# Integration problem

1. Oct 2, 2004

### tandoorichicken

Hello everybody
I'm working on a section for integration by substitution and I came across an integral that I don't know how to do a substitution for

$$\int_{1}^{2} x\sqrt{x-1} \,dx$$

How can I do this problem?

2. Oct 2, 2004

### vsage

This isn't simple substitution. You can multiply the equation fancily by 1 and then use substitution. Tell me if this is too vague.

Edit: It seems someone else found it just by substitution. Oh well, I like my way better :p

Last edited by a moderator: Oct 2, 2004
3. Oct 2, 2004

### Clausius2

$$t=\sqrt{x-1}$$
$$x=t^2+1$$
$$dx=2tdt$$

$$\int_{make it you}^{make it you} (t^2+1)t^2 2dt$$

4. Oct 2, 2004

### arildno

Another choice is:
$$\int_{1}^{2}x\sqrt{x-1}dx=\int_{1}^{2}(x-1)^{\frac{3}{2}}dx+\int_{1}^{2}\sqrt{x-1}dx$$

5. Oct 2, 2004

### tandoorichicken

Thanks, everybody. Would you all care to check my work please?

$\int_{1}^{2} x\sqrt{x-1} \,dx$
$u = \sqrt{x-1}$
$x = u^2 + 1$
$\,du = \frac{1}{2\sqrt{x-1}}\,dx$
$\,dx = 2\sqrt{x-1} \,du = 2u\,du$
$\int_{1}^{2} x\sqrt{x-1} \,dx = \int_{0}^{1} (u^2 + 1)u 2u\,du = \int_{0}^{1} 2u^4 + 2u^2 \,du = 2\int_{0}^{1} u^4 \,du + 2\int_{0}^{1} u^2 \,du = 2 + 2 = 4$

Is this correct?

6. Oct 2, 2004

### arildno

Do the following integrals once more:
$$\int_{0}^{1}u^{4}du,\int_{0}^{1}u^{2}du$$

7. Oct 2, 2004

### tandoorichicken

Oh, whoops. Thanks for pointing that out.

$$\int u^4 \,du = \frac{u^5}{5} , \int u^2 \,du = \frac{u^3}{3}$$
So then it becomes
$$\frac{2}{5} + \frac{2}{3} = \frac{16}{15}$$

8. Oct 2, 2004

### arildno

Seems much better!