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Math Help - Evaluate the Integral (different equation)

  1. #1
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    Question Evaluate the Integral (different equation)

    Is the antiderivative of \int_2^1{(1+(z/2))\,dx} is (z+(z^2/2))?
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  2. #2
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    Quote Originally Posted by katchat64 View Post
    Is the antiderivative of \int_2^1{(1+(z/2))\,dx} is (z+(z^2/2))?
    Almost. It's  -1 - \frac{z}{2} .

    \int_2^1{1+\frac{z}{2}\,dx}

    = [x+\frac{xz}{2}]^1_2

    = 1+\frac{z}{2} - 2 - \frac{2z}{2}

     = -1 - \frac{z}{2}
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  3. #3
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    Question

    Wait, you're not evaluating it right, your just taking the antiderivative of 1+ (z/2)?
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  4. #4
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    Quote Originally Posted by katchat64 View Post
    Wait, you're not evaluating it right, your just taking the antiderivative of 1+ (z/2)?
    You wrote  dx, implying that the integrand was to be integrated with respect to x, not z.

    Also, you put in limits, so your final answer should be numerical, not algebraic.

    Also, are you sure the limits are the right way around? Did you mean

     \int_1^2{1+\frac{z}{2}\,dz} ???

    If so then:

     \int_1^2{1+\frac{z}{2}\,dz}

     \int_1^2{1\,dz}+\int_1^2{\frac{z}{2}\,dz}

     \int_1^2{1\,dz}+\frac{1}{2}\int_1^2{z\,dz}

     [z]^2_1+\frac{1}{2}[\frac{z^2}{2}]^2_1

     [2-1]+\frac{1}{2}[\frac{2^2}{2} - \frac{1^2}{2}]

     1+\frac{1}{2}[\frac{4}{2} - \frac{1}{2}]

     1+\frac{1}{2}[\frac{3}{2}]

     1+\frac{3}{4}

     \frac{4}{4}+\frac{3}{4}

     \frac{7}{4}
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  5. #5
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    Thumbs up

    Yes, this is what I meant... sorry about the dx... I understand it now and that's the same exact answer I got.
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