Integral area coefficient

horia

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Hello, I am trying to find coefficients for a 2nd degree polynome -> f(x) = ax² + bx + c ==>
gif.latex
I am looking for alpha so that this equation is always true.

a and b are the roots of my function, and I want the area under the curve to always be equal the difference between root 1 and root 2.

:)) thanks
 
Hello, I am trying to find coefficients for a 2nd degree polynome -> f(x) = ax² + bx + c ==>
gif.latex
I am looking for alpha so that this equation is always true.

a and b are the roots of my function, and I want the area under the curve to always be equal the difference between root 1 and root 2.

:)) thanks
Please share your work/thought regarding this assignment - so that we know where to begin to help you.
 
Hey I've started integrating f(x) and found
gif.latex
(the primitive is between b and a ofc).
Thing is, whatever I find it is false. :'( Is it something to do with absolute values ?
 
Hey I've started integrating f(x) and found
gif.latex
(the primitive is between b and a ofc).
Thing is, whatever I find it is false. :'( Is it something to do with absolute values ?
What do you get when you just work with the left-hand-side of the equation? Complete the definite integral part.
 
I do get
gif.latex
But when inputing values like : (x,y) : (1,2) ; (1,3) . I don't get back what I expect, being :
(1,2) => Area = 1
and
(1,3) => Area = 2

Should I be doing this by another method?
 
I do get
gif.latex
But when inputing values like : (x,y) : (1,2) ; (1,3) . I don't get back what I expect, being :
(1,2) => Area = 1
and
(1,3) => Area = 2

Should I be doing this by another method?
Do not write the "equality" equation at each step. Just work with the "integral" part.

Factor out (a-b) from the expression of the integral.
 
Yes sorry I was just doing it for more clarity throughout the equation, but keeping in or out the integral part isn't too hard, it's just that I'm not finding a good solution.
 
Hello, I am trying to find coefficients for a 2nd degree polynome -> f(x) = ax² + bx + c ==>
gif.latex
I am looking for alpha so that this equation is always true.

a and b are the roots of my function, and I want the area under the curve to always be equal the difference between root 1 and root 2.

:)) thanks

A couple things need to be clarified. First, it doesn't seem that "f(x) = ax² + bx + c" is relevant at all; certainly a and b are not the roots of that polynomial, and you haven't used f anywhere. Second, when you say "always true", do you mean that alpha has to be a fixed number such that the equation is true for all a and b, or can alpha be a function of a and b? The latter seems necessary, based on what I find to be true; but that doesn't seem to be what you are saying.

If you haven't quoted the exact problem as given to you (or if this is only part of a larger problem), please state it word for word so we can be sure what you need to do.
 
A couple things need to be clarified. First, it doesn't seem that "f(x) = ax² + bx + c" is relevant at all; certainly a and b are not the roots of that polynomial, and you haven't used f anywhere. Second, when you say "always true", do you mean that alpha has to be a fixed number such that the equation is true for all a and b, or can alpha be a function of a and b? The latter seems necessary, based on what I find to be true; but that doesn't seem to be what you are saying.

If you haven't quoted the exact problem as given to you (or if this is only part of a larger problem), please state it word for word so we can be sure what you need to do.

Hello thank you for your response, no the f(x) is not relevant per se, I am just being explicit that f() is 2nd degree polynomial. The actual function I'm using is alpha(x-a)(x-b).
Anyway, alpha is not a fixed number, it depends of a,b.
like mentioned above alpha should be so that : if (a,b) : (1,2) or (2,4) THEN INTEGRAL(f()) = (1) or (2).
I hope I am being clear enough ! :)
 
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Hello thank you for your response, no the f(x) is not relevant per se, I am just being explicit that f() is 2nd degree polynomial. The actual function I'm using is a(x-a)(x-b).
Anyway, alpha is not a fixed number, it depends of a,b.
like mentioned above alpha should be so that : if (a,b) : (1,2) or (2,4) THEN INTEGRAL(f()) = (1) or (2).
I hope I am being clear enough ! :)

Good. That means that what Subhotosh Khan has been advising will lead to an answer. Have you done it? Just simplify your expression for the integral, either by expanding it to a standard cubic, or factoring out (b-a) from every term first.
 
Good. That means that what Subhotosh Khan has been advising will lead to an answer. Have you done it? Just simplify your expression for the integral, either by expanding it to a standard cubic, or factoring out (b-a) from every term first.

I don't see how to factor out b-a from every term, and I can't really manage to simplify it more than here :
gif.latex

If I factor stuff out it still isn't very clear to me how it's not solving :
gif.latex
 
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I don't see how to factor out b-a from every term, and I can't really manage to simplify it more than here :
gif.latex

If I factor stuff out it still isn't very clear to me how it's not solving :
gif.latex

That isn't factoring; it's dividing. You were told not to work on the entire equation yet.

What you've been told to do is to just simplify the expression alpha(b3/3 - a3/3 + (b + a)(a2/2 - b2/2) + (ab)(b - a)). You can expand (some say "foil") by distributing (b + a)(a2/2 - b2/2), and (ab)(b - a), then combine like terms.

Or, you can observe that both (b3 - a3) and (a2 - b2) can be factored, one factor being (b - a). That is then a common factor of all terms.
 
That isn't factoring; it's dividing. You were told not to work on the entire equation yet.

What you've been told to do is to just simplify the expression alpha(b3/3 - a3/3 + (b + a)(a2/2 - b2/2) + (ab)(b - a)). You can expand (some say "foil") by distributing (b + a)(a2/2 - b2/2), and (ab)(b - a), then combine like terms.

Or, you can observe that both (b3 - a3) and (a2 - b2) can be factored, one factor being (b - a). That is then a common factor of all terms.


Oh, okok. But my problem is not figuring out the final expression, my problem is that the formular I'm getting isn't working. When inputing test values into the formula (although it isn't simplified and this doesn't change anything) the resulting value is not correct.
I'm thinking maybe the method isn't correct. Again I want a 2nd degree polynomial expression where given 2 points (a,0) and (b,0) will give a surface (integral) of b-a.
I have been trying the method above but it doesn't seem to be correct as the values are false :/.

Simplified or not it doesn't matter I'm not looking for the better looking form of the result I already have and doesn't work.
 
That isn't factoring; it's dividing. You were told not to work on the entire equation yet.

What you've been told to do is to just simplify the expression alpha(b3/3 - a3/3 + (b + a)(a2/2 - b2/2) + (ab)(b - a)). You can expand (some say "foil") by distributing (b + a)(a2/2 - b2/2), and (ab)(b - a), then combine like terms.

Or, you can observe that both (b3 - a3) and (a2 - b2) can be factored, one factor being (b - a). That is then a common factor of all terms.


Here is the simplified form if you will :

gif.latex
 
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Here is the simplified form if you will :

gif.latex

So, if this is equal to b-a, what does alpha have to be? Don't just stop here.

But first, finish simplifying (distribute and combine like terms). Simplifying is not just something teachers demand of you because they feel like it; it makes the next steps easier, and also makes the result look a lot neater.

Oh, okok. But my problem is not figuring out the final expression, my problem is that the formular I'm getting isn't working. When inputing test values into the formula (although it isn't simplified and this doesn't change anything) the resulting value is not correct.
I'm thinking maybe the method isn't correct. Again I want a 2nd degree polynomial expression where given 2 points (a,0) and (b,0) will give a surface (integral) of b-a.
I have been trying the method above but it doesn't seem to be correct as the values are false :/.

Simplified or not it doesn't matter I'm not looking for the better looking form of the result I already have and doesn't work.

Until you find alpha and put it into the integrand, you have no formula to check! You haven't answered the question yet.

And when you do find alpha, then you will necessarily get the b-a that you expect, because that's how you are choosing alpha.
 
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