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Thread: Area between two curves: y=2x^(1/3), y=(1/8)x^2, 0<=x<=6

  1. #1

    Area between two curves: y=2x^(1/3), y=(1/8)x^2, 0<=x<=6

    Sketch the area between two curves and then find the area. This is using summations.

    y=2x1/3, y=(1/8)x2, 0<=x<=6

    So I got the sketch okay, but I feel I am making some mistake somewhere in the algebraish part.

    IGNORE THIS WORK! CHECK THE COMMENTS. I posted a new version with the right integration and still got the wrong answer. Look at that!

    1. Subtract the bottom curve from the other. [2x1/3-(1/8)x2]

    2. Get the antiderivtive. [(3/4)(2x4/3)-(1/24x3)]=[6/4x4/3-(1/24)x3]

    3. Plug the values 6 and 0 in and solve

    [(6/4)(64/3)-(1/24)(63)]-[(6/4)(04/3)-(1/24)(03)]

    =[(6/4)(64/3)-9]-0

    The book says the answer is (47/3)-[(9/2)(121/3) which is between 5 and 6. However, when I plug the value above into a calculator, I get between 7 and 8. Where am I going wrong?
    Last edited by crybloodwing; 12-03-2017 at 10:35 PM.

  2. #2
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    Quote Originally Posted by crybloodwing View Post
    Sketch the area between two curves and then find the area. This is using summations.

    y=2x1/3, y=(1/8)x2, 0<=x<=6

    So I got the sketch okay, but I feel I am making some mistake somewhere in the algebraish part.

    1. Subtract the bottom curve from the other. [2x1/3-(1/8)x2]

    2. Get the antiderivtive. [(3/4)(2x4/3)-(1/24x3)]=[6/4x4/3-(1/24)x3]

    3. Plug the values 6 and 0 in and solve

    [(6/4)(64/3)-(1/24)(63)]-[(6/4)(04/3)-(1/24)(03)]

    =[(6/4)(64/3)-9]-0

    The book says the answer is (47/3)-[(9/2)(121/3) which is between 5 and 6. However, when I plug the value above into a calculator, I get between 7 and 8. Where am I going wrong?
    Since you have sketched the curves, you do see that the curves intersect at some point before x = 6.

    You would need to do this problem in two parts - before and after the point of intersection.

    Try it and let us know what you found.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Quote Originally Posted by Subhotosh Khan View Post
    Since you have sketched the curves, you do see that the curves intersect at some point before x = 6.

    You would need to do this problem in two parts - before and after the point of intersection.

    Try it and let us know what you found.

    So intersection at (4,2)

    [(6/4)x4/3-[(1/24)x3] from 0-4, and [(1/24)x3-(6/4)x4/3] from 4-6

    Then [(6/4)(44/3)-(8/3)] -0 + [9-(6/4)(64/3)]-[(8/3)-(6/4)(44/3)]

    [(6/4)(44/3)-8/3] + [(19/3)-(6/4)(64/3)+(6/4)(44/3)]

    (12/4)(44/3)-(6/4)(64/3)+(11/3)

    3(6.3496)-1.5(10.903)+3.66666667=

    19.0488- 16.3545+3.66666667= 6.36096 but the answer is (47/3)-(9/2)(121/3​) or around 5.364

  4. #4
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    Quote Originally Posted by crybloodwing View Post
    So intersection at (4,2)

    [(6/4)x4/3-[(1/24)x3] from 0-4, and [(1/24)x3-(6/4)x4/3] from 4-6

    Then [(6/4)(44/3)-(8/3)] -0 + [9-(6/4)(64/3)]-[(8/3)-(6/4)(44/3)]

    [(6/4)(44/3)-8/3] + [(19/3)-(6/4)(64/3)+(6/4)(44/3)]

    (12/4)(44/3)-(6/4)(64/3)+(11/3)

    3(6.3496)-1.5(10.903)+3.66666667=

    19.0488- 16.3545+3.66666667= 6.36096 but the answer is (47/3)-(9/2)(121/3​) or around 5.364
    How did you get that??

    Please show your work for that.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
    Quote Originally Posted by Subhotosh Khan View Post
    How did you get that??

    Please show your work for that.
    1. Y=(2(4))1/3 y=81/3 y=2

    2. y=1/8(4)2 y=1/8(16) y=2

    So putting 4 into both equations gives 2.

    So the curves intersect at (4,2) so I then get the area from 0-4 and add it to the area from 4-6.
    Last edited by crybloodwing; 11-30-2017 at 10:30 AM.

  6. #6
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    Quote Originally Posted by crybloodwing View Post
    1. Y=(2(4))1/3 y=81/3 y=2

    2. y=1/8(4)2 y=1/8(16) y=2

    So putting 4 into both equations gives 2.

    So the curves intersect at (4,2) so I then get the area from 0-4 and add it to the area from 4-6.
    In your original post you had:

    y=2x1/3 → y = 2 * x1/3

    That is very different from:

    y = (2x)1/3............... (the form you have used later) .............. which one is correct among these two?

    That could be source of your troubles!!
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  7. #7
    Quote Originally Posted by Subhotosh Khan View Post
    In your original post you had:

    y=2x1/3 → y = 2 * x1/3

    That is very different from:

    y = (2x)1/3............... (the form you have used later) .............. which one is correct among these two?

    That could be source of your troubles!!

    It is (2x)1/3 that is not my problem because I have that equation memorized very well after doing the problem 3 times already. And on the actually page, I used extra parentheses and wrote it out the way with the sign. That is just too hard on here. I can send a picture of the whole page of work.

    Actually, I think I know what I did. I will test it.

    Still wrong, but closer. I was able to get the cubed root of 12 in the answer....

    NEW WORK RIGHT HERE

    From 0-4

    [3/4(2x)4/3-(1/24)x3] - [3/4(0)-(1/24)(0)]
    [3/4(84/3)-(1/24)(64)]
    [3/4(16)-(8/3)]
    [(48/4)-(8/3)]
    [12-(8/3)]
    [(36/3)-(8/3)]
    (28/3)

    From 4-6

    [1/24(x3)-(3/4)(2x)4/3]
    [1/24(63)-(3/4)(124/3)] - [1/24(43)-3/4(44/3)]
    [1/24(216)-(3/4)(124/3)] - [(8/3)-(36/3)]
    [9-(3/4)(124/3)]-(-(28/3)]
    [(27/3+28/3)-(3/4)(124/3)]
    ((55/3)-(3/4)(124/3)]

    Added together

    [(28/3)+(55/3)-(3/4)(124/3)]

    (83/3)-3/4(124/3)

    124/3 =(20736)1/3 = (2)(25921/3)=4(3241/3)=12(121/3)

    (83/3)-(36/4)(121/3)

    (83/3)-9(121/3)

    27.66666- 20.604= 7.06....still not the answer.
    Last edited by crybloodwing; 12-03-2017 at 10:35 PM.

  8. #8
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    Quote Originally Posted by crybloodwing View Post
    It is (2x)1/3 that is not my problem because I have that equation memorized very well after doing the problem 3 times already. And on the actually page, I used extra parentheses and wrote it out the way with the sign. That is just too hard on here. I can send a picture of the whole page of work.

    Actually, I think I know what I did. I will test it
    Oh for goodness sake: it is too hard to write (2x)^(1/3) rather than than 2x^(1/3). Whether you know what you mean does not help us know what you mean.

    In any case the anti derivative of

    [tex](2x)^{1/3}[/tex] is [tex]\dfrac{3}{4} * \sqrt[3]{2} * x^{4/3} \ne \dfrac{3}{4} * 2 * x^{4/3} = \dfrac{6}{4} * x^{4/3}.[/tex]

    So if the problem involves [tex](2x)^{1/3}[/tex], then all your work shown above is wrong.

    In that case, your finding it too much work to put in a left and a right parenthesis has caused you to waste hours of time. My grandfather used to say, "The lazy man works the hardest."

  9. #9
    Quote Originally Posted by JeffM View Post
    Oh for goodness sake: it is too hard to write (2x)^(1/3) rather than than 2x^(1/3). Whether you know what you mean does not help us know what you mean.

    In any case the anti derivative of

    [tex](2x)^{1/3}[/tex] is [tex]\dfrac{3}{4} * \sqrt[3]{2} * x^{4/3} \ne \dfrac{3}{4} * 2 * x^{4/3} = \dfrac{6}{4} * x^{4/3}.[/tex]

    So if the problem involves [tex](2x)^{1/3}[/tex], then all your work shown above is wrong.

    In that case, your finding it too much work to put in a left and a right parenthesis has caused you to waste hours of time. My grandfather used to say, "The lazy man works the hardest."
    The last time I commented, I did all that work with the cubed root of 2x. So I had that derivative right. So simply look through the work of my last post, and see if there are any errors. That is what I am asking. I am not asking about the whole problem right now. I already fixed the error with the derivative, so now I think it is algebra or something. If there were no errors in my algebra, I will look at the derivative again and see if I wrote that wrong. It is almost like you did not look through all the work I wrote in the last reply.....you will see that I kept the 3/4th out of the (2x4/3​) the whole time.
    Last edited by crybloodwing; 11-30-2017 at 07:15 PM.

  10. #10
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    Quote Originally Posted by crybloodwing View Post
    It is (2x)1/3 that is not my problem.
    That is exactly your problem. You did not integrate your function properly because you did not know how to cope with the coefficient.

    [tex]\displaystyle \int \ (2x)^{1/3} \ dx = \int \ 2^{1/3} * x^{1/3} \ dx = \dfrac{3 * 2^{1/3}x^{4/3}}{4} + C \ne \dfrac{3(2x)^{4/3}}{4} + C.[/tex]

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