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Thread: Help required with this definite integration equation!

  1. #1
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    Help required with this definite integration equation!

    Integrate s/square root of (s squared + 4 squared) ds.

    Definite integral has range 0 - 3
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  2. #2
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    Quote Originally Posted by Steve Mckenna View Post
    Integrate s/square root of (s squared + 4 squared) ds.

    Definite integral has range 0 - 3
    For $\displaystyle \int_0^3 \frac{s}{\sqrt{s^2+16}}\, ds $ try letting $\displaystyle u = s^2 +16.$
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Steve Mckenna View Post
    Integrate s/square root of (s squared + 4 squared) ds.

    Definite integral has range 0 - 3
    $\displaystyle \int {\frac{s}
    {{\sqrt {{s^2} + {4^2}} }}ds} = \frac{1}
    {2}\int {\frac{{2s}}
    {{\sqrt {{s^2} + 16} }}ds} =$$\displaystyle \left\{ \begin{gathered}
    \sqrt {{s^2} + 16} = t, \hfill \\
    {s^2} = {t^2} - 16, \hfill \\
    2sds = 2tdt \hfill \\
    \end{gathered} \right\} =$

    $\displaystyle = \frac{1}
    {2}\int {\frac{{2t}}
    {t}dt} = \int {dt} = t + C = \sqrt {{s^2} + 16} + C.$

    $\displaystyle \int\limits_0^3 {\frac{s}
    {{\sqrt {{s^2} + 16} }}ds = } \left. {\sqrt {{s^2} + 16} } \right|_0^3 = \sqrt {25} - \sqrt {16} = 5 - 4 = 1.$
    Last edited by DeMath; Jul 26th 2009 at 11:35 PM. Reason: typo
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  4. #4
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    Quote Originally Posted by DeMath View Post
    $\displaystyle \int {\frac{s}
    {{\sqrt {{s^2} + {4^2}} }}ds} = \frac{1}
    {2}\int {\frac{{2s}}
    {{\sqrt {{s^2} + 16} }}ds} =$$\displaystyle \left\{ \begin{gathered}
    \sqrt {{s^2} + 4} = t, \hfill \\
    {s^2} = {t^2} - 16, \hfill \\
    2sds = 2tdt \hfill \\
    \end{gathered} \right\} =$

    $\displaystyle = \frac{1}
    {2}\int {\frac{{2t}}
    {t}dt} = \int {dt} = t + C = \sqrt {{s^2} + 16} + C.$

    $\displaystyle \int\limits_0^3 {\frac{s}
    {{\sqrt {{s^2} + 16} }}ds = } \left. {\sqrt {{s^2} + 16} } \right|_0^3 = \sqrt {25} - \sqrt {16} = 5 - 4 = 1.$
    Many thanks
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