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agentredlum
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The question is in the title.
This is not a homework question, just curiosity on my part.
This is not a homework question, just curiosity on my part.
How? I thought the whole purpose of the absolute value stemmed back to distances along a number line, therefore you couldn't get a negative value out of the absolute value of a number or a function?HallsofIvy said:If x< 0, then |x|= -x
Bohrok said:I never knew wikipedia showed how to integrate it; just noticed it right now. I personally did it once because I wanted to derive it myself, and for higher odd powers of x such as |x3| and there is a pattern.[tex]\int|x^3|dx = \frac{1}{4}x^3|x| + c, \int|x^5|dx = \frac{1}{6}x^5|x| + c, \text{etc.}[/tex]So in general,[tex]\int|x^n|dx = \frac{1}{n + 1}x^n|x| + c \text{ for odd n}[/tex]
agentredlum said:Oh, this is interesting. Is there any difference to the answer if you represent it as |x^n|x/(n+1) + c for odd n ?
Bohrok said:No difference; it's the same way as how I have it. I should also add positive to the restriction on n: for positive odd n.
agentredlum said:What is the integral of |x^3 + x| ?
Bohrok said:I thought this would be difficult and involve integration by parts, but it turned out quite simple actually:
|x3 + x| = |x(x2 + 1)| = |x||x2 + 1| = |x|(x2 + 1) = |x|x2 + |x| = |x3| + |x|, using the facts that x2 + 1 = |x2 + 1| and x2 = |x2| since the former in each case is never negative, and |ab| = |a||b|, so
?|x3 + x| dx = ?|x3| dx + ?|x| dx
Once you integrate those, you can combine them and make the result look more condensed.
If you had |x3 - x|, that would be a different story!
disregardthat said:Split the integral as such:
[tex]\int^t_{1} |x||x^2-1| dx = \int^t_{1} |x|(x^2-1) dx[/tex] for t >= 1,
[tex]\int^t_{-1} |x||x^2-1| dx = \int^t_{-1} |x|(x^2-1) dx[/tex] for t <= -1
[tex]\int^t_{0} |x||x^2-1| dx = \int^t_{0} |x|(1-x^2) dx[/tex] for -1 < t < 1
These three integrals are solved by the method above. For each domain, scale by a constant to get an answer on the form F(x) + C, but it is not necessary.
agentredlum said:I'm still working on integral of |x^3 - x| and trying to make it co-operate with the shortcut, any ideas?
Bohrok said:I've been trying to work it out, but so far it hasn't turned out very good. However, Wolframalpha was able to integrate it (but the steps aren't much help) and it looks correct comparing the graph of the integral with its derivative. I probably need to sleep on it...
agentredlum said:According to wolframalpha...sqrt(abs(x)) has the same graph as abs(sqrt(x)) but that can't be right since domains are different
The first one has domain all real x
The second one has domain all rea NON-NEGATIVE x
What is going on here?
http://www.wolframalpha.com/input/?...,abs(x^(1/2))&asynchronous=false&equal=Submit
In the input interpretation it makes the correct interpretation but in the plots it doesn't.
Maybe it decided to consider them as complex functions, in which case they are (identical) functions whose domain is the entire complex plane.agentredlum said:According to wolframalpha...sqrt(abs(x)) has the same graph as abs(sqrt(x)) but that can't be right since domains are different
The first one has domain all real x
The second one has domain all rea NON-NEGATIVE x
Bohrok said:I've still had no luck in finding ∫|x3 - x| dx. I think it's great that WA found the integral in a "closed form" in the link I gave without breaking it up into different cases with different intervals, but it doesn't show the steps. Not even a time out, just "step-by-step results unavailable." It just seems impossible since the integral doesn't look very similar to |x3 - x|, yet differentiating it and then a lot of simplifying gives you |x3 - x| back.
Bohrok said:Sometimes WA gives results like that which don't make sense. I've tried getting graphs of functions where I tried to limit the domain to nonnegative numbers by using the square root but it would use negatives too...
I've still had no luck in finding ?|x3 - x| dx. I think it's great that WA found the integral in a "closed form" in the link I gave without breaking it up into different cases with different intervals, but it doesn't show the steps. Not even a time out, just "step-by-step results unavailable." It just seems impossible since the integral doesn't look very similar to |x3 - x|, yet differentiating it and then a lot of simplifying gives you |x3 - x| back.
While I can't remember if you're actually allowed to factor out piecewise constant functions like that, I do know that if you are, you have to be very careful to choose a piecewise constant of integration so the result is continuous.asw said:The trick is that sgn(A) = |A|/A for A≠0.
[itex]\displaystyle\int|x^3-x|\;\mathrm{d}x=\mathrm{sgn}(x^3-x)\int x^3-x\;\mathrm{d}x[/itex]
And this is not; it has jump discontinuities at 1 and -1.[itex]=\mathrm{sgn}(x^3-x).[\tfrac14x^4-\tfrac12x^2][/itex]
disregardthat said:If you follow the method of breaking the integral up into the three parts as shown you will get an antiderivative F(x) = f(x) + 1/2 when x >= 1, -f(x) when -1<x<1, and f(x)-1/2 when x <= -1, where f(x) = |x|x(x^2-2)/4.
asw said:The trick is that sgn(A) = |A|/A for A≠0.
[itex]\displaystyle\int|x^3-x|\;\mathrm{d}x=\mathrm{sgn}(x^3-x)\int x^3-x\;\mathrm{d}x=\mathrm{sgn}(x^3-x).[\tfrac14x^4-\tfrac12x^2][/itex]
[itex]\displaystyle =\frac{x^2(x^2-2)}{4}\cdot\frac{|x^3-x|}{x^3-x}=\frac{x(x^2-2)}{4}\cdot\frac{\sqrt{x^2(x^2-1)^2}}{x^2-1}[/itex]
The integral is a mathematical concept that represents the area under a curve in a graph. It is used to find the total value of a function over a given interval.
The absolute value function, denoted as |x|, is a mathematical function that returns the positive value of a number. It essentially ignores the negative sign of a number and returns the positive equivalent.
A definite integral has specific limits of integration, meaning that it calculates the area under a curve between two specific points. An indefinite integral, on the other hand, does not have specific limits and instead returns a general solution as a function of x.
The integral of |x| can be found by breaking it into two separate integrals and using the properties of absolute value. The integral of |x| from 0 to infinity is equal to the integral of x from 0 to infinity minus the integral of -x from 0 to infinity. After solving each integral individually, you can combine them to get the final answer.
The integral of |x| is significant because it represents the total value of the function |x| over a given interval. In other words, it gives us the area under the curve of |x|, which can be useful in various applications in physics, engineering, and economics.