# Intergration by substitution

1. Feb 24, 2006

### Hootenanny

Staff Emeritus
The question is to find the following intergal:
$$\int x\cdot u^{\frac{1}{2}$$ where $u = 2x -1$.
$$= \int x\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du$$
$$u = 2x -1 \Rightarrow x = \frac{u+1}{2}$$
$$= \int \frac{1}{2}(u+1)\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du \;\; = \int \frac{1}{4}\left( u^{\frac{3}{2}} + u^{\frac{1}{2}}\right) \;\; du$$
$$= \frac{2}{20} u^{\frac{5}{2}} + \frac{2}{12} u^{\frac{3}{2}} = \frac{1}{10} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}}$$

But this isnt going to give me the correct answer. Can anybody see where I've gone wrong? Thank's

2. Feb 24, 2006

### HallsofIvy

Staff Emeritus
If u= 2x- 1 then 2x= u+1 so x= (1/2)(u+1). Substitute THAT for x and the integral becomes
$$\frac{1}{2}\int(u+1)u^{\frac{1}{2}}du= \frac{1}{2}\int (u^{\frac{3}{2}}+ u^{\frac{1}{2}})du$$
That should be easy to integrate.

3. Feb 24, 2006

### Hootenanny

Staff Emeritus
$$\frac{1}{2}\int (u^{\frac{3}{2}}+ u^{\frac{1}{2}})du = \frac{1}{2} \left[ \frac{2}{5} u ^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}} \right]$$

Is that correct?

4. Feb 24, 2006

### Hootenanny

Staff Emeritus
Or would i have to make a further substitution for $(u^{\frac{3}{2}}+ u^{\frac{1}{2}})$ ?

5. Feb 24, 2006

### TD

That's correct, but don't forget the constant of integration

6. Feb 24, 2006

### Hootenanny

Staff Emeritus
Thank's TD +c

7. Feb 24, 2006

### TD

You're welcome.

To get back to your last question, there is no need for an extra substitution since the integral is lineair, which means that:

$$\int {\alpha f\left( x \right) + \beta g\left( x \right)dx} = \alpha \int {f\left( x \right)dx} + \beta \int {g\left( x \right)dx}$$

So in your case, that gives:

$$\int {u^{\frac{3}{2}} + u^{\frac{1}{2}} du} = \int {u^{\frac{3}{2}} du} + \int {u^{\frac{1}{2}} du}$$

Then you can apply the standard exponent-rule, as you did.

8. Feb 24, 2006

### 0rthodontist

What? He had it completely right the first time. 1/4 is the correct thing to multiply by because dx = 1/2 du. I'm assuming there was a dx in the original integral because otherwise he wouldn't have included the 1/2 du in his second step.

Last edited: Feb 24, 2006