Evaluate

**anothernewbie** · January 14th 2010, 04:36 AM

Hi I am in the middle of this question and im not sure if a. im right or b. where to go from here.

Q. Evaluate:

$\int \frac{s}{\sqrt{s^2+4^2}}$ between the limits of 3 and 0

So far I have:

$\int \frac{s}{\sqrt{s^2+4^2}}$

$let.P =\sqrt{s^2+4^2}$ then $dp = \frac{s}{\sqrt{s^2+16}} ds$

when S = 3 then P = 5

and

when S = 0 then P = 4

Am I right so far, if so what is the next step?

thanks

**General** · January 14th 2010, 04:38 AM

You forgot the $ds$ in your integral.
I think the substitution $u=s^2+4^2$ is better.

Do not forget to change the limits of the integration according to this substitution.

**anothernewbie** · January 14th 2010, 04:46 AM

Ok so, a few changes:

$\int \frac{s}{\sqrt{s^2+4^2}}ds$

$u =\sqrt{s^2+4^2}$ then $du = \frac{s}{\sqrt{s^2+16}} ds$

when S = 3 then u = 5

and

when S = 0 then u = 4

With the limits now between 5 and 4

Ok, that looks a bit better but im still not sure how to take this further, its been ages since ive done anything like this and my mind is blank.

**General** · January 14th 2010, 04:47 AM

No.
I said $u=s^2+4^2$
Without the square root.
This does not mean your solution is wrong.
but this substitution is easier.

**anothernewbie** · January 14th 2010, 04:49 AM

Sorry you have lost me, what happens to the sqrt then?

**General** · January 14th 2010, 04:55 AM

Originally Posted by anothernewbie

Ok so, a few changes:

$\int \frac{s}{\sqrt{s^2+4^2}}ds$

$u =\sqrt{s^2+4^2}$ then $du = \frac{s}{\sqrt{s^2+16}} ds$

when S = 3 then u = 5

and

when S = 0 then u = 4

With the limits now between 5 and 4

Ok, that looks a bit better but im still not sure how to take this further, its been ages since ive done anything like this and my mind is blank.

You changed the limits correctly.
The next step is changing your integral to another integral in terms of u according to your substitution.
To do that.
First you should find $ds$ in terms of $u$
You need to solve your substitution for s then differentiate it with respect to s.
$u=\sqrt{s^2+4^2}$
$s^2=u^2-4^2$
$s=\sqrt{u^2-4^2}$
Differentiate with respect to s.
$ds=\frac{u}{\sqrt{u^2-4^2}}$
So, For the original integral, we found ds in terms of u.
the denominator is u(Which is just our the substitution you did).
we have s in the numerator.
Actually, we found it above.
$s=\sqrt{u^2-4^2}$
You have everything in the original integral in terms of u.
Can you replace them and try to solve the resulting integral ?

-------

Originally Posted by anothernewbie

Sorry you have lost me, what happens to the sqrt then?

Sir, You are using the substitution method to solve this integral, Right?

I said your substitution $u=\sqrt{u^2+4^2}$ will make it hard a little bit.

So, I gave you another substitution to solve it, There is no relation between them.

**anothernewbie** · January 14th 2010, 05:03 AM

Originally Posted by General

Sir, You are using the substitution method to solve this integral, Right?

I said your substitution $u=\sqrt{u^2+4^2}$ will make it hard a little bit.

So, I gave you another substitution to solve it, There is no relation between them.

Yes, thanks I realised this I just wasn't sure how you would use your method. Thanks for the help I should be able to finish it now.

**anothernewbie** · January 14th 2010, 05:22 AM

Right, sorry im still not 100%, does this mean that it works out to be:

$\frac {\sqrt{5^2-16}}{\sqrt{\sqrt{(5^2-16)^2}+16}}$ - $\frac {\sqrt{4^2-16}}{\sqrt{\sqrt{(4^2-16)^2}+16}}$

Or am I totally wrong, again!

**General** · January 14th 2010, 05:32 AM

Originally Posted by anothernewbie

Right, sorry im still not 100%, does this mean that it works out to be:

$\frac {\sqrt{5^2-16}}{\sqrt{\sqrt{(5^2-16)^2}+16}}$ - $\frac {\sqrt{4^2-16}}{\sqrt{\sqrt{(4^2-16)^2}+16}}$

Or am I totally wrong, again!

Show your steps for evaluating the resulting integral which bacame from the substitution.
And also, specify which substitution did you use.
And doubling posts is not allowed in this forum.
For the substitution $u=s^2+4^2$
$u(0)=16$ and $u(3)=25$
So the new Upper/Lower limits the integral are 16 and 25.
And we have $du=2sds$
i.e. $\frac{1}{2}du=sds$
Integral will be :
$\frac{1}{2}\int_{16}^{25}u^{\frac{-1}{2}}du=\sqrt{25}-\sqrt{16} = 5 - 4 = 1$ .

**anothernewbie** · January 14th 2010, 05:41 AM

Thanks again, but

Originally Posted by General

And doubling posts is not allowed in this forum.

I don't see where I did that?????? All my posts are genuine

**General** · January 14th 2010, 05:49 AM

Originally Posted by anothernewbie

Thanks again, but

I don't see where I did that?????? All my posts are genuine

You did before my last post.
You wrote 2 posts.
Anyway, Forget it.

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