MHB Can you convert (s+1) to (u-2) in the integration of (3+s)^(1/2)(s+1)^2ds?

Click For Summary
The discussion focuses on converting the integral ∫(3+s)^(1/2)(s+1)^2ds into a more manageable form using the substitution u=s+3. This substitution leads to the integral ∫u^(1/2)(u-2)^2du, which can be expanded and integrated term by term. An alternative method is suggested, where the integrand is expressed solely in terms of s+3, allowing for integration without substitution. The key point is that the expression (s+1) transforms to (u-2) under the substitution u=s+3. The conversation emphasizes different approaches to simplifying and solving the integral.
paulmdrdo1
Messages
382
Reaction score
0
3.) ∫(3+s)1/2(s+1)2ds
 
Physics news on Phys.org
I would let:

$$u=s+3\,\therefore\,du=ds$$

and now we have:

$$\int u^{\frac{1}{2}}(u-2)^2\,du$$

Now, expand, distribute, and then apply the power rule term by term.
 
MarkFL said:
I would let:

$$u=s+3\,\therefore\,du=ds$$

and now we have:

$$\int u^{\frac{1}{2}}(u-2)^2\,du$$

Now, expand, distribute, and then apply the power rule term by term.

how do you get (u-2)2?
 
Hi everyone, :)

An alternative method without using substitutions is to write the integrand only using \(s+3\).

\begin{eqnarray}

\int(s+3)^{1/2}(s+1)^2\,ds&=&\int(s+3)^{1/2}(s+3-2)^2\,ds\\

&=&\int(s+3)^{1/2}\left((s+3)^2-4(s+3)+4\right)\,ds\\

&=&\int(s+3)^{5/2}\,d(s+3)-4\int(s+3)^{3/2}\,d(s+3)+4\int(s+3)^{1/2}\,d(s+3)\\

\end{eqnarray}

Hope you can continue. :)
 
paulmdrdo said:
how do you get (u-2)2?

Hi paulmdrdo, :)

The \(s+1\) in the integrand becomes \(u-2\). That is, \(u=s+3\Rightarrow u-2=s+1\).
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
3K
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K