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Math Help - Integration questions

  1. #1
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    Integration questions

    Hi
    Can someone look where i have made a mistake because it doesn't match the answers in the book.

    1) \int_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})

    [\frac{2a^\frac{3}{2}}{3}-\frac{2x^\frac{3}{2}}{3}]

    [\frac{2a^\frac{3}{2}}{3}-\frac{8a^\frac{3}{2}}{3}]-[\frac{2a^\frac{3}{2}}{3}-\frac{2a^\frac{3}{2}}{3}]

    \frac{-6a^\frac{3}{2}}{3}

    the book's answer is \frac{-5a^\frac{3}{2}}{3}

    2) \int_1^4\frac{3}{\sqrt{x}}-5\sqrt{x}-x^\frac{-3}{2}

    6\sqrt{x}-\frac{10x^\frac{3}{2}}{3}+\frac{2}{\sqrt{x}}

    final answer is: \frac{-38}{3}
    but book answer is \frac{-55}{3}

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone look where i have made a mistake because it doesn't match the answers in the book.

    1) \int_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})

    [\frac{2a^\frac{3}{2}}{3}-\frac{2x^\frac{3}{2}}{3}]

    [\frac{2a^\frac{3}{2}}{3}-\frac{8a^\frac{3}{2}}{3}]-[\frac{2a^\frac{3}{2}}{3}-\frac{2a^\frac{3}{2}}{3}]

    \frac{-6a^\frac{3}{2}}{3}

    the book's answer is \frac{-5a^\frac{3}{2}}{3}

    2) \int_1^4\frac{3}{\sqrt{x}}-5\sqrt{x}-x^\frac{-3}{2}

    6\sqrt{x}-\frac{10x^\frac{3}{2}}{3}+\frac{2}{\sqrt{x}}

    final answer is: \frac{-38}{3}
    but book answer is \frac{-55}{3}

    P.S
    1. I suppose you were integrating with respect to x. Note that a is a constant, independent of x, so:

    \int_{a}^{4a} a^{\frac{1}{2}}- x^{\frac{1}{2}} ~ dx = a^{\frac{1}{2}}\int_{a}^{4a} 1 ~ dx - \int_{a}^{4a} x^{\frac{1}{2}}~dx = a^{\frac{1}{2}}(4a-a) - (\frac{2}{3}(4^{\frac{3}{2}}a^{\frac{3}{2}}) - \frac{2}{3} a^{\frac{3}{2}} = \frac{9}{3}a^{\frac{3}{2}} - \frac{14}{3}a^{\frac{3}{2}} = \frac{-5}{3}a^{\frac{3}{2}}

    2. Your integral is correct, I guess you miscalculated.

    6\sqrt{4}-\frac{10 \cdot 4^{\frac{3}{2}}}{3} + \frac{2}{\sqrt{4}} = 6\cdot 2 - \frac{80}{3} + 1 = \frac{-41}{3}

    6\sqrt{1} - \frac{10}{3} + \frac{2}{\sqrt{1}} = \frac{18}{3} - \frac{10}{3} + \frac{6}{3} = \frac{14}{3}

    \frac{-41}{3} - \frac{14}{3} = \frac{-55}{3}
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  3. #3
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone look where i have made a mistake because it doesn't match the answers in the book.

    1) \int_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})


    You must ALWAYS add to the integral dx\,\,\,or\,\,\,dy or whatever, so that all can know with respect to what you're integrating. Here, I assume you're integrating wrt x (i.e., it must be \int\limits_a^{4a}(a^\frac{1}{2}-x^\frac{1}{2})\,dx ), and thus the following step is WRONG because you did NOT take a^{1\slash 2} as a constant, as you should...


    [\frac{2a^\frac{3}{2}}{3}-\frac{2x^\frac{3}{2}}{3}]


    If you're indeed integrating wrt x, here it must be \left[a^{1\slash 2}x-\frac{2}{3}x^{3\slash 2}\right]_a^{4a} =3a\cdot a^{1\slash 2}-\frac{2}{3}(8a^{3\slash 2}-a^{3\slash 2}) =-\frac{5}{3}a^{3\slash 2} , as written in your book.

    [\frac{2a^\frac{3}{2}}{3}-\frac{8a^\frac{3}{2}}{3}]-[\frac{2a^\frac{3}{2}}{3}-\frac{2a^\frac{3}{2}}{3}]

    \frac{-6a^\frac{3}{2}}{3}

    the book's answer is \frac{-5a^\frac{3}{2}}{3}

    2) \int_1^4\frac{3}{\sqrt{x}}-5\sqrt{x}-x^\frac{-3}{2}

    6\sqrt{x}-\frac{10x^\frac{3}{2}}{3}+\frac{2}{\sqrt{x}}

    final answer is: \frac{-38}{3}
    but book answer is \frac{-55}{3}

    6(\sqrt{4}-\sqrt{1})-\frac{10}{3}(4^{3\slash 2}-1)+2\left(\frac{1}{\sqrt{4}}-1\right)=6\cdot 1-\frac{10}{3}\cdot 7-1=5-\frac{70}{3}=-\frac{55}{3} , so your book = 2, you = 0.

    Tonio


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  4. #4
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    Quote Originally Posted by tonio View Post
    .
    thanks for the help
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