# integral

• May 11th 2009, 07:33 PM
jahichuanna
integral
1) a. $1$ b. $1.5$ c. no d. discontinuous @ $|g-1|<2$ and $g(3)$
e. the discontinuity at $g(3)$ can be removed by assigning $g(3)=1$. All the other discontinuities can be removed by extending the domain from $-1\leq x\leq3$ to $-\infty\leq x\leq\infty$.

2) a. asymptotes @ $x=\{-2,0\}$
b. @ $-2, \lim_{x \to -2}\frac{x-1}{x^2(x+2)}=\frac{-3}{0} \implies \lim_{x \to -2}\frac{x-1}{x^3+x2} = \frac{-3}{0} \implies -\infty$
@ $-2, \lim_{x \to 0}\frac{x-1}{x^2(x+2)}=\frac{-1}{0} \implies \lim_{x \to 0}\frac{x-1}{x^3+x2} = \frac{-1}{0} \implies -\infty$

3) $14\ln\left(-\cos\left(\frac{x}{2}+1.67\right)\right)+34 \implies 14/\left(\cos\left(\frac{x}{2}+1.67\right)\right) * \sin\left(\frac{x}{2} + 1.67\right) * \frac{1}{2}$ $= -7\tan\left(\frac{x}{2}+1.67\right)$
a. @ $t=0, -7\tan\left(1.67\right) = 70.33$
@ $t=5, -7\tan\left(\frac{5}{2}+1.67\right)= -11.6$
b. i $0=-7\tan\left(\frac{x}{2}+1.67\right) \implies \tan^{-1}(0)=\frac{x}{2}+1.67 \implies \pi-1.67$ $=\frac{x}{2} \implies 2.94 \text{ seconds}\approx x$
ii $0=14\ln\left(-\cos\left(\frac{x}{2}+1.67\right)\right)+34 \implies \ln\left(-\cos\left(\frac{x}{2}+1.67\right)\right)=-\frac{17}{7}$ $\implies
\cos\left(\frac{x}{2} +1.67\right)$
$=-e^{\frac{-17}{7}} \implies \frac{x}{2}+1.67=\pi\pm1.483$ $\implies x=2\left(\pi\pm1.483-1.67\right) \implies
x=\{5.91, -.023\} \implies x=5.91\text{ seconds}$

4) a. $V=\frac{\pi h}{3}\left(r_2^2 + r_2r_1 +r_1^2\right)$
b. $\text{similar triangles}:\frac{r}{h} = \frac{6}{8} \therefore r=\frac{3h}{4}\text{ and }h=\frac{4r}{3}$
i $V=\frac{\pi}{3}\left(\frac{3h}{4}\right)^2h \implies \frac{dV}{dt}=\frac{9\pi h^2}{16}\frac{dh}{dt}$
$r = 5 \therefore h=\frac{20}{3}$
$15=\frac{9\pi\left(\frac{20}{3}\right)^2}{16}\frac {dh}{dt} \implies \frac{dh}{dt}=\frac{3}{5\pi}$
ii $V=\frac{\pi}{3}r^2\left(\frac{4r}{3}\right)=\frac{ 4\pi r^3}{9} \implies \frac{dV}{dt}=\frac{4\pi r^2}{3}\frac{dr}{dt}$
$15=\frac{4\pi(5)^2}{3}\frac{dr}{dt} \implies \frac{dr}{dt}=\frac{9}{20\pi}$

5) i $y=\frac{11^3\cos\left(\frac{11}{2}\right)}{2}-11 = \ 460,000$
ii $y=\int^{11.5}_0 \frac{x^3\cos\left(\frac{x}{2}\right)}{2}-x \,dx \implies \text{(integration by parts 3 times)}$ $\implies \left.x^3\sin\left(\frac{x}{2}\right)+6x^2\cos\lef t(\frac{x}{2}\right)-24x\sin\left(\frac{x}{2}\right)-48cos\left(\frac{x}{2}\right)-\frac{x^2}{2}\right|^{11}_0$ $= -332.833- -48 = \ -284,833$ Meaning that they lost $284,833 over the first 11 years. 6) $h=\frac{\pi-0}{4}, T_4=\frac{h}{2}\left(y_0+2y_1+2y_2+2y_3+y_4\right)$ $T: \frac{\pi}{8}\left(0+\frac{\pi^2\sqrt{2}}{16}+\fra c{\pi^2}{2}+\frac{9\pi^2\sqrt{2}}{16}+0\right)$ $=\frac{\pi}{8}\left(\frac{10\pi^2\sqrt{2}+8\pi^2}{ 16}\right)$ $=\frac{5\pi^3\sqrt{2}+4\pi^3}{64}$ $\int^{\pi}_0 x^2\sin x(x) \,dx \implies \text{(integration by parts 2 times)} 2\pi-1$ $\frac{\frac{5\pi^3\sqrt{2}+4\pi^3}{64}-(2\pi-1)}{2\pi-1}=-.339\%$ 7) a. $100 N * 40 m = 4000 J$ b. $.8 \frac{N}{m} \left(40 m-x\right)=32 N -x \implies \int^{40}_0 32-.8x \,dx = \left.32x-.4x^2 \right|^{40}_0$ $= 32(40)-.4(40)^2 - (32(0)-.4(0)^2) = 640 J$ c. $4000 J + 640 J = 4640 J$ 8) $\lim_{x \to \pi} \frac{\ln\left(\sin\left(\frac{x}{2}\right)\right) }{x} = \frac{0}{0} \implies \lim_{x \to \pi} \frac{\frac{1}{\sin\left(\frac{x}{2}\right)*\cos\l eft(\frac{x}{2}\right)*\frac{1}{2}}}{1} = 0/1 = 0$ 9) $x^3-2x^2+5 \implies 3x^2-4x \implies 6x-4 \implies 6$ $(-1)^3-2(-1)^2+5 \implies (3(-1)^2-4(-1))(x+1) \implies \frac{(6(-1)-4)(x+1)^2}{2!}$ $\implies \frac{6(x+1)^3}{3!} = 2 + 7(x+1) - 5(x+1)^2 + (x+1)^3$ $\text{General term for }n\geq4 = 0$ 10) $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{1}{2}\left(r_2^2 - r_1^2\right) \,d\theta \implies 2\int^{0}_{-\frac{\pi}{2}} \frac{1}{2}\left(2cos\theta+cos^2\theta\right) \,d\theta$ $\implies 2\left.\left(2sin\theta-\frac{1}{3}sin^3\theta\right)\right|^{\frac{\pi}{2 }}_0 = 2-\frac{\pi}{4}$ • May 16th 2009, 01:50 AM CaptainBlack Quote: Originally Posted by jahichuanna 1) a. $1$ b. $1.5$ c. no d. discontinuous @ $|g-1|<2$ and $g(3)$ e. the discontinuity at $g(3)$ can be removed by assigning $g(3)=1$. All the other discontinuities can be removed by extending the domain from $-1\leq x\leq3$ to $-\infty\leq x\leq\infty$. 2) a. asymptotes @ $x=\{-2,0\}$ b. @ $-2, \lim_{x \to -2}\frac{x-1}{x^2(x+2)}=\frac{-3}{0} \implies \lim_{x \to -2}\frac{x-1}{x^3+x2} = \frac{-3}{0} \implies -\infty$ @ $-2, \lim_{x \to 0}\frac{x-1}{x^2(x+2)}=\frac{-1}{0} \implies \lim_{x \to 0}\frac{x-1}{x^3+x2} = \frac{-1}{0} \implies -\infty$ 3) $14\ln\left(-\cos\left(\frac{x}{2}+1.67\right)\right)+34 \implies 14/\left(\cos\left(\frac{x}{2}+1.67\right)\right) * \sin\left(\frac{x}{2} + 1.67\right) * \frac{1}{2}$ $= -7\tan\left(\frac{x}{2}+1.67\right)$ a. @ $t=0, -7\tan\left(1.67\right) = 70.33$ @ $t=5, -7\tan\left(\frac{5}{2}+1.67\right)= -11.6$ b. i $0=-7\tan\left(\frac{x}{2}+1.67\right) \implies \tan^{-1}(0)=\frac{x}{2}+1.67 \implies \pi-1.67$ $=\frac{x}{2} \implies 2.94 \text{ seconds}\approx x$ ii $0=14\ln\left(-\cos\left(\frac{x}{2}+1.67\right)\right)+34 \implies \ln\left(-\cos\left(\frac{x}{2}+1.67\right)\right)=-\frac{17}{7}$ $\implies$ $ \cos\left(\frac{x}{2} +1.67\right)" alt=" \cos\left(\frac{x}{2} +1.67\right)" /> $=-e^{\frac{-17}{7}} \implies \frac{x}{2}+1.67=\pi\pm1.483$ $\implies x=2\left(\pi\pm1.483-1.67\right) \implies$ $ x=\{5.91, -.023\} \implies x=5.91\text{ seconds}" alt=" x=\{5.91, -.023\} \implies x=5.91\text{ seconds}" /> 4) a. $V=\frac{\pi h}{3}\left(r_2^2 + r_2r_1 +r_1^2\right)$ b. $\text{similar triangles}:\frac{r}{h} = \frac{6}{8} \therefore r=\frac{3h}{4}\text{ and }h=\frac{4r}{3}$ i $V=\frac{\pi}{3}\left(\frac{3h}{4}\right)^2h \implies \frac{dV}{dt}=\frac{9\pi h^2}{16}\frac{dh}{dt}$ $r = 5 \therefore h=\frac{20}{3}$ $15=\frac{9\pi\left(\frac{20}{3}\right)^2}{16}\frac {dh}{dt} \implies \frac{dh}{dt}=\frac{3}{5\pi}$ ii $V=\frac{\pi}{3}r^2\left(\frac{4r}{3}\right)=\frac{ 4\pi r^3}{9} \implies \frac{dV}{dt}=\frac{4\pi r^2}{3}\frac{dr}{dt}$ $15=\frac{4\pi(5)^2}{3}\frac{dr}{dt} \implies \frac{dr}{dt}=\frac{9}{20\pi}$ 5) i $y=\frac{11^3\cos\left(\frac{11}{2}\right)}{2}-11 = \ 460,000$ ii $y=\int^{11.5}_0 \frac{x^3\cos\left(\frac{x}{2}\right)}{2}-x \,dx \implies \text{(integration by parts 3 times)}$ $\implies \left.x^3\sin\left(\frac{x}{2}\right)+6x^2\cos\lef t(\frac{x}{2}\right)-24x\sin\left(\frac{x}{2}\right)-48cos\left(\frac{x}{2}\right)-\frac{x^2}{2}\right|^{11}_0$ $= -332.833- -48 = \ -284,833$ Meaning that they lost$284,833 over the first 11 years.

6) $h=\frac{\pi-0}{4}, T_4=\frac{h}{2}\left(y_0+2y_1+2y_2+2y_3+y_4\right)$
$T: \frac{\pi}{8}\left(0+\frac{\pi^2\sqrt{2}}{16}+\fra c{\pi^2}{2}+\frac{9\pi^2\sqrt{2}}{16}+0\right)$ $=\frac{\pi}{8}\left(\frac{10\pi^2\sqrt{2}+8\pi^2}{ 16}\right)$ $=\frac{5\pi^3\sqrt{2}+4\pi^3}{64}$
$\int^{\pi}_0 x^2\sin x(x) \,dx \implies \text{(integration by parts 2 times)} 2\pi-1$
$\frac{\frac{5\pi^3\sqrt{2}+4\pi^3}{64}-(2\pi-1)}{2\pi-1}=-.339\%$

7) a. $100 N * 40 m = 4000 J$
b. $.8 \frac{N}{m} \left(40 m-x\right)=32 N -x \implies \int^{40}_0 32-.8x \,dx = \left.32x-.4x^2 \right|^{40}_0$ $= 32(40)-.4(40)^2 - (32(0)-.4(0)^2) = 640 J$
c. $4000 J + 640 J = 4640 J$

8) $\lim_{x \to \pi} \frac{\ln\left(\sin\left(\frac{x}{2}\right)\right) }{x} = \frac{0}{0} \implies \lim_{x \to \pi} \frac{\frac{1}{\sin\left(\frac{x}{2}\right)*\cos\l eft(\frac{x}{2}\right)*\frac{1}{2}}}{1} = 0/1 = 0$

9) $x^3-2x^2+5 \implies 3x^2-4x \implies 6x-4 \implies 6$
$(-1)^3-2(-1)^2+5 \implies (3(-1)^2-4(-1))(x+1) \implies \frac{(6(-1)-4)(x+1)^2}{2!}$ $\implies \frac{6(x+1)^3}{3!} = 2 + 7(x+1) - 5(x+1)^2 + (x+1)^3$
$\text{General term for }n\geq4 = 0$

10) $\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{1}{2}\left(r_2^2 - r_1^2\right) \,d\theta \implies 2\int^{0}_{-\frac{\pi}{2}} \frac{1}{2}\left(2cos\theta+cos^2\theta\right) \,d\theta$ $\implies 2\left.\left(2sin\theta-\frac{1}{3}sin^3\theta\right)\right|^{\frac{\pi}{2 }}_0 = 2-\frac{\pi}{4}$

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CB