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Thread: integral

  1. #1
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    integral

    1) a. $\displaystyle 1$ b. $\displaystyle 1.5$ c. no d. discontinuous @ $\displaystyle |g-1|<2$ and $\displaystyle g(3)$
    e. the discontinuity at $\displaystyle g(3)$ can be removed by assigning $\displaystyle g(3)=1$. All the other discontinuities can be removed by extending the domain from $\displaystyle -1\leq x\leq3$ to $\displaystyle -\infty\leq x\leq\infty$.

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

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

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

    5) i $\displaystyle y=\frac{11^3\cos\left(\frac{11}{2}\right)}{2}-11 = \$ 460,000$
    ii $\displaystyle y=\int^{11.5}_0 \frac{x^3\cos\left(\frac{x}{2}\right)}{2}-x \,dx \implies \text{(integration by parts 3 times)}$ $\displaystyle \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 $ $\displaystyle = -332.833- -48 = \$ -284,833$ Meaning that they lost $284,833 over the first 11 years.

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

    7) a. $\displaystyle 100 N * 40 m = 4000 J$
    b. $\displaystyle .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 $ $\displaystyle = 32(40)-.4(40)^2 - (32(0)-.4(0)^2) = 640 J$
    c. $\displaystyle 4000 J + 640 J = 4640 J$

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

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

    10) $\displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{1}{2}\left(r_2^2 - r_1^2\right) \,d\theta \implies 2\int^{0}_{-\frac{\pi}{2}} \frac{1}{2}\left(2cos\theta+cos^2\theta\right) \,d\theta$ $\displaystyle \implies 2\left.\left(2sin\theta-\frac{1}{3}sin^3\theta\right)\right|^{\frac{\pi}{2 }}_0 = 2-\frac{\pi}{4}$
    Last edited by Chris L T521; May 11th 2009 at 07:28 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. $\displaystyle 1$ b. $\displaystyle 1.5$ c. no d. discontinuous @ $\displaystyle |g-1|<2$ and $\displaystyle g(3)$
    e. the discontinuity at $\displaystyle g(3)$ can be removed by assigning $\displaystyle g(3)=1$. All the other discontinuities can be removed by extending the domain from $\displaystyle -1\leq x\leq3$ to $\displaystyle -\infty\leq x\leq\infty$.

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

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

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

    5) i $\displaystyle y=\frac{11^3\cos\left(\frac{11}{2}\right)}{2}-11 = \$ 460,000$
    ii $\displaystyle y=\int^{11.5}_0 \frac{x^3\cos\left(\frac{x}{2}\right)}{2}-x \,dx \implies \text{(integration by parts 3 times)}$ $\displaystyle \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 $ $\displaystyle = -332.833- -48 = \$ -284,833$ Meaning that they lost $284,833 over the first 11 years.

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

    7) a. $\displaystyle 100 N * 40 m = 4000 J$
    b. $\displaystyle .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 $ $\displaystyle = 32(40)-.4(40)^2 - (32(0)-.4(0)^2) = 640 J$
    c. $\displaystyle 4000 J + 640 J = 4640 J$

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

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

    10) $\displaystyle \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{1}{2}\left(r_2^2 - r_1^2\right) \,d\theta \implies 2\int^{0}_{-\frac{\pi}{2}} \frac{1}{2}\left(2cos\theta+cos^2\theta\right) \,d\theta$ $\displaystyle \implies 2\left.\left(2sin\theta-\frac{1}{3}sin^3\theta\right)\right|^{\frac{\pi}{2 }}_0 = 2-\frac{\pi}{4}$
    If you really need help try posting this in reasonably comprehensible English, one question per thread.

    CB
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