====Differential Equations to Difference Equations====

===Answers for selected problems===

**Chapter 4**

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**Section 4.1**

1. (a) The area is between $ \dfrac{77}{60} $ and $ \dfrac{25}{12} $. 

(b) The area is between $ 14 $ and $ 30 $. 

%%(c)%% The area is between $ 6 $ and $ 14 $. 

(d) The area is between $ \dfrac{\sqrt{2}\pi}{4} $ and $ \dfrac{(2 + \sqrt{2})\pi}{4} $. 



2. (a) $ \displaystyle{\frac{81}{4} \le \int_1^4 3xdx \le \frac{99}{4}}$ 


%%(c)%% $ \displaystyle{-\frac{2\pi}{3} \le \int_{-\pi}^\pi \cos(x)dx \le \frac{2\pi}{3}}$ 


(e) $ \displaystyle{-0.3072 \le \int_0^1 (x^3 - x)dx \le -0.1813}$ 



3. (a) $ A_R = 30 $, $ A_L = 14 $ 


%%(c)%% $ A_R = \dfrac{496}{315} $, $ A_L = \dfrac{352}{105} $ 


(d) $ A_R = \dfrac{25}{64} $, $ A_L = \dfrac{9}{64} $ 



4. (a) $ A_R = 0.33835 $, $ A_L = 0.32835 $ 

%%(c)%% $ A_R = 4.08040 $, $ A_L = 3.92040 $ 

(e) $ A_R = 0.69065 $, $ A_L = 0.69565 $ 

(g) $ A_R = 1.76281 $, $ A_L = 1.76281 $ 


5. (a) $ A_R = \dfrac{15}{4} $, $ A_L = \dfrac{7}{4} $ 

%%(c)%% $ A_R = 0 $, $ A_L = 0 $ 

(e) $ A_R = -\dfrac{5}{4} $, $ A_L = -\dfrac{5}{4} $ 


6. (a) $ A_R = 9.13545 $, $ A_L = 8.86545 $ 

%%(c)%% $ A_R = 6.27654 $, $ A_L = 6.27654 $ 

(e) $ A_R = -1.33320 $, $ A_L = -1.33320 $ 

(g) $ A_R = -0.34407 $, $ A_L = -0.34907 $ 


7. (a) $ \displaystyle{\int_0^4 xdx = 8}$ 

%%(c)%% $ \displaystyle{\int_0^3 \sqrt{9 - x^2} \ dx = \frac{9\pi}{4}}$ 

(d) $ \displaystyle{\int_{-2}^2 4\sqrt{4-t^2} \ dt = 8\pi}$ 

(e) $ \displaystyle{\int_{-2}^2 x^3dx = 0}$ 

8. (a) $ \displaystyle{\int_0^1 f(x)dx = \frac{3}{2}}$ 

%%(c)%% $ \displaystyle{\int_0^3 f(x)dx = \frac{19}{2}}$ 


12. %%(c)%% $ F'(x) = f(x) $

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**Section 4.2**

1. (a) $ A_T = \dfrac{11}{32} $, $ A_M = \dfrac{21}{64} $ 

%%(c)%% $ A_T = \dfrac{101}{60} $, $ A_M = \dfrac{496}{315} $ 

(e) $ A_T = 30 $, $ A_M = 29 $ 


2. (a) $ A_S = \dfrac{1}{3} $ 

%%(c)%% $ A_S = \dfrac{6089}{3780} $ 

(e) $ A_S = \dfrac{88}{3} $ 


3. (a) $ A_T = 0.33340 $, $ A_M = 0.33330 $ 

%%(c)%% $ A_T = 166.92480 $, $ A_M = 166.78760 $ 

(e) $ A_T = -6.28525 $, $ A_M = -6.28215 $ 

(g) $ A_T = 2.74670 $, $ A_M = 2.74685 $ 


4. (a) $ A_S = 0.33333 $ 

%%(c)%% $ A_S = 166.83333 $ 

(e) $ A_S = -6.28318 $ 

(g) $ A_S = 2.74680 $ 


5. (a) Stop when $ n = 4 $, $ A_S = 21 $ 

%%(c)%% Stop when $ n = 4 $, $ A_S = 1.57080 $ 

(e) Stop when $ n = 32 $, $ A_S = 7.64040 $ 


8. (a) $ A_S = 428 $ 

(b) $ A $ represents the average temperature over that $ 12 $ hour period. 

%%(c)%% Rounded to two decimal places, $ A = 35.67 $ and 
$\displaystyle{
\frac{1}{25}\sum_{t=0}^{24} T\left(\frac{t}{2}\right) = 36.04.
}$


9. Area $ \displaystyle{= \int_0^\pi \sin^2(x)dx = 1.57080} $, rounded to five decimal places. 

10. Area $ \displaystyle{= \int_0^1 x^2dx + \int_1^2 (x - 2)^2dx = 0.66667} $, rounded to five decimal places.

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**Section 4.3**

1. (a) $ \displaystyle{\int_0^1 xdx = \frac{1}{3}}$ 

%%(c)%% $ \displaystyle{\int_0^2 x^3dx = 4}$ 

(e) $ \displaystyle{\int_0^2 (2x^3 + 3x^2 +x - 4)dx = 10}$ 

(g) $ \displaystyle{\int_1^4 \frac{1}{\sqrt{t}} \ dt = 2}$ 

(i) $ \displaystyle{\int_0^{\frac{\pi}{2}} \sin(x)dx = 1}$ 

(j) $ \displaystyle{\int_{-\pi}^\pi \cos(z)dz = 0}$ 


2. (a) $ \displaystyle{\int_0^2 (x + 1)^2 dx = \frac{26}{3}}$ 

%%(c)%% $ \displaystyle{\int_0^4 \sqrt{1 + 2t} \ dt = \frac{26}{3}}$ 

(e) $ \displaystyle{\int_0^\pi \sin(2x)dx = 0}$ 

(g) $ \displaystyle{\int_0^{\frac{\pi}{3}} 4\sin(3x)dx = \frac{8}{3}}$ 

(i) $ \displaystyle{\int_0^{\sqrt{\pi}} 2x\sin(x^2)dx = 2}$ 

(k) $ \displaystyle{\int_0^{\sqrt{\pi}} x\sin(x^2)dx = 1}$ 


4. (a) $ F'(x) = \sin^2(4x) $ 

%%(c)%% $ F'(x) = -\cos^3(x) $ 

(e) $ \displaystyle{f'(x) = \frac{2x}{1 + x^4}}$ 

(f) $ \displaystyle{h'(z) = 3\sqrt{1 + 9z^2} - \sqrt{1 + z^2}}$ 

5. (a) $ \displaystyle{\frac{d}{dx}\int_1^x \frac{1}{1 + t^2} \ dt = \frac{1}{1 + x^2}}$ 

%%(c)%% $ \displaystyle{\frac{d}{dt}\int_{t^2}^5 \sin^2(3x)dx = -2t\sin^2(3t^2)}$ 


6. Area $ \displaystyle{= \int_0^{\frac{\pi}{2}} 3\sin(2x)dx = 3}$


7. Area $ \displaystyle{= \int_0^1 x^2dx + \int_1^2 (x - 2)^2dx = \frac{2}{3}}$ 


8. Area $ \displaystyle{= \int_0^1 (x - x^2)dx = \frac{1}{6}}$ 

9. Area $ \displaystyle{= \int_{-1}^1 (2 - 2x^2)dx = \frac{8}{3}}$

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**Section 4.4**

1. (a) $ \displaystyle{\int (x^3 + 3x - 6)dx = \frac{1}{4}x^4 + \frac{3}{2}x^2 - 6x + c}$ 

%%(c)%% $ \displaystyle{\int \frac{1}{x^4} \ dx = -\frac{1}{3x^3} + c}$ 

(e) $ \displaystyle{\int \frac{12}{\sqrt{t}} \ dt = 24\sqrt{t} + c}$ 

(g) $ \displaystyle{\int 7\sqrt{x + 5} \ dx = \frac{14}{3}(x  + 5)^{\frac{3}{2}} + c}$ 


2. (a) $ \displaystyle{\int \sin(3x)dx = -\frac{1}{3}\cos(3x) + c}$ 

%%(c)%% $ \displaystyle{\int \sqrt{3t - 1} \ dt = \frac{2}{9}(3t - 1)^{\frac{3}{2}} + c}$ 

(e) $ \displaystyle{\int 7\sec^2(2x)dx = \frac{7}{2}\tan(2x) + c}$ 

(g) $ \displaystyle{\int 2\csc^2(7x)dx = -\frac{2}{7}\cot(7x) + c}$ 


3. $ \displaystyle{\int 6x\sqrt{1 + 3x^2} \ dx = \frac{2}{3}(1 + 3x^2)^{\frac{3}{2}} + c}$ 

%%(c)%% $ \displaystyle{\int x^2(3 + x^3)^{10} dx = \frac{1}{33}(3 + x^3)^{11} + c}$ 

(e) $ \displaystyle{\int 4t\sin(t^2)dt = -2\cos(t^2) + c}$ 

(g) $ \displaystyle{\int \sin^3(t)\cos(t)dt = \frac{1}{4}\sin^4(t) + c}$ 


4. (a) $ \displaystyle{\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \ dx = -2\cos(\sqrt{x}) + c}$ 

%%(c)%% $ \displaystyle{\int \sec^3(4x)\tan(4x)dx = \frac{1}{12}\sec^3(4x) + c}$ 

(e) $ \displaystyle{\int \frac{\sin(x)}{\cos^2(x)} \ dx = \sec(x) + c}$ 

(f) $ \displaystyle{\int t\sqrt{t - 2} \ dt = \frac{2}{5}(t - 2)^{\frac{5}{2}} + \frac{4}{3}(t - 2)^{\frac{3}{2}} + c}$ 


5. $ \displaystyle{\int_0^1 (4x^2 - 3x - 5)dx = -\frac{31}{6}}$ 

%%(c)%% $ \displaystyle{\int_0^{\frac{\pi}{4}} 3\sin(2x))dx = \frac{3}{2}}$ 

(e) $ \displaystyle{\int_{-\frac{\pi}{12}}^0 6\sec(3t)\tan(3t)dt = 2 - 2\sqrt{2}}$ 

(g) $ \displaystyle{\int_0^2 x\sqrt{x^2 + 1} \ dx = \frac{5\sqrt{5} - 1}{3}}$ 


6. (a) $ \displaystyle{\int_{-1}^1 \frac{5x^2}{(x^3 + 2)^2} \ dx = \frac{10}{9}}$ 

%%(c)%% $ \displaystyle{\int_0^{\sqrt{\pi}} 3x\sin(x^2)dx = 3}$ 

(e) $ \displaystyle{\int_0^{\frac{\pi}{2}} \sin^3(2t)\cos(2t)dt = 0}$ 

(g) $ \displaystyle{\int_{-1}^1 x(1 + x^2)^{25} dx = 0}$ 


7. (a) $ \displaystyle{\int_0^3 \frac{x}{\sqrt{x+1}} \ dx = \frac{8}{3}}$ 

%%(c)%% $ \displaystyle{\int_{-\frac{\pi}{12}}^{\frac{\pi}{24}} \tan^4(4x)\sec^2(4x)dx = \frac{61}{45\sqrt{3}}}$ 

(e) $ \displaystyle{\int_0^1 4x(1 + x)^{25}dx = \frac{3,355,443,202}{351}}$ 

(g) $ \displaystyle{\int_1^5 5u\sqrt{2u - 1} \ du = \frac{428}{3}}$ 


8. Area $ \displaystyle{= \int_0^{\frac{\pi}{6}} 4\sin(6t)dt = \frac{4}{3}}$

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**Section 4.5**

1. (a) $ \displaystyle{\int 3x\sin(x)dx = -3x\cos(x) + 3\sin(x) + c}$ 

%%(c)%% $ \displaystyle{\int 4x\sin(3x)dx = -\frac{4}{3}x\cos(3x) + \frac{4}{9}\sin(3x) + c}$ 

(e) $ \displaystyle{\int 2x^2\sin(4x)dx = -\frac{1}{2}x^2\cos(4x) + \frac{1}{4}x\sin(4x) + \frac{1}{16}\cos(4x) + c}$ 

(g) $ \displaystyle{\int 3x^3\sin(2x)dx = -\frac{3}{2}x^3\cos(2x) + \frac{9}{4}x^2\sin(2x) + \frac{9}{4}x\cos(2x)}$
$ \displaystyle{- \frac{9}{8}\sin(2x) + c}$ 

(h) $ \displaystyle{\int x\sqrt{1+x} \ dx = \frac{2}{3}x(1 + x)^{\frac{3}{2}} - \frac{4}{15}(1 + x)^{\frac{5}{2}} + c}$ 


2. (a) $ \displaystyle{\int_0^\pi 4x\sin(x)dx = 4\pi}$ 

%%(c)%% $ \displaystyle{\int_0^{\frac{\pi}{3}} 2t\sin(3t)dt = \frac{2\pi}{9}}$ 

(e) $ \displaystyle{\int_0^{\frac{\pi}{4}} 2x^2\sin(2x)dx = \frac{\pi}{4} -\frac{1}{2}}$ 


3. (a) $ \displaystyle{\int \sin^2(2x)dx = \frac{x}{2} - \frac{1}{8}\sin(4x) + c}$ 

%%(c)%% $ \displaystyle{\int 5\sin^2(2t)\cos^2(2t)dt = \frac{5}{8}t - \frac{5}{64}\sin(8t) + c}$ 

(e) $ \displaystyle{\int 6\cos^3(2z)dz = 3\sin(2z) - \sin^3(2z) + c}$ 

(g)  $ \displaystyle{\int \cos^5(2x)dx = \frac{1}{2}\sin(2x) - \frac{1}{3}\sin^3(2x) + \frac{1}{10}\sin^5(2x) + c}$ 


4. (a) $ \displaystyle{\int_0^\pi \sin^2(x)dx = \frac{\pi}{2}}$ 

%%(c)%% $ \displaystyle{\int_0^{\frac{\pi}{2}} 3\sin^2(z)\cos^2(z)dz = \frac{3\pi}{16}}$ 

(e) $ \displaystyle{\int_0^\pi \sin^3(3t)dt = \frac{4}{9}}$ 


5. Using [[http://wxmaxima.sourceforge.net/wiki/index.php/Main_Page|wxMaxima]]:
(a) $ \displaystyle{\int \cos^6(x)dx = \frac{5}{16}x + \frac{1}{4}\sin(2x) - \frac{1}{48}\sin^3(2x) + \frac{3}{64}\sin(4x) + c}$ 

%%(c)%% $ \displaystyle{\int \sin^4(2t)\cos^4(3t)dt = \frac{9}{64}t - \frac{3}{64}\sin(2t) - \frac{23}{512}\sin(4t) + \frac{1}{32}\sin(6t) + \frac{1}{512}\sin(8t) - \frac{1}{80}\sin(10t) + \frac{1}{256}\sin(12t) + \frac{1}{448}\sin(14t) - \frac{1}{512}\sin(16t) + \frac{1}{2560}\sin(20t) + c}$ 

(e) $ \displaystyle{\int_{-1}^1 \sqrt{1 - x^2} \ dx = \frac{\pi}{2}}$ 

(g) $ \displaystyle{\int_0^{2\pi} \sin^8(2t)dt = \frac{35\pi}{64}}$ 


6. Using [[http://wxmaxima.sourceforge.net/wiki/index.php/Main_Page|wxMaxima]]:
(a) $ \displaystyle{\int_{\pi}^\pi \sin^4(x)dx = \frac{3\pi}{4}}$ 

%%(c)%% $ \displaystyle{\int_0^5 \sin(3x^2)dx = 0.33117}$ (approximation, rounded to five decimal places)

(e) $ \displaystyle{\int_1^2 \frac{1}{x} \ dx = \log(2)}$ 

(g) $ \displaystyle{\int_0^\pi \sqrt{5 - 3\sin^2(t)} \ dt = 5.80675}$ (approximation, rounded to five decimal places)


7. (a) $ T = 1.52309 $, rounded to fice decimal places. 

%%(c)%% The approximation is $ 1.41923 $, rounded to five decimal places.

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**Section 4.6**

1. (a) $ \displaystyle{\int_1^\infty \frac{1}{x^3} \ dx = \frac{1}{2}}$ 

(b) $ \displaystyle{\int_4^\infty \frac{3}{x^7} \ dx = \frac{1}{8192}}$ 

%%(c)%% $ \displaystyle{\int_{10}^\infty \frac{1}{5x^4} \ dx = \frac{1}{15000}}$ 

(d) This integral diverges. 

(e) $ \displaystyle{\int_0^\infty \frac{3}{2x + 3)^2} \ dx = \frac{1}{2}}$ 

(f) This integral diverges. 


2. (a) $ \displaystyle{\int_{-\infty}^{-2} \frac{3}{x^2} \ dx = \frac{3}{2}}$ 

(b) $ \displaystyle{\int_{-\infty}^\infty \frac{x}{(x^2 + 4)^4} \ dx = 0}$ 

%%(c)%% This integral diverges. 

(d) This integral diverges. 

3. (a) Converges 

(b) Converges 

%%(c)%% Diverges 

(d) Converges 

(e) Converges 

(f) Converges 


4. (a) $ \displaystyle{\int_0^8 \frac{1}{x^{\frac{1}{3}}} \ dx = 6}$ 

(b) This integral diverges.

%%(c)%% $ \displaystyle{\int_0^1 \frac{1}{\sqrt{1-x}} \ dx = 2}$ 

(d) $ \displaystyle{\int_0^5 \frac{5}{(t - 2)^{\frac{2}{5}}} \ dt = \frac{25}{3}\left(2^{\frac{3}{5}} + 3^{\frac{3}{5}}\right)}$ 

(e) This integral diverges. 

(f) $ \displaystyle{\int_{-1}^2 \frac{3}{x^\frac{1}{3}} \ dx = \frac{9}{2}\left(2^\frac{2}{3} - 1\right)}$ 


5. (a) $ \displaystyle{\int_1^\infty \frac{1}{x^p} \ dx = \frac{1}{p - 1}}$ 


6. (a) $ \displaystyle{\int_0^1 \frac{1}{x^p} \ dx = \frac{1}{1 - p}}$ 


8. (a) $ \displaystyle{P(x) = \left(\frac{\sigma}{x}\right)^\alpha}$ 

(b) $ \displaystyle{A = \frac{\alpha}{\alpha - 1}\sigma}$ 

(d) $ A = 60,000 $, $ P(A) = 0.1165 $, $ P(2A) = 0.0507 $

These values tell us that only $ 11.65 $% of the population have incomes exceeding the average income of $ 60,000 $ and only $ 5.07 $% of the population have incomes exceeding $ 120,000 $. 

(e) $ \displaystyle{P(A) = \left(\frac{\alpha - 1}{\alpha}\right)^\alpha}$, $ \alpha > 1 $

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**Section 4.7**

1. Area $ \displaystyle{= \int_0^1 (x - x^2)dx = \frac{1}{6}}$ 


3. Area $ \displaystyle{\int_{-1}^2 (x + 2 - x^2)dx = \frac{9}{2}}$ 


5. Using Newton's method, and rounding to five decimal places, $ r = 0.87673 $ is the nonzero solution of the equation $ \sin(x) = x^2 $. Hence, rounded to five decimal places,
\[
\text{Area} = \int_0^r (\sin(x) - x^2)dx = 0.13570.
\]


7. Area $ \displaystyle{= \int_0^{\frac{\pi}{4}} (\cos(x) - \sin(x))dx = \sqrt{2} - 1}$ 


9. (a) Area $ \displaystyle{= \int_{-2}^1 (2 - x - x^2)dx}$ 

(b) Area $ \displaystyle{= \int_0^1 2\sqrt{y}dy + \int_1^4 (2 - y + \sqrt{y})dy}$ 

%%(c)%% Area $ = \dfrac{9}{2} $ 


11. Area $ \displaystyle{= \int_{-3}^2 (6 - y - y^2)dy = \frac{125}{6}}$ 


13. Area $ \displaystyle{\int_{-1}^0 (3x^3 - x^4 + 4x^2)dx + \int_0^4 (3x^3 - x^4 + 4x^2)dx}$
$ \displaystyle{= \int_{-1}^4(3x^3 - x^4 + 4x^2)dx = \frac{875}{4}}$ 


14. Area $ \approx 178,667 $ square yards

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**Section 4.8**

1. (a) Distance $ \displaystyle{= \int_0^3 |32t|dt = 144}$ 

%%(c)%% Distance $ \displaystyle{= \int_0^2 |t^2 - t - 6|dt = \frac{34}{3}}$ 

(e) Distance $ \displaystyle{= \int_0^\pi |2\sin(2t)|dt = 4}$ 


3. (a) $ \displaystyle{x(t) = \int_0^t (3s^2 - 6)ds + 5 = t^3 - 6t + 5}$ 


4. $ \displaystyle{x(t) = -\int_0^t x_0\sqrt{\frac{k}{m}}\sin\left(\sqrt{\frac{k}{m}}s\right)ds + x_0 = x_0\cos\left(\sqrt{\frac{k}{m}} \ t\right)}$ 


6. (a) $ \displaystyle{x(t) = \int_2^t (3s^3 + 6s - 17)ds = \frac{3}{4}t^4 + 3t^2 - 17t + 14}$ 

%%(c)%% $ \displaystyle{x(t) = \int_0^t \sin^2(2s)ds = \frac{1}{2}t - \frac{1}{8}\sin(4t) + 2}$ 

(e) $ \displaystyle{x(t) = \int_4^t \sqrt{1 + 2s} \ ds + 3 = \frac{1}{3}(1 + 2t)^\frac{3}{2} - 6}$ 


8. (a) $ \displaystyle{x(t) = -16t^2 + 20t + 100}$ feet

%%(c)%% Maximum height $ \displaystyle{= x\left(\frac{5}{8}\right) = 106.25}$ feet


9. (a) Length $ \displaystyle{= \int_0^2 \sqrt{1 + 9x} \ dx = \frac{2}{27}(19\sqrt{19} - 1)}$ 

%%(c)%% Length $ \displaystyle{= \int_{-1}^1 \sqrt{1 + 9x^4} \ dx = 3.09573}$, rounded to five decimal places 

(e) Length $ \displaystyle{= \int_0^\pi \sqrt{1 + \sin^2(2t)} \ dt = 3.82020}$, rounded to five decimal places 


10. The flat sheet would need to be $ 14.64 $ feet long.

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