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Math Help - 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 \int_0^3 \frac{s}{\sqrt{s^2+16}}\, ds try letting 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
    \int {\frac{s}<br />
{{\sqrt {{s^2} + {4^2}} }}ds}  = \frac{1}<br />
{2}\int {\frac{{2s}}<br />
{{\sqrt {{s^2} + 16} }}ds}  = \left\{ \begin{gathered}<br />
  \sqrt {{s^2} + 16}  = t, \hfill \\<br />
  {s^2} = {t^2} - 16, \hfill \\<br />
  2sds = 2tdt \hfill \\ <br />
\end{gathered}  \right\} =

    = \frac{1}<br />
{2}\int {\frac{{2t}}<br />
{t}dt}  = \int {dt}  = t + C = \sqrt {{s^2} + 16}  + C.

    \int\limits_0^3 {\frac{s}<br />
{{\sqrt {{s^2} + 16} }}ds = } \left. {\sqrt {{s^2} + 16} } \right|_0^3 = \sqrt {25}  - \sqrt {16}  = 5 - 4 = 1.
    Last edited by DeMath; July 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
    \int {\frac{s}<br />
{{\sqrt {{s^2} + {4^2}} }}ds}  = \frac{1}<br />
{2}\int {\frac{{2s}}<br />
{{\sqrt {{s^2} + 16} }}ds}  = \left\{ \begin{gathered}<br />
  \sqrt {{s^2} + 4}  = t, \hfill \\<br />
  {s^2} = {t^2} - 16, \hfill \\<br />
  2sds = 2tdt \hfill \\ <br />
\end{gathered}  \right\} =

    = \frac{1}<br />
{2}\int {\frac{{2t}}<br />
{t}dt}  = \int {dt}  = t + C = \sqrt {{s^2} + 16}  + C.

    \int\limits_0^3 {\frac{s}<br />
{{\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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