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Math Help - Help with integral with upper/lower limts

  1. #1
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    Help with integral with upper/lower limts

    Q. Evaluate \int \frac{6}{x^2}dx

    A. \int \frac{6}{x^2}dx=-\frac{6}{x}=???


    The evaluation equation has an upper limit of 3 and a lower limit of 1. I couldn't figure out how to show that with LaTEX. As you can see by my answer above, I have it partially completed. I'm stuck on how to finish it.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Yogi_Bear_79
    Q. Evaluate \int \frac{6}{x^2}dx

    A. \int \frac{6}{x^2}dx=-\frac{6}{x}=???


    The evaluation equation has an upper limit of 3 and a lower limit of 1. I couldn't figure out how to show that with LaTEX. As you can see by my answer above, I have it partially completed. I'm stuck on how to finish it.
    Click on the first term to see how to set the limits:

    \int_{1}^{3} \frac{6}{x^2}dx <br />
=\left[-\frac{6}{x} \right]_1^3<br />
<br />
=(-6/3)-(-6/1)=-2+6=4<br />

    RonL
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  3. #3
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    To see if I understand, I have posted a second question that I worked brom begining to end, please validate.


    Q. Evaluate \int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx

    A. \int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx = \left{ \frac{1}{3} \sqrt x(x+12 \sqrt x +48 \right]_{4}^{1}=51 -20 = 30
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Yogi_Bear_79
    To see if I understand, I have posted a second question that I worked brom begining to end, please validate.


    Q. Evaluate \int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx

    A. \int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx = \left{ \frac{1}{3} \sqrt x(x+12 \sqrt x +48 \right]_{4}^{1}=51 -20 = 30
    First you TeX needs tidying up

    \int_{4}^{1} \frac {(4+\sqrt x)^2}{2\sqrt x}dx <br />
 = \left[ \frac{1}{3} \sqrt x(x+12 \sqrt x +48) \right]_{4}^{1}
    <br />
=(1+12+48)/3 -(4+12.2+48)2/3

    =61/3-152/3=91/3
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