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Math Help - area between curves

  1. #1
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    area between curves

    I am struggling with a area between the curves problem. I need to find the area between y=tanx and y=2sinx over the interval and .

    Since the curves cross at 0 and are symmetrical, I'll just go from to 0 and double the answer. So I get



    I can't get the interval on the integral to work with a fraction and Latex.

    Anyway, that yields:

    -\ln(\cos(x))-2\cos(x)

    The answer in the book is 2+2\ln(2), after you multiply by 2 to account for the 2nd half of the interval.

    I agree with much of the answer, except that I get 2+2\ln(\frac{1}{2}), because \cos(\frac{\pi}{3}) (positive or negative) is 1/2.

    Anything that I could be doing wrong?
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  2. #2
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    Re: area between curves

    Quote Originally Posted by baldysm View Post
    I am struggling with a area between the curves problem. I need to find the area between y=tanx and y=2sinx over the interval and .

    Since the curves cross at 0 and are symmetrical, I'll just go from to 0 and double the answer.
    There's your problem! The graphs are NOT symmetric. They are both odd functions so the integral from [tex]-\pi/3[tex] to \pi/3 is 0.

    So I get



    I can't get the interval on the integral to work with a fraction and Latex.

    Anyway, that yields:

    -\ln(\cos(x))-2\cos(x)

    The answer in the book is 2+2\ln(2), after you multiply by 2 to account for the 2nd half of the interval.

    I agree with much of the answer, except that I get 2+2\ln(\frac{1}{2}), because \cos(\frac{\pi}{3}) (positive or negative) is 1/2.

    Anything that I could be doing wrong?
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  3. #3
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    Re: area between curves

    Quote Originally Posted by baldysm View Post
    I am struggling with a area between the curves problem. I need to find the area between y=tanx and y=2sinx over the interval and .
    Since the curves cross at 0 and are symmetrical, I'll just go from to 0 and double the answer. So I get
    Quote Originally Posted by HallsofIvy View Post
    There's your problem! The graphs are NOT symmetric. They are both odd functions so the integral from [tex]-\pi/3[tex] to \pi/3 is 0.
    I beg to differ with Prof Ivey. The function \tan(x)-2\sin(x) is symmetric about (0,0) as are all odd functions.
    Look at this graph..
    From that graph you can see that the area you seek is 2\int_0^{\pi /3} {\left[ {2\sin (x) - \tan (x)} \right]dx}.
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    Re: area between curves

    I do believe they are symmetric, but it's possible I am wrong, but the graph looks real close.

    If they are symmetric,

    2\int_0^{\pi /3} {\left[ {2\sin (x) - \tan (x)} \right]dx} should equal 2\int_{-\pi /3}^0 {\left[ \tan (x) - {2\sin (x)} \right]dx}

    The part I am struggling with is:

    \int_0^{\pi/3} \tan(x)

    is -\ln(\cos(x)) although you can use some trig identities to change the form around a bit.

    Cos(x) can never be larger than 1 for any real number, so how does -\ln(\cos(x)) become -\ln(2)?

    I have Maple 17, and that gives the answer as -\ln(2) as well.
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  5. #5
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    Re: area between curves

    Ah, 'symmetric about the origin'. But Baldysm said "Since the curves cross at 0 and are symmetrical, I'll just go from to 0 and double the answer" implying that he thought the function was symmetric about the y-axis.
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  6. #6
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    Re: area between curves

    [QUOTE=baldysm;791936]I do believe they are symmetric, but it's possible I am wrong, but the graph looks real close.

    If they are symmetric,

    2\int_0^{\pi /3} {\left[ {2\sin (x) - \tan (x)} \right]dx} should equal 2\int_{-\pi /3}^0 {\left[ \tan (x) - {2\sin (x)} \right]dx}
    Cos(x) can never be larger than 1 for any real number, so how does -\ln(\cos(x)) become -\ln(2)?

    I have Maple 17, and that gives the answer as -\ln(2) as well.[/QUOTE]

    It appears that Mathematica agrees with your Maple. See this link.
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  7. #7
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    Re: area between curves

    Quote Originally Posted by baldysm View Post
    I am struggling with a area between the curves problem. I need to find the area between y=tanx and y=2sinx over the interval and .

    Since the curves cross at 0 and are symmetrical, I'll just go from to 0 and double the answer. So I get



    I can't get the interval on the integral to work with a fraction and Latex.

    Anyway, that yields:

    -\ln(\cos(x))-2\cos(x)

    The answer in the book is 2+2\ln(2), after you multiply by 2 to account for the 2nd half of the interval.

    I agree with much of the answer, except that I get 2+2\ln(\frac{1}{2}), because \cos(\frac{\pi}{3}) (positive or negative) is 1/2.

    Anything that I could be doing wrong?
    To evaluate this, notice that in your interval, \displaystyle \begin{align*} \tan{(x)} = 2\sin{(x)} \end{align*} when \displaystyle \begin{align*} x = \left\{ -\frac{\pi}{3}, 0 , \frac{\pi}{3} \right\} \end{align*}. When \displaystyle \begin{align*} x \in \left( -\frac{\pi}{3}, 0 \right) , \, \tan{(x)} > 2\sin{(x)} \end{align*}, and when \displaystyle \begin{align*} x \in \left( 0, \frac{\pi}{3} \right) , \, 2\sin{(x)} > \tan{(x)} \end{align*}. So the area between the two curves is given by:

    \displaystyle \begin{align*} A &= \int_{-\frac{\pi}{3}}^0{\tan{(x)}\,dx} - \int_{-\frac{\pi}{3}}^0{2\sin{(x)}\,dx} + \int_0^{\frac{\pi}{3}}{2\sin{(x)}\,dx} - \int_0^{\frac{\pi}{3}}{\tan{(x)}\,dx} \\ &= \left[ -\ln{ \left| \cos{(x)} \right| } \right]_{-\frac{\pi}{3}}^0 - \left[ -2\cos{(x)} \right]_{-\frac{\pi}{3}}^0 + \left[ -2\cos{(x)} \right]_0^{\frac{\pi}{3}} - \left[ -\ln{ \left| \cos{(x)} \right| } \right] _0^{\frac{\pi}{3}} \\ &= \left\{ \left[ -\ln{ \left| \cos{(0)} \right| } \right] - \left[ -\ln{ \left| \cos{ \left( -\frac{\pi}{3} \right) } \right| } \right] \right\} - \left\{ \left[ -2\cos{(0)} \right] - \left[ -2\cos{ \left( -\frac{\pi}{3} \right) } \right] \right\} + \left\{ \left[ -2\cos{ \left( \frac{\pi}{3} \right) } \right] - \left[ -2\cos{ (0)} \right] \right\} - \left\{ \left[ -\ln{ \left| \cos{ \left( \frac{\pi}{3} \right) } \right| } \right] - \left[ -\ln{ \left| \cos{(0)} \right| } \right] \right\} \end{align*}

    \displaystyle \begin{align*} &= -0 + \ln{ \left( \frac{1}{2} \right) } + 2 - 1 - 1 + 2 + \ln{ \left( \frac{1}{2} \right) } - 0 \\ &= 2 + 2\ln{ \left( \frac{1}{2} \right) } \\ &= 2 - 2\ln{(2)} \end{align*}
    Thanks from baldysm
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  8. #8
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    Re: area between curves

    You're awesome guys. The last 2 lines of Proveit's response gave me what I needed. After a Google of log properties, I found the reciprocal property of logs and was able to take it to the next step.
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