Solutions

1.
By definition,
$\displaystyle {\frac{{d}}{{dx}}}$$\displaystyle \left(\vphantom{ x^{8}}\right.$x8$\displaystyle \left.\vphantom{ x^{8}}\right)$ = $\displaystyle \lim_{{h\rightarrow 0}}^{}$$\displaystyle {\frac{{\left( x+h\right) ^{8}-x^{8}}}{{h}}}$  
  = $\displaystyle \lim_{{h\rightarrow 0}}^{}$$\displaystyle {\frac{{8x^{7}h+28x^{6}h^{2}+\cdot \cdot \cdot +28x^{2}h^{6}+8xh^{7}+h^{8}}}{{h}}}$  
  = $\displaystyle \lim_{{h\rightarrow 0}}^{}$$\displaystyle \left(\vphantom{ 8x^{7}+28x^{6}h+\cdot \cdot \cdot +28x^{2}h^{5}+8xh^{6}+h^{7}}\right.$8x7 +28x6h + ⋅⋅⋅ +28x2h5 +8xh6 + h7$\displaystyle \left.\vphantom{ 8x^{7}+28x^{6}h+\cdot \cdot \cdot +28x^{2}h^{5}+8xh^{6}+h^{7}}\right)$  
  =  8x7  

BITMAPSETProbSolvHint0.2006in0.243in0inq1

2.
Defining g by g(x) = x - f (x)/f′(x), from the Calculus submenu, choose Iterate to get

$\displaystyle \left[\vphantom{ \begin{array}{c} .5 \\ -.75 \\ .29167 \\ -1.5684 \\ -.4654 \\ .84164 \end{array} }\right.$$\displaystyle \begin{array}{c} .5 \\ -.75 \\ .29167 \\ -1.5684 \\ -.4654 \\ .84164 \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{c} .5 \\ -.75 \\ .29167 \\ -1.5684 \\ -.4654 \\ .84164 \end{array} }\right]$

If this result seems to be headed nowhere, it is doing so for good reason. The function f is always positive, so it has no zeroes. Newton's method is searching for something that does not exist.BITMAPSETProbSolvHint0.2006in0.243in0inq2

3.
It is sufficient to find three numbers a, b, and m that satisfy f′(a) = m, f′(b) = m, and ${\frac{{f(b)-f(a)}}{{% b-a}}}$ = m. Put these three equations inside a 3×1 matrix, and from the Solve submenu, choose Exact to get the solutions

m = 4b3 -30b2 + 54b - 18, b = b, a = b

and
m = -8  
a = 5 - ρ  
b = ρ  

where ρ is a root of $\left(\vphantom{ Z^{2}-5Z+1}\right.$Z2 - 5Z + 1$\left.\vphantom{ Z^{2}-5Z+1}\right)$. The first solution is not allowed, because the problem requires a≠b.

Leave the insertion point in the equation Z2 - 5Z + 1 = 0, and from the Solve submenu, choose Exact to get the solutions

Z = $\displaystyle {\frac{{5}}{{2}}}$ + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$  
Z = $\displaystyle {\frac{{5}}{{2}}}$ - $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$  

Choosing ρ = ${\frac{{5}}{{2}}}$ + ${\frac{{1}}{{2}}}$$\sqrt{{21}}$, we have

a = $\displaystyle {\frac{{5}}{{2}}}$ - $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$  
b =  $\displaystyle {\frac{{5}}{{2}}}$ + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$  

Evaluating and expanding,
f (a) =  $\displaystyle \left(\vphantom{ \frac{5}{2}-\frac{1}{2}\sqrt{21}}\right.$$\displaystyle {\frac{{5}}{{2}}}$ - $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$$\displaystyle \left.\vphantom{ \frac{5}{2}-\frac{1}{2}\sqrt{21}}\right)$$\displaystyle \left(\vphantom{ \frac{3}{2}-% \frac{1}{2}\sqrt{21}}\right.$$\displaystyle {\frac{{3}}{{2}}}$ - $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$$\displaystyle \left.\vphantom{ \frac{3}{2}-% \frac{1}{2}\sqrt{21}}\right)$$\displaystyle \left(\vphantom{ -\frac{1}{2}-\frac{1}{2}\sqrt{21}}\right.$ - $\displaystyle {\frac{{1}}{{2}}}$ - $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$$\displaystyle \left.\vphantom{ -\frac{1}{2}-\frac{1}{2}\sqrt{21}}\right)$$\displaystyle \left(\vphantom{ -\frac{7}{2}-\frac{1}{2}\sqrt{21}}\right.$ - $\displaystyle {\frac{{7}}{{2}}}$ - $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$$\displaystyle \left.\vphantom{ -\frac{7}{2}-\frac{1}{2}\sqrt{21}}\right)$  
  =   -21 + 4$\displaystyle \sqrt{{21}}$  

so that
y = f (a) + m(x - a)  
  = -21 + 4$\displaystyle \sqrt{{21}}$ -8(x - $\displaystyle {\frac{{5}}{{2}}}$ + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \sqrt{{21}}$  
  =   - 1 - 8x  

Plot the two curves x$\left(\vphantom{ x-1}\right.$x - 1$\left.\vphantom{ x-1}\right)$$\left(\vphantom{ x-3}\right.$x - 3$\left.\vphantom{ x-3}\right)$$\left(\vphantom{ x-6}\right.$x - 6$\left.\vphantom{ x-6}\right)$ and -1 - 8x, just for visual verification. Use a viewing window with domain interval -1≤x≤6.5 to generate the following picture. BITMAPSETProbSolvHint0.2006in0.243in0inq3dtbpF3in2.0003in0pt

4.
The series is given by $\sqrt{{1-k\sin ^{2}t}}$ =  1 + $\left(\vphantom{ -\frac{1}{2}k}\right.$ - ${\frac{{1}}{{2}}}$k$\left.\vphantom{ -\frac{1}{2}k}\right)$t2 + $\left(\vphantom{ \frac{1}{6}k-\frac{1}{8}k^{2}}\right.$${\frac{{1}}{{6}}}$k - ${\frac{{1}}{{8}}}$k2$\left.\vphantom{ \frac{1}{6}k-\frac{1}{8}k^{2}}\right)$t4 + O$\left(\vphantom{ t^{6}}\right.$t6$\left.\vphantom{ t^{6}}\right)$. Thus, an estimate for E is given by
E $\displaystyle \approx$ $\displaystyle \int_{{0}}^{{\pi /2}}$$\displaystyle \left[\vphantom{ 1+\left( -\frac{1}{2}k\right) t^{2}+\left( \frac{1}{6}k-\frac{1}{8}k^{2}\right) t^{4}}\right.$1 + $\displaystyle \left(\vphantom{ -\frac{1}{2}k}\right.$ - $\displaystyle {\frac{{1}}{{2}}}$k$\displaystyle \left.\vphantom{ -\frac{1}{2}k}\right)$t2 + $\displaystyle \left(\vphantom{ \frac{1}{6}k-\frac{1}{8}k^{2}}\right.$$\displaystyle {\frac{{1}}{{6}}}$k - $\displaystyle {\frac{{1}}{{8}}}$k2$\displaystyle \left.\vphantom{ \frac{1}{6}k-\frac{1}{8}k^{2}}\right)$t4$\displaystyle \left.\vphantom{ 1+\left( -\frac{1}{2}k\right) t^{2}+\left( \frac{1}{6}k-\frac{1}{8}k^{2}\right) t^{4}}\right]$dt  
  =   - $\displaystyle {\frac{{1}}{{1280}}}$k2π5 + $\displaystyle {\frac{{1}}{{2}}}$π + $\displaystyle {\frac{{1}}{{960}}}$kπ5 - $\displaystyle {\frac{{1}}{{48}}}$kπ3  

As a check, k = 1 yields 1.0045 compared with the exact value

$\displaystyle \int_{{0}}^{{\pi /2}}$$\displaystyle \sqrt{{1-\sin ^{2}t}}$dt = 1 

and k = 0 yields ${\frac{{1}}{{2}}}$π, which agrees precisely with BITMAPSETProbSolvHint0.2006in0.243in0inq4

$\displaystyle \int_{{0}}^{{\pi /2}}$dt =  $\displaystyle {\frac{{1}}{{2}}}$π

5.
Compute natural logs on both sides and separate variables to get

$\displaystyle {\frac{{\ln x}}{{x}}}$ = $\displaystyle {\frac{{\ln y}}{{y}}}$

Plot the graph of ${\frac{{\ln x}}{{x}}}$ on the interval 1≤x≤10 and locate the extreme values of ${\frac{{\ln x}}{{x}}}$ by solving

$\displaystyle {\frac{{d}}{{dx}}}$$\displaystyle \left(\vphantom{ \frac{\ln x}{x}}\right.$$\displaystyle {\frac{{\ln x}}{{x}}}$$\displaystyle \left.\vphantom{ \frac{\ln x}{x}}\right)$ = 0

Note that 2 is the only integer between 1 and e, and verify that 4 = 42 is true.BITMAPSETProbSolvHint0.2006in0.243in0inq5

6.
The flow is given by the integral

$\displaystyle \int_{{0}}^{{R}}$α(R2 - r2)2πrdr =  $\displaystyle {\frac{{1}}{{2}}}$απR4

If R is reduced by one-half, then R4 is reduced to ${\frac{{1}}{{16}}}$ of the original amount.BITMAPSETProbSolvHint0.2006in0.243in0inq6

7.
The series expansion is given by

m0$\displaystyle \left(\vphantom{ 1-\frac{v^{2}}{c^{2}}}\right.$1 - $\displaystyle {\frac{{v^{2}}}{{c^{2}}}}$$\displaystyle \left.\vphantom{ 1-\frac{v^{2}}{c^{2}}}\right)^{{-1/2}}_{}$ =  m0 + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle {\frac{{% m_{0}}}{{c^{2}}}}$v2 + $\displaystyle {\frac{{3}}{{8}}}$$\displaystyle {\frac{{m_{0}}}{{c^{4}}}}$v4 + O$\displaystyle \left(\vphantom{ v^{6}}\right.$v6$\displaystyle \left.\vphantom{ v^{6}}\right)$

If ${\frac{{v}}{{c}}}$ is small, then the model

m $\displaystyle \approx$  m0 + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle {\frac{{m_{0}}}{{c^{2}}}}$v2

is useful for estimating the increased mass.BITMAPSETProbSolvHint0.2006in0.243in0inq7

8.
The following sequence requires Evaluate, Simplify, Combine Trig Functions, Factor, and Simplify. BITMAPSETProbSolvHint0.2006in0.243in0inq8
$\displaystyle \dint$2xcos bxdx = $\displaystyle {\dfrac{{\frac{\ln 2}{b^{2}+\ln ^{2}2}2^{x}+2^{1+x}% \frac{b}{b^... ...{b^{2}+\ln ^{2}2}% 2^{x}\tan ^{2}\dfrac{1}{2}bx}}{{1+\tan ^{2}\frac{1}{2}bx}}}$  
  = $\displaystyle {\dfrac{{2^{1+x}\ln 2\cos ^{2}\frac{1}{2}bx+2^{1+x}b\sin \frac{1}{2}bx\cos \frac{1}{2}bx-\left( \ln 2\right) 2^{x}}}{{b^{2}+\ln ^{2}2}}}$  
  = $\displaystyle {\dfrac{{1}}{{2\left( b^{2}+\ln ^{2}2\right) }}}$21+xln 2 cos bx + $\displaystyle {\dfrac{{1}}{{% 2\left( b^{2}+\ln ^{2}2\right) }}}$21+xln 2  
         + $\displaystyle {\dfrac{{1}}{{2\left( b^{2}+\ln ^{2}2\right) }}}$21+xb sin bx - $\displaystyle {\dfrac{{1}}{{% b^{2}+\ln ^{2}2}}}$2xln 2  
  = $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle {\dfrac{{2^{1+x}\ln 2\cos bx+2^{1+x}\ln 2+2^{1+x}b\sin bx-2\left( \ln 2\right) 2^{x}}}{{b^{2}+\ln ^{2}2}}}$  
  = 2x$\displaystyle {\dfrac{{\cos bx\ln 2+b\sin bx}}{{b^{2}+\ln ^{2}2}}}$  

9.
The integral

$\displaystyle \int_{{-\pi }}^{{\pi }}$$\displaystyle {\frac{{1+\sin x}}{{\left( x-\cos x\right) ^{2}}}}$dx

is improper, because x - cos x = 0 has a root ( $\approx$ .73909) between - π and π . Evaluate gives
$\displaystyle \int_{{-\pi }}^{{.73909}}$$\displaystyle {\frac{{1+\sin x}}{{\left( x-\cos x\right) ^{2}}}}$dx =   -1.2277×105 - $\displaystyle {\frac{{1}}{{\pi -1}}}$  
$\displaystyle \int_{{.73909}}^{{\pi }}$$\displaystyle {\frac{{1+\sin x}}{{\left( x-\cos x\right) ^{2}}}}$dx =   - $\displaystyle {\frac{{1}}{{\pi +1}}}$ +1.2277×105  

Change Digits Used in Display in the Settings dialog box to 10. Solving cos x = x numerically gives x = .7390851332. Using this as a limit, Evaluate gives

$\displaystyle \int_{{-\pi }}^{{.7390851332}}$$\displaystyle {\frac{{1+\sin x}}{{\left( x-\cos x\right) ^{2}}}}$dx =  3.941192327×1010 - $\displaystyle {\frac{{1}}{{\pi -1}}}$  
$\displaystyle \int_{{.7390851332}}^{{\pi }}$$\displaystyle {\frac{{1+\sin x}}{{\left( x-\cos x\right) ^{2}}}}$dx =   - $\displaystyle {\frac{{1}}{{\pi +1}}}$ -3.941192327×1010  

providing some evidence that both integrals diverge.BITMAPSETProbSolvHint0.2006in0.243in0inq9

10.
Scientific Notebook returns

$\displaystyle \lim_{{h\rightarrow 0^{+}}}^{}$$\displaystyle \int_{{0}}^{{\infty }}$sin(x1+h) dx = 1

This result is reasonable, because the integral f (h) = $\int_{{0}}^{{\infty }}$sin(x1+h) dx can be viewed as a convergent alternating series for h > 0, and

g(y) = $\displaystyle \int_{{0}}^{{y}}$sin x dx = 1 - cos y

ranges in value between 0 and 2, with an average value of 1. BITMAPSETProbSolvHint0.2006in0.243in0inq10

11.
We need to show that for each of the three functions f (x) = x3, f (x) = xex, and f (x) = sin2x cos x,BITMAPSETProbSolvHint0.2006in0.243in0inq11

a.
If g is defined by g(x) = $\int_{{a}}^{{x}}$f (t)dt for x∈$\left[\vphantom{ a,b}\right.$a, b$\left.\vphantom{ a,b}\right]$, then g′(x) = f (x), and

b.
If F is any antiderivative of f, then. $\int_{{a}}^{{b}}$f (x)dx = F(b) - F(a).

For f (x) = x3,

g(x) = $\displaystyle \int_{{a}}^{{x}}$t3dt = $\displaystyle {\frac{{1}}{{4}}}$x4 - $\displaystyle {\frac{{1}}{{4}}}$a4

and

g′(x) = $\displaystyle {\frac{{d}}{{dx}}}$$\displaystyle \left(\vphantom{ \frac{1}{4}x^{4}-\frac{1}{4}a^{4}}\right.$$\displaystyle {\frac{{1}}{{4}}}$x4 - $\displaystyle {\frac{{1}}{{4}}}$a4$\displaystyle \left.\vphantom{ \frac{1}{4}x^{4}-\frac{1}{4}a^{4}}\right)$ = x3

The antiderivatives of f are of the form

F(x) = $\displaystyle \int$x3dx = $\displaystyle {\frac{{1}}{{4}}}$x4 + C

for different constants C. Now

F(b) - F(x) = $\displaystyle \left[\vphantom{ \allowbreak \frac{1}{4}x^{4}+C}\right.$$\displaystyle {\frac{{1}}{{4}}}$x4 + C$\displaystyle \left.\vphantom{ \allowbreak \frac{1}{4}x^{4}+C}\right]_{{x=a}}^{{x=b}}$ = $\displaystyle {\frac{{1}}{{4}}}$b4 - $\displaystyle {\frac{{1}}{{4}}}$a4

which is the same as

$\displaystyle \int_{{a}}^{{b}}$x3dx = $\displaystyle {\frac{{1}}{{4}}}$b4 - $\displaystyle {\frac{{1}}{{4}}}$a4

For f (x) = xex,

g(x) = $\displaystyle \int_{{a}}^{{x}}$tetdt = xex - ex - aea + ea

and

g′(x) = $\displaystyle {\frac{{d}}{{dx}}}$$\displaystyle \left(\vphantom{ xe^{x}-e^{x}-ae^{a}+e^{a}}\right.$xex - ex - aea + ea$\displaystyle \left.\vphantom{ xe^{x}-e^{x}-ae^{a}+e^{a}}\right)$ = xex

The antiderivatives of f are of the form

F(x) = $\displaystyle \int$xexdx = xex - ex + C

for different constants C. Now

F(b) - F(x) = $\displaystyle \left[\vphantom{ xe^{x}-e^{x}+C}\right.$xex - ex + C$\displaystyle \left.\vphantom{ xe^{x}-e^{x}+C}\right]_{{x=a}}^{{x=b}}$ = beb - eb - aea + ea

which is the same as

$\displaystyle \int_{{a}}^{{b}}$xexdx = beb - eb - aea + ea

For f (x) = sin2x cos x,

g(x) = $\displaystyle \int_{{a}}^{{x}}$sin2t cos tdt = $\displaystyle {\frac{{1}}{{3}}}$sin x - $\displaystyle {\frac{{1}}{{3}% }}$sin x cos2x - $\displaystyle {\frac{{1}}{{3}}}$sin a + $\displaystyle {\frac{{1}}{{3}}}$sin a cos2a

and
g′(x) = $\displaystyle {\frac{{d}}{{dx}}}$$\displaystyle \left(\vphantom{ \frac{1}{3}\sin x-\frac{1}{3}\sin x\cos ^{2}x-\frac{1}{3}\sin a+\frac{1}{3}\sin a\cos ^{2}a}\right.$$\displaystyle {\frac{{1}}{{3}}}$sin x - $\displaystyle {\frac{{1}}{{3}}}$sin x cos2x - $\displaystyle {\frac{{1}}{{3}}}$sin a + $\displaystyle {\frac{{1}}{{3}}}$sin a cos2a$\displaystyle \left.\vphantom{ \frac{1}{3}\sin x-\frac{1}{3}\sin x\cos ^{2}x-\frac{1}{3}\sin a+\frac{1}{3}\sin a\cos ^{2}a}\right)$  
  = cos x - cos3x = $\displaystyle \left(\vphantom{ 1-\cos ^{2}x}\right.$1 - cos2x$\displaystyle \left.\vphantom{ 1-\cos ^{2}x}\right)$$\displaystyle \left(\vphantom{ \cos x}\right.$cos x$\displaystyle \left.\vphantom{ \cos x}\right)$ = sin2x cos x  

The antiderivatives of f are of the form

F(x) = $\displaystyle \int$sin2x cos xdx = $\displaystyle {\frac{{1}}{{3}}}$sin3x + C

for different constants C. Now

F(b) - F(x) = $\displaystyle \left[\vphantom{ \frac{1}{3}\sin ^{3}x+C}\right.$$\displaystyle {\frac{{1}}{{3}}}$sin3x + C$\displaystyle \left.\vphantom{ \frac{1}{3}\sin ^{3}x+C}\right]_{{x=a}}^{{x=b}}$ = $\displaystyle {\frac{{1}}{{3}}}$sin3b - $\displaystyle {\frac{{1}}{{3}}}$sin3a

while

$\displaystyle \int_{{a}}^{{b}}$sin2x cos xdx = $\displaystyle {\frac{{1}}{{3}}}$sin b - $\displaystyle {\frac{{1}}{{3}}}$sin b cos2b - $\displaystyle {\frac{{1}}{{3}}}$sin a + $\displaystyle {\frac{{1}}{{3}}}$sin a cos2a

Applying Combine + Trig Functions to both of these expression shows that they are equal.
$\displaystyle {\frac{{1}}{{3}}}$sin b - $\displaystyle {\frac{{1}}{{3}}}$sin b cos2b - $\displaystyle {\frac{{1}}{{3}}}$sin a + $\displaystyle {\frac{{1}}{{3}}}$sin a cos2a = $\displaystyle {\frac{{1}}{{4}}}$sin b - $\displaystyle {\frac{{1}}{{12}% }}$sin 3b - $\displaystyle {\frac{{1}}{{4}}}$sin a + $\displaystyle {\frac{{1}}{{12}}}$sin 3a  
$\displaystyle {\frac{{1}}{{3}}}$sin3b - $\displaystyle {\frac{{1}}{{3}}}$sin3a = $\displaystyle {\frac{{1}}{{4}}}$sin b - $\displaystyle {\frac{{1}}{{12}}}$sin 3b - $\displaystyle {\frac{{1}}{{4}}}$sin a + $\displaystyle {\frac{{1}}{{12}}}$sin 3a  

Another way to demonstrate this equality is to replace cos2x by - sin2x and apply Expand to get

$\displaystyle {\frac{{1}}{{3}}}$sin b - $\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \left(\vphantom{ \sin b}\right.$sin b$\displaystyle \left.\vphantom{ \sin b}\right)$$\displaystyle \left(\vphantom{ 1-\sin ^{2}b}\right.$1 - sin2b$\displaystyle \left.\vphantom{ 1-\sin ^{2}b}\right)$ - $\displaystyle {\frac{{1}}{{3}}}$sin a + $\displaystyle {\frac{{1}}{{3}}}$$\displaystyle \left(\vphantom{ \sin a}\right.$sin a$\displaystyle \left.\vphantom{ \sin a}\right)$$\displaystyle \left(\vphantom{ 1-\sin ^{2}a}\right.$1 - sin2a$\displaystyle \left.\vphantom{ 1-\sin ^{2}a}\right)$ = $\displaystyle {\frac{{1}}{{3}}}$sin3b - $\displaystyle {\frac{{1}}{{3}}}$sin3a



Subsections