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Math Help - Integration

  1. #1
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    Integration

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  2. #2
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    Quote Originally Posted by r_maths View Post
    Hello,

    all your considerations are OK.

    To integrate the square root you have to use integration by substitution.

    I've attached an image with my calculations:
    Attached Thumbnails Attached Thumbnails Integration-tragflaeche.gif  
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  3. #3
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    Hello, r_maths!

    Did you find the limits of integration?


    I sketched the curves.
    . . . . . . . . . . . . . . . . . . . ______
    The upper curve is: .y .= .√2x + 1
    . . This is the upper half of a parabola with vertex (-,0), opening to the right.

    The lower curve is: .y .= .(2x + 1)(x - 2)/6
    . . This is an up-opening parabola with x-intercepts: - and 2

    . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _____
    To find their intersections: .(2x + 1)(x - 2)/6 .= .√2x + 1

    Square both sides: .(2x + 1)(x - 2)/36 .= .2x + 1

    Divide by (2x + 1): . (2x + 1)(x - 2)/36 .= .1 .**

    . . which simplifies to the cubic: .2x - 7x + 4x - 32 .= .0

    . . which factors: .(x - 4)(2x + x + 8) .= .0

    . . and has one real root: .x .= .4


    Therefore, the limits are: .x = - and x = 4

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **
    From the graph, we know they intersect at x = -
    Then assuming (2x + 1) ≠ 0, we can divide by it.

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  4. #4
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    The limits were kindly given from the question.
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