Chapter 2

$12$ units²

$\frac{3}{10}$ unit²

$2+2\sqrt{2}$ units²

$\frac{5}{3}$ units²

$\frac{5}{3}$ units²

$\frac{\pi }{2}$

$8\pi$ units³

$21\pi$ units³

$\frac{10\pi }{3}$ units³

$60\pi$ units³

$\frac{15\pi }{2}$ units³

$8\pi$ units³

$12\pi$ units³

$\frac{11\pi }{6}$ units³

$\frac{\pi }{6}$ units³

Use the method of washers; $V={\displaystyle {\int }_{\mathrm{-1}}^1\pi }\left[{\left(2-x^2\right)}^2-{\left(x^2\right)}^2\right]dx$

$\frac{1}{6}\left(5\sqrt{5}-1\right)\approx 1.697$

$\text{Arc Length}\approx 3.8202$

$\text{Arc Length}=3.15018$

$\frac{\pi }{6}\left(5\sqrt{5}-3\sqrt{3}\right)\approx 3.133$

$12\pi$

$70\text{/}3$

$24\pi$

$8$ ft-lb

Approximately $\mathrm{43,255.2}$ ft-lb

$\mathrm{156,800}$ N

Approximately 7,164,520,000 lb or 3,582,260 t

$M=24,\bar{x}=\frac{2}{5}\text{m}$

$\left(\mathrm{-1},\mathrm{-1}\right)$ m

The centroid of the region is $\left(3\text{/}2,6\text{/}5\right).$

The centroid of the region is $\left(1,13\text{/}5\right).$

The centroid of the region is $\left(0,2\text{/}5\right).$

$6{\pi }^2$ units³

1. $\frac{d}{dx}\text{ln}\left(2x^2+x\right)=\frac{4x+1}{2x^2+x}$
2. $\frac{d}{dx}{\left(\text{ln}\left(x^3\right)\right)}^2=\frac{6\text{ln}\left(x^3\right)}{x}$

${\displaystyle \int \frac{x^2}{x^3+6}}dx=\frac{1}{3}\text{ln}\left|x^3+6\right|+C$

$4\text{ln}2$

1. $\frac{d}{dx}\left(\frac{e^{x^2}}{e^{5x}}\right)=e^{x^2-5x}\left(2x-5\right)$
2. $\frac{d}{dt}{\left(e^{2t}\right)}^3=6e^{6t}$

${\displaystyle \int \frac{4}{e^{3x}}dx}=-\frac{4}{3}{e}^{\mathrm{-3}x}+C$

1. $\frac{d}{dt}{4}^{t^4}=4^{t^4}\left(\text{ln}4\right)\left(4t^3\right)$
2. $\frac{d}{dx}{\text{log}}_3\left(\sqrt{x^2+1}\right)=\frac{x}{\left(\text{ln}3\right)\left(x^2+1\right)}$

${\displaystyle \int x^2{2}^{x^3}dx}=\frac{1}{3\text{ln}2}{2}^{x^3}+C$

There are $\mathrm{81,377,396}$ bacteria in the population after $4$ hours. The population reaches $100$ million bacteria after $244.12$ minutes.

At $5\text{\%}$ interest, she must invest $\text{\$}\mathrm{223,130.16}.$ At $6\text{\%}$ interest, she must invest $\text{\$}\mathrm{165,298.89}.$

$38.90$ months

The coffee is first cool enough to serve about $3.5$ minutes after it is poured. The coffee is too cold to serve about $7$ minutes after it is poured.

A total of $94.13$ g of carbon remains. The artifact is approximately $\mathrm{13,300}$ years old.

1. $\frac{d}{dx}\left(\text{tanh}\left(x^2+3x\right)\right)=\left({\text{sech}}^2\left(x^2+3x\right)\right)\left(2x+3\right)$
2. $\frac{d}{dx}\left(\frac{1}{{\left(\text{sinh}x\right)}^2}\right)=\frac{d}{dx}{\left(\text{sinh}x\right)}^{\mathrm{-2}}=\mathrm{-2}{\left(\text{sinh}x\right)}^{\mathrm{-3}}\text{cosh}x$

1. ${\displaystyle \int {\text{sinh}}^{3}x\text{cosh}xd}x=\frac{{\text{sinh}}^{4}x}{4}+C$
2. ${\displaystyle \int {\text{sech}}^2\left(3x\right)d}x=\frac{\text{tanh}\left(3x\right)}{3}+C$

1. $\frac{d}{dx}\left({\text{cosh}}^{\mathrm{-1}}\left(3x\right)\right)=\frac{3}{\sqrt{9x^2-1}}$
2. $\frac{d}{dx}{\left({\text{coth}}^{\mathrm{-1}}x\right)}^3=\frac{3{\left({\text{coth}}^{\mathrm{-1}}x\right)}^2}{1-x^2}$

1. ${\displaystyle \int \frac{1}{\sqrt{x^2-4}}d}x={\text{cosh}}^{\mathrm{-1}}\left(\frac{x}{2}\right)+C$
2. ${\displaystyle \int \frac{1}{\sqrt{1-e^{2x}}}d}x=\text{−}{\text{sech}}^{\mathrm{-1}}\left(e^x\right)+C$

$52.95\text{ft}$

$\frac{32}{3}$

$\frac{13}{12}$

$36$

243 square units

4

$\frac{2{\left(e-1\right)}^2}{e}$

$\frac{1}{3}$

$\frac{34}{3}$

$\frac{5}{2}$

$\frac{1}{2}$

$\frac{9}{2}$

$\frac{9}{2}$

$\frac{3\sqrt{3}}{2}$

$e\mathrm{-2}$

$\frac{27}{4}$

$\frac{4}{3}-\text{ln}(3)$

$\frac{1}{2}$

$\frac{1}{2}$

$\mathrm{-2}\left(\sqrt{2}-\pi \right)$

$1.067$

$1.605$

$7.523$

$\frac{3\pi -4}{12}$

$1.429$

$\text{\$}\mathrm{33,333.33}$ total profit for $200$ cell phones sold

$3.32$ mi represents how far ahead the hare is from the tortoise

$\frac{343}{24}$

$4\sqrt{3}$

$\pi -\frac{32}{25}$

8 units³

$\frac{32}{3\sqrt{2}}$ units³

$\frac{7}{24}\pi r^{2}h$ units³

$\frac{\pi }{24}$ units³

$2$ units³

$\frac{\pi }{240}$ units³

$\frac{4096\pi }{5}$ units³

$\frac{8\pi }{9}$ units³

$\frac{\pi }{2}$ units³

$207\pi$ units³

$\frac{4\pi }{5}$ units³

$\frac{16\pi }{3}$ units³

$\pi$ units³

$\frac{16\pi }{3}$ units³

$\frac{72\pi }{5}$ units³

$\frac{108\pi }{5}$ units³

$\frac{3\pi }{10}$ units³

$2\sqrt{6}\pi$ units³

$9\pi$ units³

$\frac{\pi }{20}\left(75-4{\text{ln}}^5\left(2\right)\right)$ units³

$\frac{m^2\pi }{3}\left(b^3-a^3\right)$ units³

$\frac{4a^{2}b\pi }{3}$ units³

$2{\pi }^2$ units³

$\frac{2a{b}^2\pi }{3}$ units³

$V=\frac{2\mathrm{\pi }}{3}(2r+h){\left(r-h\right)}^2$ units³

$\frac{\pi }{3}\left(h+R\right){\left(h-2R\right)}^2$ units³

$54\pi$ units³

$81\pi$ units³

$\frac{512\pi }{7}$ units³

$2\pi$ units³

$\frac{2\pi }{3}$ units³

$2\pi$ units³

$\frac{4\pi }{5}$ units³

$\frac{64\pi }{3}$ units³

$\frac{32\pi }{5}$ units³

$\frac{7\pi }{6}$

$48\pi$

$\frac{308\pi }{5}$

$\frac{512\pi }{7}$

$\frac{8\pi }{5}$

$\frac{28\pi }{15}$ units³

$\frac{3\pi }{10}$ units³

$\mathrm{\pi }\left(\frac{6·2^{2/3}}{5}-\frac{11}{10}\right)=\frac{\mathrm{\pi }}{10}\left(12·2^{2/3}-11\right)\approx 2.5286$ units³

$0.9876$ units³

$3\sqrt{2}$ units³

$\frac{496\pi }{15}$ units³

$\frac{398\pi }{15}$ units³

$15.9074$ units³

$\frac{1}{3}\pi r^{2}h$ units³

$\pi r^{2}h$ units³

$\pi a^2$ units³

$2\sqrt{26}$

$2\sqrt{17}$

$\frac{\pi }{6}\left(17\sqrt{17}-5\sqrt{5}\right)$

$\frac{13\sqrt{13}-8}{27}$

$\frac{4}{3}$

$2.0035$

$\frac{123}{32}$

$10$

$\frac{20}{3}$

$\frac{1}{675}\left(229\sqrt{229}-8\right)$

$\frac{1}{8}\left(4\sqrt{5}+\text{ln}\left(9+4\sqrt{5}\right)\right)$

$1.201$

$15.2341$

$\frac{49\pi }{3}$

$70\pi \sqrt{2}$

$8\pi$

$120\pi \sqrt{26}$

$\frac{\pi }{6}\left(17\sqrt{17}-1\right)$

$9\sqrt{2}\pi$

$\frac{10\sqrt{10}\pi }{27}\left(73\sqrt{73}-1\right)$

$25.645$

$\mathrm{\pi }(2+\mathrm{\pi })$

$10.5017$

$24$ ft

$2$

Answers may vary

For more information, look up Gabriel’s Horn.

$150$ ft-lb

$200\text{J}$

$1$ J

$\frac{39}{2}$

$\text{ln}\left(243\right)$

$\frac{332\pi }{15}$

$100\pi$

$20\pi \sqrt{15}$

$6$ J

$5$ cm

$36$ J

$\mathrm{18,750}$ ft-lb

Weight $=\frac{32}{3}\times {10}^9\text{lb}$

Work $=\frac{4}{3}\times {10}^{12}\text{ft-lb}$

$9.71\times {10}^8\text{N m}$

a. $\mathrm{3,000,000}$ lb, b. $\mathrm{750,000}$ lbs

$23.25\pi$ million ft-lb

$\frac{A\rho H^2}{2}$

Answers may vary

$\frac{5}{4}$

$\left(\frac{2}{3},\frac{2}{3}\right)$

$\left(\frac{7}{4},\frac{3}{2}\right)$

$\frac{3L}{4}$

$\frac{\pi }{2}$

$\frac{e^2+1}{e^2-1}$

$\frac{{\pi }^2-4}{\pi }$

$\frac{1}{4}\left(1+e^2\right)$

$\left(\frac{a}{3},\frac{b}{3}\right)$

$\left(0,\frac{\pi }{8}\right)$

$(0,3)$

$\left(0,\frac{4}{\pi }\right)$

$\left(\frac{5}{8},\frac{1}{3}\right)$

$\frac{2m\pi }{3}$

$\pi a^{2}b$

$\left(\frac{4}{3\pi },\frac{4}{3\pi }\right)$

$\left(\frac{1}{2},\frac{2}{5}\right)$

$\left(0,\frac{28}{9\pi }\right)$

Center of mass: $(\frac{3a}{8},\frac{2a^2}{5}),$ volume: $\frac{\pi a^4}{2}$

Volume: $2{\pi }^2{a}^2\left(b+a\right)$

$\frac{1}{x}$

$-\frac{1}{x{(\text{ln}x)}^2}$

$\text{ln}\left(x+1\right)+C$

$\text{ln}\left(x\right)+1$

$\text{cot}\left(x\right)$

$\frac{7}{x}$

$\text{csc}\left(x\right)\text{sec}x$

$\mathrm{-2}\text{tan}x$

$\frac{1}{2}\text{ln}\left(\frac{5}{3}\right)$

$2-\frac{1}{2}\text{ln}\left(5\right)$

$\frac{1}{\text{ln}\left(2\right)}-1$

$\frac{1}{2}\text{ln}\left(2\right)$

$\frac{1}{3}{\left(\text{ln}x\right)}^3$

$\frac{2x^3}{\sqrt{x^2+1}\sqrt{x^2-1}}$

$x^{\mathrm{-2}-(1\text{/}x)}\left(\text{ln}x-1\right)$

$e{x}^{e-1}$

$1$

$-\frac{1}{x^2}$

$\pi -\text{ln}\left(2\right)$

$\frac{1}{x}$

$e^5-6{\text{units}}^2$

$\text{ln}\left(4\right)-1{\text{units}}^2$

$2.8656$

$3.1502$

True

True; $k=\frac{\text{ln}\left(2\right)}{t}$

$20$ hours

No. The relic is approximately $871$ years old.

$73$ years

$5$ days $6$ hours $27$ minutes

$12$

$8.618\text{\%}$

$\text{\$}6766.76$

$9$ hours $13$ minutes

$\mathrm{239,179}$ years

$P\prime \left(t\right)=43e^{0.01604t}.$ The population is always increasing.

The population reaches $10$ billion people in $2032.$

$P\prime \left(t\right)=2.259e^{0.06407t}.$ The population is always increasing.

$e^x\text{and}{e}^{\text{−}x}$

Answers may vary

Answers may vary

Answers may vary

$3\text{sinh}\left(3x+1\right)$

$\text{−}\text{tanh}\left(x\right)\text{sech}\left(x\right)$

$4\text{cosh}\left(x\right)\text{sinh}\left(x\right)$

$\frac{x{\text{sech}}^2\left(\sqrt{x^2+1}\right)}{\sqrt{x^2+1}}$

$6{\text{sinh}}^5\left(x\right)\text{cosh}\left(x\right)$

$\frac{1}{2}\text{sinh}\left(2x+1\right)+C$

$\frac{1}{2}\text{sinh}\left(x^2\right)+C$

$\frac{1}{3}{\text{cosh}}^3\left(x\right)+C$

$\text{ln}\left(1+\text{cosh}(x)\right)+C$

$\text{cosh}\left(x\right)+\text{sinh}\left(x\right)+C$

$\frac{4}{1-16x^2}$

$\frac{\text{sinh}\left(x\right)}{\sqrt{{\text{cosh}}^2\left(x\right)+1}}$

$\text{−}\text{csc}\left(x\right)$

$-\frac{1}{\left(x^2-1\right){\text{tanh}}^{\mathrm{-1}}\left(x\right)}$

$\frac{1}{a}{\text{tanh}}^{\mathrm{-1}}\left(\frac{x}{a}\right)+C$

$\sqrt{x^2+1}+C$

${\text{cosh}}^{\mathrm{-1}}\left(e^x\right)+C$

Answers may vary

$37.30$

$y=\frac{1}{c}\text{cosh}(cx)$

$\mathrm{-0.521095}$

$10$

False

False

$16\sqrt{3} {\mathrm{units}}^3$

$\frac{162\pi }{5}$

a. $4,$ b. $\frac{128\pi }{7},$ c. $\frac{64\pi }{5}$