Integrating Square Roots - Absolute Value Needed?

In summary: One thing that you might find interesting and useful is to graph the function f(x) = \sqrt{x^2}.So, what does the absolute value function look like?It's a v-shaped graph, with the vertex at (0,0).What's the domain of this function?The domain is all real numbers, because you can square any real number and still have a real number.What's the range of this function?The range is all non-negative real numbers.In summary, When integrating a polynomial under a square root function, it is important to simplify the square root, cancel out exponents, use absolute value signs, and integrate using the appropriate bounds. This is because the square root of any number squared
  • #1
Qube
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Homework Statement



http://i.minus.com/i61zvy2BbtqkI.png

Homework Equations



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

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?
 
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  • #2
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
Redbelly98 said:
But you ended up with (x-1) which is negative, not positive. Can you think of how to fix that?

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
Qube said:
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?

Yes. On [0,1] how can you express |x-1| without the absolute value signs?
 
  • #5
In general, if you have
[tex]\sqrt{a^2} = |a|[/tex]
for real a.

You can see this with a couple of specific examples:
a = -3
[tex]\sqrt{(-3)^2} = \sqrt{9} = 3 = |a|[/tex]
a = 3
[tex]\sqrt{3^2} = \sqrt{9} = 3 = |a|[/tex]
 
  • #6
LCKurtz said:
Yes. On [0,1] how can you express |x-1| without the absolute value signs?

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
Qube said:
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?

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

it was the square root of the square of a negative number.
 
  • #8
Qube said:
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?

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:
[tex] \int_0^2 \sqrt{x^2 - 2x + 1} \; dx. [/tex]

RGV
 
  • #9
Ray Vickson said:
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:
[tex] \int_0^2 \sqrt{x^2 - 2x + 1} \; dx. [/tex]

RGV

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:

[tex] \int_0^2 \sqrt{x^2 - 2x + 1} \; dx. [/tex]

= [tex] \int_0^2 \left| (x - 1) \right|\; dx. [/tex]

(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.

= [tex] \int_0^1 \ {(1 - x)} \; dx. [/tex] + [tex] \int_1^2 \ {(x - 1)} \; dx. [/tex]

= [itex]\frac{1}{2}[/itex] + 1 + ([itex]\frac{-1}{2}[/itex] + 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?
 
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  • #10
I think you've got that right.
 

FAQ: Integrating Square Roots - Absolute Value Needed?

What is the purpose of integrating square roots?

The purpose of integrating square roots is to find the area under a curve represented by a square root function. This allows us to solve problems related to motion, work, and other real-world scenarios.

Why is absolute value needed when integrating square roots?

Absolute value is needed when integrating square roots because the area under the curve can be negative in some cases. Absolute value ensures that we get a positive value for the area and allows us to accurately calculate the total area under the curve.

How do you integrate square roots?

To integrate square roots, we use the power rule for integration, where we raise the exponent by 1 and divide the coefficient by the new exponent. For example, the integral of √x is 2/3x^(3/2) + C. We can also use substitution or other integration techniques depending on the complexity of the square root function.

Are there any special cases when integrating square roots?

Yes, there are special cases when integrating square roots. One example is when the square root is inside a trigonometric function, such as √(1-sin(x)^2). In this case, we use trigonometric identities to simplify the function before integrating.

What are the applications of integrating square roots in real life?

Integrating square roots has various applications in real life, such as calculating the work done by a variable force, determining the displacement of an object under a non-uniform acceleration, and finding the speed of an object at a particular time. It is also used in engineering, physics, and other fields to solve complex problems involving motion and work.

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