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Math Help - Definite Integral

  1. #1
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    Definite Integral

    I need help with definite integrals guys. The probelm asks for me to evaluate the integral wich is </^1(on top) 2(on bottom) (1+(z/2) dz Im supposed to use these 8 rules for definitive integrals to solve wich are "order of integration", "zero width interval" "additivity" amond others. And I also use these three rules wich are </^ b(on top) a(on bottom) xdx + b^2/2 -a^2/2 and theres two others for use of a c (constant) and when it is x^2. Please I need help guys, I cant seem to solve it and i got a test tomorrow.
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  2. #2
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    1) Remember that no one can see into your brain or read your book. You MUST learn to communicate clearly. Please work a little harder on your notation.

    2) One must use 8 rules, but you supply only 3? This goes to #1. We can't read your mind. If you want all 8, you may have to list them, at least.

    3) I'll take a little shot at the first piece:

     <br />
\int_{1}^{2}(1+\frac{z}{2})dz<br />

    This might be the Additive Rule

     <br />
\int_{1}^{2}1 dz+\int_{1}^{2}\frac{z}{2} dz<br />
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  3. #3
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    Sorry its hard to be clear with this stuff but i know i skipped over alot =D I understand that part, for the left side I use c(b-a) so thats -1. However for the right side if i take 1/2 as the constant, and do the same thing c(b-a) I get -.5. So -1+.-5 = -1.5 not -7/4!

    Btw the 1 and 2 are switched.
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  4. #4
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    Quote Originally Posted by jmoh View Post
    Sorry its hard to be clear with this stuff but i know i skipped over alot =D I understand that part, for the left side I use c(b-a) so thats -1. However for the right side if i take 1/2 as the constant, and do the same thing c(b-a) I get -.5. So -1+.-5 = -1.5 not -7/4!

    Btw the 1 and 2 are switched.
    \int_2^1 dz = z|_2^1 = 1 - 2 = -1

    and
    \int_2^1  \frac{z}{2} dz = \frac{1}{2} \int_2^1 z dz = \frac{1}{2} \cdot \frac{1}{2}z^2|_2^1

    = \frac{1}{4}(1 - 4) = -\frac{3}{4}

    So you have \int_2^1 dz + \int_2^1 \frac{z}{2} dz = -1 - \frac{3}{4} = -\frac{7}{4}

    as advertised.

    -Dan
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  5. #5
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    Ah! I see I was upside down.

    This might be the Rule of the Zero Width Interval?

    \int_{2}^{1}dx = \int_{2}^{\frac{3}{2}}dx + \int_{\frac{3}{2}}^{1}dx

    The important point would be that 3/2 is in both integrals, but it doesn't hurt anyone's feelings. We haven't actually counted it twice.
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