# Integrating Square Roots - Absolute Value Needed?

1. Mar 31, 2012

### Qube

1. The problem statement, all variables and given/known data

http://i.minus.com/i61zvy2BbtqkI.png [Broken]

2. Relevant equations

One can factor the polynomial to (x-1)^2

3. The attempt at a solution

After factoring the polynomial, I integrate (x-1) given the bounds of 0 and 1. I get -1/2. The solution manual says the answer is positive 1/2. What am I doing wrong?

Last edited by a moderator: May 5, 2017
2. Mar 31, 2012

### Redbelly98

Staff Emeritus
The square root of any quantity is taken to be a positive value, by convention.

But you ended up with (x-1) which is negative, not positive. Can you think of how to fix that?

3. Mar 31, 2012

### Qube

x-1 is not necessarily negative. I assume that your point is that (x-1) is negative over the the open interval of 0 to 1 (0 to 1 also happen to be the bounds of integration).

To rectify this problem, it seems, I would have to integrate the absolute value of (x-1) over the bounds of integration, right?

4. Mar 31, 2012

### LCKurtz

Yes. On [0,1] how can you express |x-1| without the absolute value signs?

5. Mar 31, 2012

### RoshanBBQ

In general, if you have
$$\sqrt{a^2} = |a|$$
for real a.

You can see this with a couple of specific examples:
a = -3
$$\sqrt{(-3)^2} = \sqrt{9} = 3 = |a|$$
a = 3
$$\sqrt{3^2} = \sqrt{9} = 3 = |a|$$

6. Mar 31, 2012

### Qube

To express (x-1) without absolute value signs, distribute a negative 1 to each value.

That results in (-x+1)

Now I just find the integral of (1-x) over the bounds of 0 to 1.

To conclude: should I always use absolute value signs when integrating a square root function over bounds which result in the square root of a negative number?

7. Mar 31, 2012

### SammyS

Staff Emeritus
It wasn't the square root of a negative number,

it was the square root of the square of a negative number.

8. Mar 31, 2012

### Ray Vickson

The use of absolute value signs is a pain; just use instead the actual forms that are >= 0 over all parts of your integration range. Just for fun, and to bring home the point, try the following:
$$\int_0^2 \sqrt{x^2 - 2x + 1} \; dx.$$

RGV

9. Mar 31, 2012

### Qube

Thank you for providing the additional practice :)! I was stumped by the problem for quite a while, but it turns out that my algebra was at fault. Here's my solution:

$$\int_0^2 \sqrt{x^2 - 2x + 1} \; dx.$$

= $$\int_0^2 \left| (x - 1) \right|\; dx.$$

(x-1) is negative when x < 1, therefore we must distribute a negative 1 to each term to get (-x+1)

(x-1) is positive when x ≥ 1.

Given these conditions, we can now split up the integral and appropriately set the bounds of integration.

= $$\int_0^1 \ {(1 - x)} \; dx.$$ + $$\int_1^2 \ {(x - 1)} \; dx.$$

= $\frac{1}{2}$ + 1 + ($\frac{-1}{2}$ + 1)

= 1

The takeaway seems to be:

When integrating a polynomial under a square root function, always:

1) Simplify the square root
2) Cancel out exponents
3) Use absolute value signs
4) Integrate using the appropriate bounds

I wish to really understand this topic. We use absolute value signs because the square root of any number squared has two solutions except for the number 0.

To account for that extra solution, we must use the absolute value signs, which forces us to split up the integral. Is this correct?

Last edited: Mar 31, 2012
10. Mar 31, 2012

### Dick

I think you've got that right.