Thread: Need help with this integral

1. 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.

2. 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:

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

Provided that $c \in (a, b)$

3. 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?

4. 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.

5. ah ok thanks. that makes it much clearer.

6. 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?

7. 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 :

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

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