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    Hello!

    I have two questions about two cases of an integration. The integration  Fa(x)=\int_{a}^{x}(2t^2 + 4t) \, dt is given.

    Fa(x) = 4/3 and a = 0 is the first case in which I should find out the value of x. That is my calculation:

     Fa(x)=\int_{0}^{x}(2t^2 + 4t) \, dt =\frac{4}{3}

     = 2\int_{0}^{x}t^2 \, dt + 4\int_{0}^{x}t \,dt=\frac{4}{3}

     = \frac{2*1*x^3}{3}+ \frac{4*1*x^2}{2} =\frac{4}{3}

    =\frac{2x^3}{3}+  2x^2 =\frac{4}{3}

    in the end I got the value of x by trials. It is x = 0,732 (ca.). My question in this case is, what I have to do to solve my calculation to x? What should I do to complete my calculation?

    In the second case x is 2 and the integration is a function of zero. The value of a is ask for. After an integration rule, namely \int_{a}^{a}f(x) \, dx = 0 I suppose that the value of a must be 2 too. Here is my calculation:

    f(x)=\int_{a}^{2}(2t^2 + 4t) \, dt = 0

    =2\int_{a}^{2}t^2 \, dt + 4\int_{a}^{2}t \, dt = 0

    =2\int_{0}^{2}t^2 \, dt - 2\int_{0}^{a}t^2 \, dt  + 4\int_{0}^{2} \, dt - 4\int_{0}^{a}t \, dt = 0

    =\frac{2*1*2^3}{3}- \frac{2*a^3}{3}+ \frac{4*2^2}{2}-\frac{4*a^2}{2}= 0

    =\frac{16}{3}- \frac{2a^3}{3}+8 - \frac{4a^2}{2} = 0

    =\frac{40}{3}- \frac{2a^3}{3}-2a^2 = 0

    And when I put a = 2 in the equation, I get the function of zero for the integration, but my question is the same: what I have to do to solve the equation to a? I would love to know the rest of the solution processes. I guess it is possible to use the solution process of a cubic function in the second case to get the value of a. And in the first case? The solution process of a cubic function makes not sense in my opinion, as the integration is not a zero of a function.

    Thanks in advance!
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    Why are the variables x and t, as well the integration operators dt and dx randomly scattered all over the place?

    Peace.
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    (Original post by Kallisto)
    \frac{2x^3}{3}+  2x^2 =\frac{4}{3}

    in the end I got the value of x by trials. It is x = 0,732 (ca.). My question in this case is, what I have to do to solve my calculation to x? What should I do to complete my calculation?

    Re-arrange your equations into the standard form of a polynomial. Multiply through by 3/2 to simplify it a bit and get integer coefficients.

    Now try factors (positive and negative) of the constant term as possible roots, and you'll find an easy one.

    Using polynomial division you get a linear factor and a quadratic one, and you can use the standard formula on the quadratic one to get the exact answer.

    Which is:
    Spoiler:
    Show

    \sqrt{3}-1


    I presume x is >=0, otherwise you'll have problems eliminating the other solutions.

    In the second case x is 2 and the integration is a function of zero. The value of a is ask for. After an integration rule, namely \int_{a}^{a}f(x) \, dx = 0 I suppose that the value of a must be 2 too.
    Yes, that is one possible value of "a". And the remainder of your working will show that there is no other possible value.

    \frac{40}{3}- \frac{2a^3}{3}-2a^2 = 0

    And when I put a = 2 in the equation, I get the function of zero for the integration, but my question is the same: what I have to do to solve the equation to a?
    Similar method to the first one. Except in this case you know one of the roots - it's 2. So, by the factor theorem, (a-2) is a factor. If you take that out you're left with a quadratic, which in this case has no roots.
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    (Original post by WhiteGroupMaths)
    Why are the variables x and t, as well the integration operators dt and dx randomly scattered all over the place?

    Peace.
    My bad! when I establish the algorithm, I have overlooked the variables in integration. I will change it. Thanks!

    (Original post by ghostwalker)
    Re-arrange your equations into the standard form of a polynomial. Multiply through by 3/2 to simplify it a bit and get integer coefficients.
    (...)
    Why I should multiply by 3/2? or did you mean 2/3? as the term is \frac{2x^2}{3} I would multiply by three and then I get  2x^3 + 6x^2 = 4 and now I subtract the equation by 4 and I get  2x^3 + 6x^2 - 4 = 0 and then it is possible to use a polynomial division which you told, I would get an quadratic function and I am able to solve the variable by pq-formula. But then I apply these steps, I would calculate a value of a zero of a function. That's why it makes no sense to use a polynomial division in my opinion.

    (Original post by ghostwalker)
    Yes, that is one possible value of "a". And the remainder of your working will show that there is no other possible value.

    Similar method to the first one. Except in this case you know one of the roots - it's 2. So, by the factor theorem, (a-2) is a factor. If you take that out you're left with a quadratic, which in this case has no roots.
    Could you be more precise, please? I don't still know what should I do exactly in both of them.
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    (Original post by Kallisto)
    Why I should multiply by 3/2? or did you mean 2/3? as the term is \frac{2x^2}{3} I would multiply by three and then I get  2x^3 + 6x^2 = 4 and now I subtract the equation by 4 and I get  2x^3 + 6x^2 - 4 = 0
    Well, multiplying by 3/2 gets you to x^3+3x^2-2=0

    and then it is possible to use a polynomial division which you told, I would get an quadratic function and I am able to solve the variable by pq-formula. But then I apply these steps, I would calculate a value of a zero of a function. That's why it makes no sense to use a polynomial division in my opinion.
    You need to find a root first by trying out factors of -2, i.e. 1,-1,2,-2.

    Once you have a factor, you can then use polynomial division.

    You are trying to find the zeros (roots) of the equation, to find the values of x that satisfy it. Polynomial division is the way to go.

    Could you be more precise, please? I don't still know what should I do exactly in both of them.
    Which part don't you understand, and need more details for?
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    (Original post by ghostwalker)
    (...) Which part don't you understand, and need more details for?
    I didn't understand why you multiply by 3/2, but then I have found out that there is no difference whether I multiply by 3/2 or divide by 2/3 which I did. Moreover I didn't understand why you use a polynomial division, as I thought I would get a value of x for a zero of a function instead off 4/3. For that reason I exclude a polynomial division from the beginning. But when I followed the steps which you told, I got the value of x for 4/3 indeed!

    First I divide by 2/3, after that I substrate by 2 and then I got the equation which you posted above:  x^3 + 3x^2 - 2 = 0 . And then I used a polynomial division by x+1:  x^3 + 3x^2 - 2 : (x+1) = x^2 + 2x - 2 .
    To get the value of x, I used a pq-fromula. I got x = 0,732. Then I put the value in the term of integration and I got 4/3!
    I have found out a solution process at last! I'm so happy that it works in the first case.

    I did the same procedure in the second one, but it doesn't work! I used a polynomial division by x+2... perhaps it would work, if I used an other term as divisor?
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    I have found out that there is no difference whether I multiply by 3/2 or divide by 2/3
    /facepalm

    I used a polynomial division by x+2...
    If x=2 is a root then (x MINUS 2)=0 and is a factor.
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    When I divide \frac {-2a^3} {3} - 2a^2 + \frac {40} {3} = 0 by -2/3, I get  a^3 + 3a^2 - 20 = 0 . And when I use a polynomial division by x-2 I get:  a^2 + 5a + 10 . I see at the first glance that a = 2 is not a zero of a function in the term. What is wrong again?
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    (Original post by Kallisto)
    When I divide \frac {-2a^3} {3} - 2a^2 + \frac {40} {3} = 0 by -2/3, I get  a^3 + 3a^2 - 20 = 0 . And when I use a polynomial division by x-2 I get:  a^2 + 5x + 10 . I see at the first glance that a = 2 is not a zero of a function in the term. What is wrong again?
    a=2 is a zero of the function  a^3 + 3a^2 - 20 = 0

    When you take out the factor (a-2), then a=2 will not be a zero of what remains  a^2 + 5a + 10

     a^3 + 3a^2 - 20\equiv (a-2) (a^2 + 5a + 10)

    The quadratic has no real roots, and hence the only zero is a=2.
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    (Original post by ghostwalker)
    (...)
    The quadratic has no real roots, and hence the only zero is a=2.
    In other words: it is useless to use the polynomial division. That is to say I have to solve the cubic function  a^3 + 3a^2 - 20 = 0 to get the variable. Am I right? then I have to use the solution process of a cubic function. That is the only possibility which makes sense in my opinion.
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    (Original post by Kallisto)
    In other words: it is useless to use the polynomial division. That is to say I have to solve the cubic function  a^3 + 3a^2 - 20 = 0 to get the variable. Am I right? then I have to use the solution process of a cubic function. That is the only possibility which makes sense in my opinion.
    No.

    You know x=2 is a solution to Fa(x)=0

    So, (x-2) is a factor.

    After polynomial division, the remaining quadratic has no roots, no zeros.

    So x=2 is the only solution.

    Done.
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    Maybe this image will help you understand?

    Name:  Screen Shot 2013-03-24 at 4.01.54 PM.png
Views: 76
Size:  19.6 KB
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    (Original post by ghostwalker)
    No.

    You know x=2 is a solution to Fa(x)=0

    So, (x-2) is a factor.

    After polynomial division, the remaining quadratic has no roots, no zeros.

    So x=2 is the only solution.

    Done.
    Then I use the integration rule which I have posted above to reason my answer and then I put a = 2 in the equation
     a^3 + 3a^2 - 20 = 0 and I'm done at last. Why I should use a polynomial division, if get a term which has no roots and no zeros?

    (Original post by aznkid66)
    Maybe this image will help you understand?

    Name:  Screen Shot 2013-03-24 at 4.01.54 PM.png
Views: 76
Size:  19.6 KB
    Thanks for effort!
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    Because you want to check if there are any remaining roots, aside from a=2. What you were getting tripped up on was that there were not any remaining roots, so the whole process seemed pointless to you. However, if you recall the first exercise, it is a quick and easy way to find the other 2 roots without testing more integrals, as we have the quadratic formula. Your original question was "what else do I have to do to solve the equation for a?" and the answer is to find the other roots, in which case none exist and the final answer is "nothing else".
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    (Original post by Kallisto)
    Then I use the integration rule which I have posted above to reason my answer.
    That will get you an answer, yes.

    But it does not show you that it is the only answer.

    Hence the work I outlined above.

    You seem very keen to solve a general cubic equation, but it is not necessary.

    I suspect your command of English is making this a lot more difficult than it needs to be.
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    I guess it is enough to use a polynomial division to get the quadratic function from the cubic one. After that I use the pq-formula to show up that there is no another solution. I realize that the solution process of a cubic function is too complicate, especially there is an easier way to solve it.
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    As there is no answer anymore and my problems seem to be solved, this thread is answered. Thanks for help!
 
 
 
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