Evaulating Definite integrals: x/sqrt(2x + 1)

[crayzeerunner]

Integral [x / (sqrt(2x + 1)] dx, from 0 to 4.

What I have done so far is:

u = 2x + 1
du = 2
x = (u -1)/2

Zeros: x = -1/2

The directions say do not forget to change the limits for u. I forget how to do this , if any one can help me I would greatly appreciate it. Thanks

 

[galactus]

\(\displaystyle x=\frac{u-1}{2}\)

\(\displaystyle \frac{u-1}{2}=4\), solve for u

\(\displaystyle \frac{u-1}{2}=0\), solve for u.

These will be your new limits of integration.

 

[crayzeerunner]

So I my new limits of integration are 3 and 1.

Do I then set up the problem Integral(x/sqrt(2x + 1))dx from 1 to 3.
(1/2)Integral(du(u-1)/2u)dx) ?

 

[pka]

\(\displaystyle \L
\frac{x}{{\sqrt {2x + 1} }} \Rightarrow \quad \frac{{u - 1}}{{2\sqrt u }} \Rightarrow \frac{{u^{\frac{1}{2}} - u^{ - \frac{1}{2}} }}{2}\)

 

[crayzeerunner]

I understand the first two steps pka has listed, but I do not understand how you get from step two to the third answer. Here is what I do.

Integral(u-1/(2sqrt(u))) from 1 to 3 =
(1/2)Integral((u-1)(u^-1/2) from 1 to 3=
(1/2)(((u^2)/2)-u)(u^1/2) from 1 to 3=
(1/2)((u^5/2)/2)-u^3/2) from 1 to 3

Is this correct or am I doing something wrong?

 

[galactus]

You had better check again. It's NOT 1 to 3.

\(\displaystyle \frac{u-1}{2}=4\). How do you get 3?.

 

[crayzeerunner]

Oops. my mistake. Its from 1 to 9.

 

[pka]

\(\displaystyle \L
u = 2x + 1\quad \Rightarrow \quad du = 2dx\quad \Rightarrow \quad \begin{array}{ccc}
x\| & 0 & 4 \\
\hline
u\| & 1 & 9 \\
\end{array}
\\

\frac{{u - 1}}{{2\sqrt u }} = \frac{{u^{1 - \frac{1}{2}} - u^{ - \frac{1}{2}} }}{2}\quad \Rightarrow \quad \frac{1}{4}\int\limits_1^9 {\left( {u^{\frac{1}{2}} - u^{ - \frac{1}{2}} } \right)du}\)

 

[galactus]

There ya' go