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    I have to find the area under the curve with the equation y=2√2x, but I can't figure out how to integrate the surd. Could anyone help me integrate the equation?
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    Rewrite the surd as a power of 2x.

    Then it's simple.
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    (Original post by fablereader)
    I have to find the area under the curve with the equation y=2√2x, but I can't figure out how to integrate the surd. Could anyone help me integrate the equation?
    √2x <-- is also 2x^1/2
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    (Original post by Wunderbarr)
    Rewrite the surd as a power of 2x.

    Then it's simple.

    That gets me 2(2x)^1/2, and I just don't know how to get from there. As a guess, I tried integrating that as 2x(2x)^3/2, which equals 4x^2(2x^1/2), but the resulting area underneath the curve was nowhere near that of the area I got when I used the trapezium rule.

    For context: The equations is y=2(√2x), and the first part of the question is to use the trapezium rule with eight strips to estimate the value of the area from x=5 to x=1. The area result is approximately 19.18. However, using the result of the integration above I got the area as over 300.

    I hope I have written this clearly and correctly.
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    (Original post by fablereader)
    x
    Is this a C2 question?
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    (Original post by Wunderbarr)
    Is this a C2 question?
    Yes.
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    (Original post by fablereader)
    Yes.
    So if you have something like (5x)½ then to integrate that, you would need to integrate it as you were if the 5x were just x, but then multiply by the reciprocal of what you get when you differentiate 5x with respect to x.

    So for this example, you would get 1/5 (5x)^(3/2) which is the same as [one over five times open bracket 5x close bracket to the power of three over two] (I am not mastered in the ways of LaTeX >_<).
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    it is not clear if it is (2√2)x or 2√(2x) which is being integrated
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    (Original post by the bear)
    it is not clear if it is (2√2)x or 2√(2x) which is being integrated
    I imagine it's what I have LaTeXd below.

    (Original post by fablereader)
    I have to find the area under the curve with the equation y=2√2x, but I can't figure out how to integrate the surd. Could anyone help me integrate the equation?
    OP, is your equation y=2\sqrt{2x} {?}. Your question is a little ambiguous.
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    (Original post by Kvothe the arcane)
    I imagine it's what I have LaTeXd below.



    OP, is your equation y=2\sqrt{2x} \ {?}. Your question is a little ambiguous.
    The equation is y=2√2x.
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    (Original post by fablereader)
    The equation is y=2√2x.
    Is that (a)2 \sqrt{2x} or (b)2\sqrt{2}x?

    I.e. is the x under the square root as well?
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    (Original post by Kvothe the arcane)
    Is that (a)2 \sqrt{2x} or (b)2\sqrt{2}x?

    I.e. is the x under the square root as well?
    The x is under the square root, so option a.

    Sorry, I initially said option b. My mind is slightly busy with lots of things/freaking out whenever I think on this problem
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    (Original post by fablereader)
    The x is under the square root, so option a.

    You have 2 \sqrt{2x}

    There is the property that \sqrt{ab}=\sqrt{a} \times \sqrt{b} so the above is equivalent to 2\sqrt{2} \times \sqrt{x}.

    You wish to evaluate \displaystyle \int_1^5 2 \sqrt{2x} \ dx. It would be easier to write this as \displaystyle 2\sqrt{2}\int_1^5 x^{\frac{1}{2}} \ dx

    That way you can multiply your definite integral by the surd at the end. I hope this helps.

    To integrate, it's the simple rule that \displaystyle \int x^a  \ dx = \dfrac{x^{a+1}}{a+1} + \mathrm{C}
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    (Original post by Kvothe the arcane)
    You have 2 \sqrt{2x}

    There is the property that \sqrt{ab}=\sqrt{a} \times \sqrt{b} so the above is equivalent to 2\sqrt{2} \times \sqrt{x}.

    You wish to evaluate \displaystyle \int_1^5 2 \sqrt{2x} \ dx. It would be easier to write this as \displaystyle 2\sqrt{2}\int_1^5 x^{\frac{1}{2}} \ dx

    That way you can multiply your definite integral by the surd at the end. I hope this helps.

    To integrate, it's the simple rule that \displaystyle \int x^a  \ dx = \dfrac{x^{a+1}}{a+1} + \mathrm{C}
    Yes! It has worked very well indeed. Finding the area under the curve via integration has gotten me 19.19, very close to the trapezium rule estimate of 19.18. Thank you very much.
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    You can write it either way and it makes no difference because the 2√2 part is constant. The value of a coefficient doesn't affect the integration with respect to x, you just multiply by it after.

    (2x)^1/2 = 2^1/2 (x^1/2) = √2 (x^1/2)

    So the integral is 1/3(4√2) x^3/2 + c , where c is constant
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    (Original post by Wunderbarr)
    So if you have something like (5x)½ then to integrate that, you would need to integrate it as you were if the 5x were just x, but then multiply by the reciprocal of what you get when you differentiate 5x with respect to x.

    So for this example, you would get 1/5 (5x)^(3/2) which is the same as [one over five times open bracket 5x close bracket to the power of three over two] (I am not mastered in the ways of LaTeX >_<.
    Inverse chain rule isn't covered in C2.
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    (Original post by fablereader)
    Yes! It has worked very well indeed. Finding the area under the curve via integration has gotten me 19.19, very close to the trapezium rule estimate of 19.18. Thank you very much.
    I'm glad I was able to help .
 
 
 
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