# Need help with this integral

• Oct 4th 2008, 12:45 AM
Hweengee
Need help with this integral
Hi guys

I've been doing some practice problems on integration and this particular one stumped me.

Int(|sqrt(x)-1|.dx)

Evaluate the integral at (0,4).

So far I've gotten this when x>1, sqrt(x)-1>0, so the integral is 2/3.x^(3/2)-x+C. I'm still stuck on the (0,1) portion.

I'm a little rusty at integration so any advice would be appreciated.

Thanks.
• Oct 4th 2008, 12:52 AM
Chop Suey
Notice that on the interval [0,1], sqrt(x) - 1 is less than 0 or equal to zero, and from [1, 4] it is greater than or equal to zero. Then you can split the integral into two, and then get rid of absolute value. Remember that:

$\displaystyle \int^{b}_{a} f(x)~dx =\int^{c}_{a} f(x)~dx + \int^{b}_{c} f(x)~dx$

Provided that $\displaystyle c \in (a, b)$
• Oct 4th 2008, 12:57 AM
Hweengee
Yea I've tried splitting them up. Would it be right to say that on (0,1) it can be integrated as -sqrt(x)+1?
• Oct 4th 2008, 01:01 AM
Chop Suey
Quote:

Originally Posted by Hweengee
Yea I've tried splitting them up. Would it be right to say that on (0,1) it can be integrated as -sqrt(x)+1?

Yes. :D
• Oct 4th 2008, 01:02 AM
Hweengee
ah ok thanks. that makes it much clearer.
• Oct 4th 2008, 01:08 AM
Hweengee
Ok this may sound weird, but what happens when x<0? the root of a negative number is complex so what happens to the absolute value of a complex number minus 1?
• Oct 4th 2008, 05:24 AM
Moo
Quote:

Originally Posted by Hweengee
Ok this may sound weird, but what happens when x<0? the root of a negative number is complex so what happens to the absolute value of a complex number minus 1?

In that case, consider the absolute values signs as modulus :

$\displaystyle |a+ib|=\sqrt{a^2+b^2}$

Otherwise, you can use the complex logarithms, but it's not what you have here ;)