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Math Help - integral

  1. #1
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    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<br />
\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 <br />
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}
    Last edited by Chris L T521; May 11th 2009 at 08:28 PM. Reason: Fixed LaTeX errors.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jahichuanna View Post
    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}
    If you really need help try posting this in reasonably comprehensible English, one question per thread.

    CB
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