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Not sure where Im getting the error (parametrically, the are under a curve) Watch

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    x = 3cost, y = 9sin2t I'm being asked to show that the equation to be integrated becomes Asin2tsint dt.and state A I have shown this and got A to be -27 the book says A to be 27 9sin2t(-3sint) = -27sin2tsintwhere dx/dt = -3sint I'm not sure where I'm going wrong???
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    Just from looking at it, I think you're right. The book is probably just wrong.
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    (Original post by Vanetti)
    x = 3cost, y = 9sin2t I'm being asked to show that the equation to be integrated becomes Asin2tsint dt.and state A I have shown this and got A to be -27 the book says A to be 27 9sin2t(-3sint) = -27sin2tsintwhere dx/dt = -3sint I'm not sure where I'm going wrong???
    Has the book swapped the limits of integration by any chance?

    Remember that if you swap the limits of an integral, then the value is multiplied by -1
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    (Original post by davros)
    Has the book swapped the limits of integration by any chance?

    Remember that if you swap the limits of an integral, then the value is multiplied by -1
    hmm...... "
    Remember that if you swap the limits of an integral, then the value is multiplied by -1
    "
    I have never ever heard about that , thank you very much it is not even in my textbook to state that. Would you be as kind as to elaborate?

    The integral is pi/2 on top and 0 on the bottom I believe this means the integrals were not swapped. Thanks for the information however how exactly does this work
    If I had 1 and 2 in that order top, bottom I would then do the integral limit 1 -2 ,as oppose to the standard 2 -1.?
    (would that then mean I put the - before I even integrate it???)
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    If they're just asking the for the area, then you take the magnitude of the integral. So even if the integral is -27, the area is 27.
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    (Original post by Vanetti)
    hmm...... "
    Remember that if you swap the limits of an integral, then the value is multiplied by -1
    "
    I have never ever heard about that , thank you very much it is not even in my textbook to state that. Would you be as kind as to elaborate?

    The integral is pi/2 on top and 0 on the bottom I believe this means the integrals were not swapped. Thanks for the information however how exactly does this work
    If I had 1 and 2 in that order top, bottom I would then do the integral limit 1 -2 ,as oppose to the standard 2 -1.?
    (would that then mean I put the - before I even integrate it???)
    If you think about what happens with the limits in integrals, you should be able to see that:

    \displaystyle\int^a_b y\ dx = -\displaystyle\int^b_a y\ dx
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    (Original post by Vanetti)
    hmm...... "
    Remember that if you swap the limits of an integral, then the value is multiplied by -1
    "
    I have never ever heard about that , thank you very much it is not even in my textbook to state that. Would you be as kind as to elaborate?
    To expand on what the poster above wrote, if

    \displaystyle \int f(x) dx = F(x) + c

    then

    \displaystyle \int^b_a f(x) dx = F(b) -F(a)

    and

    \displaystyle \int^a_b f(x) dx = F(a) -F(b)

    so it's an immediate consequence of the definition of an integral that if you swap the limits over then your answer gets multiplied by -1 - you wouldn't necessarily expect the textbook to state this explicitly.

    Normally when you integrate to find an area you work left-to-right, so the lower limit is the smaller x-value and the upper limit is the greater x-value. You need to be careful with parametric equations, especially when there are trig functions involved, because sometimes the smaller x-value corresponds to the bigger parameter value, and vice versa, so you should never assume that the smaller number goes at the bottom of the integral.

    I'd really need to see a screenshot of your book / markscheme to know if they've made an error in their working, or if you've misunderstood what they've asked you to do.
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    (Original post by davros)
    To expand on what the poster above wrote, if

    \displaystyle \int f(x) dx = F(x) + c

    then

    \displaystyle \int^b_a f(x) dx = F(b) -F(a)

    and

    \displaystyle \int^a_b f(x) dx = F(a) -F(b)

    so it's an immediate consequence of the definition of an integral that if you swap the limits over then your answer gets multiplied by -1 - you wouldn't necessarily expect the textbook to state this explicitly.

    Normally when you integrate to find an area you work left-to-right, so the lower limit is the smaller x-value and the upper limit is the greater x-value. You need to be careful with parametric equations, especially when there are trig functions involved, because sometimes the smaller x-value corresponds to the bigger parameter value, and vice versa, so you should never assume that the smaller number goes at the bottom of the integral.

    I'd really need to see a screenshot of your book / markscheme to know if they've made an error in their working, or if you've misunderstood what they've asked you to do.
    If this is clear enough?
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    (Original post by Vanetti)
    Name:  DSC_0006.jpg
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    If this is clear enough?
    Yes (apart from having to rotate my head sideways!).

    I agree with the book. A = 27 for the integral given.

    Remember what I said earlier about the order of the limits! If you were integrating y as a function of x, you would be integrating from left to right, starting at x = 0 and ending at x = 3.

    What value of t corresponds to a value of x = 0, y = 0?
    What value of t corresponds to a value of x = 3, y = 0?

    So what are the original lower and upper limits of your integral, and how does your integral compare to the one given in the book?
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    (Original post by davros)
    Yes (apart from having to rotate my head sideways!).

    I agree with the book. A = 27 for the integral given.

    Remember what I said earlier about the order of the limits! If you were integrating y as a function of x, you would be integrating from left to right, starting at x = 0 and ending at x = 3.

    What value of t corresponds to a value of x = 0, y = 0?
    What value of t corresponds to a value of x = 3, y = 0?

    So what are the original lower and upper limits of your integral, and how does your integral compare to the one given in the book?
    OH, I can't believe I did not realize this thanks for pointing it out, this makes a lot of sense I'm thankful for this.
 
 
 
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