Help required with this definite integration equation!

**Steve Mckenna** · July 26th 2009, 08:54 AM

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

Definite integral has range 0 - 3

**Danny** · July 26th 2009, 09:37 AM

Originally Posted by Steve Mckenna

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 $u = s^2 +16.$

**DeMath** · July 26th 2009, 09:44 AM

Originally Posted by Steve Mckenna

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

Definite integral has range 0 - 3

$\int {\frac{s} {{\sqrt {{s^2} + {4^2}} }}ds} = \frac{1} {2}\int {\frac{{2s}} {{\sqrt {{s^2} + 16} }}ds} =$ $\left\{ \begin{gathered} \sqrt {{s^2} + 16} = t, \hfill \\ {s^2} = {t^2} - 16, \hfill \\ 2sds = 2tdt \hfill \\ \end{gathered} \right\} =$

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

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

**Steve Mckenna** · July 26th 2009, 11:32 PM

Originally Posted by DeMath

$\int {\frac{s} {{\sqrt {{s^2} + {4^2}} }}ds} = \frac{1} {2}\int {\frac{{2s}} {{\sqrt {{s^2} + 16} }}ds} =$ $\left\{ \begin{gathered} \sqrt {{s^2} + 4} = t, \hfill \\ {s^2} = {t^2} - 16, \hfill \\ 2sds = 2tdt \hfill \\ \end{gathered} \right\} =$

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

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

Many thanks

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