \frac{\pi}{3}\); (b) \(\frac{4\pi}{9}\leq I \leq \frac{8\pi}{9}\).
12. (a) \(17+ 18 + \cdots + 24+ 25 = 64 + 125\) or \(189 =189\);
    (c) \(82+83+\cdots + 99+100 = 81 + 100\); \( \sum_{9^2+1}^{10^2} i = 9^3+10^3\) (\(10\)-th row);
    (d) \( \sum_{(n-1)^2+1}^{n^2} i = (n-1)^3+n^3\) (\(n\)-th row);
    (e) use the summation formulas.