# Thread: Help required with this definite integration equation!

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

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

Definite integral has range 0 - 3

2. 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 $u = s^2 +16.$

3. Originally Posted by Steve Mckenna
Integrate s/square root of (s squared + 4 squared) ds.

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

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

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

4. 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