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Math Help - Evaluate the integral

  1. #1
    Member dokrbb's Avatar
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    Evaluate the integral

    So, here I am, beginning the calc II and beginning to make new mistakes as well , I have this problem, which states:
    -Evaluate the integral by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry
    of \displaystyle \int_0^2{\mid{6x - 7}\mid dx.

    What I have done is:

    -evaluated the function
    a) 6x - 7 < 0 for 0\leq x < \frac{7}{6} and 6x - 7 > 0 for \frac{7}{6} < x \leq 2

    we have {\mid{6x - 7}\mid = -(6x-7) = 7-6x,  for 0\leq x < \frac{7}{6} and {\mid{6x - 7}\mid = (6x-7) for \frac{7}{6} < x \leq 2

    - then, by evaluating the integral, I have got:

    \displaystyle \int_0^2{\mid{6x - 7}\mid dx  =  \displaystyle \int_{0}^{\frac{7}{6}}(7 - 6x)dx + \displaystyle \int_{\frac{7}{6}}^{2}(6x - 7)dx

    \displaystyle (7x - 3x^2)\right|_{0}^{\frac{7}{6}} + (3x^2 - 7x)\right|_{\frac{7}{6}}^{2}

     [ (\frac{49}{6} - \frac{147}{6}) - (0)] + [(12-14) - (\frac{147}{6} - \frac{49}{6})] = (-\frac{98}{6}) + (- \frac{110}{6}) = - \frac{208}{6} ,

    and I must have made a mistake somewhere, because it say that it's wrong, or, maybe I should solve it by another method, but I have no idea how...

    thank you, for all your input,
    dokrbb
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  2. #2
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    Re: Evaluate the integral

    Quote Originally Posted by dokrbb View Post
    So, here I am, beginning the calc II and beginning to make new mistakes as well , I have this problem, which states:
    -Evaluate the integral by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry
    of \displaystyle \int_0^2{\mid{6x - 7}\mid dx.

    What I have done is:

    -evaluated the function
    a) 6x - 7 < 0 for 0\leq x < \frac{7}{6} and 6x - 7 > 0 for \frac{7}{6} < x \leq 2

    we have {\mid{6x - 7}\mid = -(6x-7) = 7-6x,  for 0\leq x < \frac{7}{6} and {\mid{6x - 7}\mid = (6x-7) for \frac{7}{6} < x \leq 2

    - then, by evaluating the integral, I have got:

    \displaystyle \int_0^2{\mid{6x - 7}\mid dx  =  \displaystyle \int_{0}^{\frac{7}{6}}(7 - 6x)dx + \displaystyle \int_{\frac{7}{6}}^{2}(6x - 7)dx

    \displaystyle (7x - 3x^2)\right|_{0}^{\frac{7}{6}} + (3x^2 - 7x)\right|_{\frac{7}{6}}^{2}

     [ (\frac{49}{6} - \frac{147}{6}) - (0)] + [(12-14) - (\frac{147}{6} - \frac{49}{6})] = (-\frac{98}{6}) + (- \frac{110}{6}) = - \frac{208}{6} ,

    and I must have made a mistake somewhere, because it say that it's wrong, or, maybe I should solve it by another method, but I have no idea how...

    thank you, for all your input,
    dokrbb
    It should be
    \displaystyle (7x - 3x^2)\right|_{0}^{\frac{7}{6}} + (3x^2 - 7x)\right|_{\frac{7}{6}}^{2}

     [ (\frac{49}{6} - \frac{147}{36}) - (0)] + [(12-14) - (\frac{147}{36} - \frac{49}{6})] = 2
    Thanks from dokrbb
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  3. #3
    Member dokrbb's Avatar
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    Re: Evaluate the integral

    now, looking for another example I just understood that I have to do otherwise:

    i have to draw a line corresponding to the function (6x-7) which intersects the x axis in the point 7/6, and I have 2 triangles I can evaluate as it asks with basics geometry;

    since it asks me the absolute value, I have (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(2) = 49/12 + 10/12 = 59/12, but is considered also wrong, what don't I get here?
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    Member dokrbb's Avatar
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    Re: Evaluate the integral

    Quote Originally Posted by dokrbb View Post
    now, looking for another example I just understood that I have to do otherwise:

    i have to draw a line corresponding to the function (6x-7) which intersects the x axis in the point 7/6, and I have 2 triangles I can evaluate as it asks with basics geometry;

    since it asks me the absolute value, I have (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(2) = 49/12 + 10/12 = 59/12, but is considered also wrong, what don't I get here?
    we were writing in the same time the posts, can you look at that now, thanks
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  5. #5
    Junior Member FelixFelicis28's Avatar
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    Re: Evaluate the integral

    Quote Originally Posted by dokrbb View Post
    now, looking for another example I just understood that I have to do otherwise:

    i have to draw a line corresponding to the function (6x-7) which intersects the x axis in the point 7/6, and I have 2 triangles I can evaluate as it asks with basics geometry;

    since it asks me the absolute value, I have (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(2) = 49/12 + 10/12 = 59/12, but is considered also wrong, what don't I get here?
    The area of triangle 2 is incorrect. Your base is correct but your height should 6x - 7 |_{x = 2}.
    Thanks from dokrbb
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    Member dokrbb's Avatar
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    Re: Evaluate the integral

    Quote Originally Posted by FelixFelicis28 View Post
    The area of triangle 2 is incorrect. Your base is correct but your height should 6x - 7 |_{x = 2}.
    (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(5) = 49/12 + 25/12 = 74/12, so here is the mistake, right?
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  7. #7
    Junior Member FelixFelicis28's Avatar
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    Re: Evaluate the integral

    Quote Originally Posted by dokrbb View Post
    (A of triangle 1) + (A of triangle 2) = 1/2(7/6)(7) + 1/2(5/6)(5) = 49/12 + 25/12 = 74/12, so here is the mistake, right?
    \frac{74}{12} = \frac{37}{6} is correct.
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