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Math Help - Mean Value Theorem for integrals w/ log

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    Mean Value Theorem for integrals w/ log

    An arched shaped monument is 570 ft high and has a 600 ft base. It can be modeled by the function  y=1140-285(e^{0.00439x}+e^{-0.00439x}) where the base of the arch is[-300,300] and x and y are measured in feet. Find the average height of the arch above the ground.

    I was able to do this but it took a while. What is the best way to to do this so I dont have to enter too many buttons on my calculator which personally increases my rate of error.

    My questions: 1. Can I use ln on the integrand and what would it look like? 2. How do I deal with symmety for a LN()?


     y=1140-285(e^{0.00439x}+e^{-0.00439x})

    avg value =\frac{1}{600}\int_{-300}^{300}(1140-285(e^{0.00439x}+e^{-0.00439x})) dx


    factor out 285 and distribute the minus sign

    avg value =\frac{1}{600}*285\int_{-300}^{300}(4-e^{0.00439x}-e^{-0.00439x}) dx


    Can I use a Natral Log? and does it look like this???

    LN(avg value) = LN[\frac{285}{600}]\int_{-300}^{300}(ln4-0.00439x+0.00439x)dx

    Specifically, I want to know do I use LN on the factored out numbers and the avgvalue?

    If all is well then My next step would be to split up the intrgrand and check for symmetry. If I didnt previously apply the Ln, then all of the integrand was an even functon. It appers that e^x is an even function but what about LN(x)

    ln(-x)= no bueno
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  2. #2
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    Re: Mean Value Theorem for integrals w/ log

    Quote Originally Posted by delgeezee View Post
    An arched shaped monument is 570 ft high and has a 600 ft base. It can be modeled by the function  y=1140-285(e^{0.00439x}+e^{-0.00439x}) where the base of the arch is[-300,300] and x and y are measured in feet. Find the average height of the arch above the ground.

    I was able to do this but it took a while. What is the best way to to do this so I dont have to enter too many buttons on my calculator which personally increases my rate of error.

    My questions: 1. Can I use ln on the integrand and what would it look like? 2. How do I deal with symmety for a LN()?


     y=1140-285(e^{0.00439x}+e^{-0.00439x})

    avg value =\frac{1}{600}\int_{-300}^{300}(1140-285(e^{0.00439x}+e^{-0.00439x})) dx


    factor out 285 and distribute the minus sign

    avg value =\frac{1}{600}*285\int_{-300}^{300}(4-e^{0.00439x}-e^{-0.00439x}) dx


    Can I use a Natral Log? and does it look like this???

    LN(avg value) = LN[\frac{285}{600}]\int_{-300}^{300}(ln4-0.00439x+0.00439x)dx

    Specifically, I want to know do I use LN on the factored out numbers and the avgvalue?

    If all is well then My next step would be to split up the intrgrand and check for symmetry. If I didnt previously apply the Ln, then all of the integrand was an even functon. It appers that e^x is an even function but what about LN(x)

    ln(-x)= no bueno
    note that the antiderivative of e^{kx} is \frac{1}{k} e^{kx} ... it does not involve a log

    let a = 0.00439

    h_{avg} =\frac{285}{600} \int_{-300}^{300}(4-e^{ax}-e^{-ax}) dx

    h_{avg} =\frac{285}{600} \left[4x- \frac{1}{a}e^{ax} + \frac{1}{a}e^{-ax} \right]_{-300}^{300}

    also note that you may take advantage of symmetry ...

    h_{avg} =\frac{570}{600} \left[4x- \frac{1}{a}e^{ax} + \frac{1}{a}e^{-ax} \right]_0^{300}
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    Re: Mean Value Theorem for integrals w/ log

    Quote Originally Posted by delgeezee View Post
    An arched shaped monument is 570 ft high and has a 600 ft base. It can be modeled by the function  y=1140-285(e^{0.00439x}+e^{-0.00439x}) where the base of the arch is[-300,300] and x and y are measured in feet. Find the average height of the arch above the ground.

    I was able to do this but it took a while. What is the best way to to do this so I dont have to enter too many buttons on my calculator which personally increases my rate of error.

    My questions: 1. Can I use ln on the integrand and what would it look like? 2. How do I deal with symmety for a LN()?


     y=1140-285(e^{0.00439x}+e^{-0.00439x})

    avg value =\frac{1}{600}\int_{-300}^{300}(1140-285(e^{0.00439x}+e^{-0.00439x})) dx


    factor out 285 and distribute the minus sign

    avg value =\frac{1}{600}*285\int_{-300}^{300}(4-e^{0.00439x}-e^{-0.00439x}) dx


    Can I use a Natral Log? and does it look like this???

    LN(avg value) = LN[\frac{285}{600}]\int_{-300}^{300}(ln4-0.00439x+0.00439x)dx
    No. ln(a+ b) is NOT equal to ln(a)+ ln(b).

    Specifically, I want to know do I use LN on the factored out numbers and the avgvalue?
    I don't see why you would want to. The logarithm is not relevant. You want to integrate
    \frac{57}{120}\left(\int_{-300}^{300}4 dx- \int_{-300}^{300}e^{0.00439x}dx- \int_{-300}^{300}e^{-0.00439x}dx
    all of which are very easy integrals.

    If all is well then My next step would be to split up the intrgrand and check for symmetry. If I didnt previously apply the Ln, then all of the integrand was an even functon. It appers that e^x is an even function but what about LN(x)

    ln(-x)= no bueno
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