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Math Help - integration (area under a curve)

  1. #1
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    integration (area under a curve)

    Hello guys! Im having trouble with this question i would appreciate the help.

    curve y=( 2x-3)^3

    a) find the x coordinates of the two points on the curve which have gradient 6

    b) the region shaded in the diagram is bounded by part of the curve and by the two axes. Find, by integration, the area of this region.

    Thank you in advance!

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  2. #2
    MHF Contributor Calculus26's Avatar
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    a. f ' = 6(x-3)^2

    now simply se this equal to 6

    b. the integration limits are 0 and the x-intercept (set f = 0 simple enough)

    the area is a naegative of the integral of f(x) on this interval
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  3. #3
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    Quote Originally Posted by Oasis1993 View Post
    Hello guys! Im having trouble with this question i would appreciate the help.

    curve y=( 2x-3)^3

    a) find the x coordinates of the two points on the curve which have gradient 6

    b) the region shaded in the diagram is bounded by part of the curve and by the two axes. Find, by integration, the area of this region.

    Thank you in advance!

    I see Calculus26 already answered you, but I wrote this all out so I'll just post it anyway.

    a) Take the derivative a set it equal to 6. \frac{d}{dx} (2x-3)^3 = 6(2x-3)^2

    solve for x when 6(2x-3)^2 = 6

    b) You are integrating \int_{0}^{x} (2x-3)^3dx

    To find the upper limit of integration you need to set your function equal to zero, because that is the x value of your x intercept. Your function equals 0 when (2x-3) = 0, or when x = 3/2

    So your integral becomes \int_{0}^{\frac{3}{2}} (2x-3)^3dx

    You'll get a negative number for the area since the curve is below the x-axis on the interval which you're integrating, so your area will be the absolute value of the answer you get by evaluating that integral.
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  4. #4
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    Thank you for your reply!
    But could you please show the steps because i cant get the correct answer...
    when i take the differentiaton of y=(2x-3)^3 i get 2(2x-3)^2 do i then have to equal it to 6?
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  5. #5
    Ted
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    Quote Originally Posted by Oasis1993 View Post
    Thank you for your reply!
    But could you please show the steps because i cant get the correct answer...
    when i take the differentiaton of y=(2x-3)^3 i get 2(2x-3)^2 do i then have to equal it to 6?
    The derivative is 6(2x-3)^2.
    Then equal it to 6.
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  6. #6
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    sorry, my mistake.
    Thank you.
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  7. #7
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    Ted,
    can i ask one more question?
    when i do 6(2x-3)^2 = 6
    i get 2. which is correct but it asks for TWO x-coordinates.
    how can i get the second one?
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  8. #8
    Ted
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    Quote Originally Posted by Oasis1993 View Post
    Ted,
    can i ask one more question?
    when i do 6(2x-3)^2 = 6
    i get 2. which is correct but it asks for TWO x-coordinates.
    how can i get the second one?
    6(2x-3)^2=6.
    devide both sides by 6:
    (2x-3)^2=1.
    expand:
    4x^2-12x+9=1
    4x^2-12x+8=0

    solve the last quadratic equation to find your 2 and the another one.
    Last edited by Ted; February 20th 2010 at 03:48 AM. Reason: another*
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  9. #9
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    Just to point out that from Ted's quadratic you may find it easier to solve if you divide both sides by 4
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