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    lets use "/" for "square root", and £ for integration sign lol..

    Question: Estimate £/(2x+1), limits 0 to 4, using the trapezium rule with four strips.

    y = 3, x = 4
    y = /7, x = 3
    y = /5, x = 2
    y = /3, x = 1
    y = 1, x = 0

    using the trapezium rule, 1/2 x 1 x [(1 + 3)(/7 + /5 + /3)]
    = 13.23

    book says that this answer is incorrect, but it has been wrong in the past. can somebody please tell me where i've gone wrong if i have?
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    ayone?
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    The answer should be close to 8.67.

    Here's what you're supposed to do:

    0.5 * 1 * (0 + 2( (3)^0.5 + (5)^0.5 + (7)^0.5 ) + 3)

    Which gives 8.11, which is clearly an under-estimation.
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    And here's the right formula:
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    (Original post by aura1947)
    And here's the right formula:
    my book says that it's simplified to "1/2 x strip width x (first + last + 2 x rest)" ?
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    (Original post by dumb maths student)
    my book says that it's simplified to "1/2 x strip width x (first + last + 2 x rest)" ?
    Same thing.

    But what you are NOT doing is this:

    1/2 x 1 x [(1 + 3)+(/7 + /5 + /3)]

    Which lead me to assume you don't know the right formula.
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    (Original post by aura1947)
    Same thing.

    But what you are NOT doing is this:

    1/2 x 1 x [(1 + 3)+(/7 + /5 + /3)]

    Which lead me to assume you don't know the right formula.
    oh right.. the book says the answer is 1.4.

    but i still get 5.3 :O
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    Oh yes, and this:

    1/2 x 1 x [(0 + 3)+(/7 + /5 + /3)]

    And the answer cannot be 1.4.

    The question is asking for the integral of (2x+1)^0.5 between the limits 0 and 4, right? In that case the answer would be 8.67.
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    (Original post by aura1947)
    Oh yes, and this:

    1/2 x 1 x [(0 + 3)+(/7 + /5 + /3)]

    And the answer cannot be 1.4.

    The question is asking for the integral of (2x+1)^0.5 between the limits 0 and 4, right? In that case the answer would be 8.67.
    wow this book is constantly lying to me.

    are you sure its 0+3 and not 1+3? when x = 0, y = 1, and when x = 4, y = 3..
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    (Original post by dumb maths student)
    wow this book is constantly lying to me.

    are you sure its 0+3 and not 1+3? when x = 0, y = 1, and when x = 4, y = 3..
    Ow. This is embarrassing.

    You were right. But you forgot to multiply it by 2!

    1/2 x 1 x [(1 + 3)+2(/7 + /5 + /3)] = 8.61

    Which is pretty close to 8.67...which is cool.

    Sorry, my bad, I should've paid more attention from the start...would have resulted in the problem being solved more quickly.
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    (Original post by aura1947)
    Ow. This is embarrassing.

    You were right. But you forgot to multiply it by 2!

    1/2 x 1 x [(1 + 3)+2(/7 + /5 + /3)] = 8.61

    Which is pretty close to 8.67...which is cool.

    Sorry, my bad, I should've paid more attention from the start...would have resulted in the problem being solved more quickly.
    haha cant believe i forgot to multiply by 2 aswell.

    thanks a lot! i would rep if i had any rep power

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    (Original post by dumb maths student)
    haha cant believe i forgot to multiply by 2 aswell.

    thanks a lot! i would rep if i had any rep power

    Hehe, you are welcome.

    And no problem, it's the thought that counts. xD
 
 
 
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