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    From WJEC June 2007;

    1. Use the trapezium rule with five ordinates to find an approximate value for the integral

    \displaystyle\int^\frac{\pi}{2}_  0 \sqrt {1+sinx}\ dx

    I am not having a problem with the trapezium rule but my y figures are not the same as the mark schemes.

    My Values
    Spoiler:
    Show
    x = 0

y = \sqrt {1+sin0} = 1

    x = \frac{1}{8}\pi
    y = \sqrt {1+sin(\frac{1}{8}\pi)} = 1.003451067

    x = \frac{1}{6}\pi
    y = \sqrt {1+sin(\frac{1}{6}\pi)} = 1.004558806

    x = \frac{1}{4}\pi
    y = \sqrt {1+sin(\frac{1}{4}\pi)} = 1.00683035

    x = \frac{1}{2}\pi
    y = \sqrt {1+sin(\frac{1}{2}\pi)} = 1.013613404


    Markscheme Values
    Spoiler:
    Show
    x = 0

y = 1

    x = \frac{1}{8}\pi
    y = 1.175875602

    x = \frac{1}{6}\pi
    y = 1.306562965

    x = \frac{1}{4}\pi
    y = 1.387039845

    x = \frac{1}{2}\pi
    y = 1.414213562


    Maybe i am inputting them into the calculator wrong, but i have tried to work out whats going on but i cant seem to figure it out.

    Any help would be appreciated.
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    (Original post by _Morsey_)
    From WJEC June 2007;

    1. Use the trapezium rule with five ordinates to find an approximate value for the integral

    \displaystyle\int^\frac{\pi}{2}_  0 \sqrt {1+sinx}\ dx

    I am not having a problem with the trapezium rule but my y figures are not the same as the mark schemes.

    My Values
    Spoiler:
    Show
    x = 0

y = \sqrt {1+sin0} = 1

    x = \frac{1}{8}\pi
    y = \sqrt {1+sin(\frac{1}{8}\pi)} = 1.003451067

    x = \frac{1}{6}\pi
    y = \sqrt {1+sin(\frac{1}{6}\pi)} = 1.004558806

    x = \frac{1}{4}\pi
    y = \sqrt {1+sin(\frac{1}{4}\pi)} = 1.00683035

    x = \frac{1}{2}\pi
    y = \sqrt {1+sin(\frac{1}{2}\pi)} = 1.013613404


    Markscheme Values
    Spoiler:
    Show
    x = 0

y = 1

    x = \frac{1}{8}\pi
    y = 1.175875602

    x = \frac{1}{6}\pi
    y = 1.306562965

    x = \frac{1}{4}\pi
    y = 1.387039845

    x = \frac{1}{2}\pi
    y = 1.414213562


    Maybe i am inputting them into the calculator wrong, but i have tried to work out whats going on but i cant seem to figure it out.

    Any help would be appreciated.
    You need to be in radians. Not degrees.
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    (Original post by _Morsey_)
    From WJEC June 2007;

    1. Use the trapezium rule with five ordinates to find an approximate value for the integral

    \displaystyle\int^\frac{\pi}{2}_  0 \sqrt {1+sinx}\ dx

    I am not having a problem with the trapezium rule but my y figures are not the same as the mark schemes.

    My Values
    Spoiler:
    Show
    x = 0

y = \sqrt {1+sin0} = 1

    x = \frac{1}{8}\pi
    y = \sqrt {1+sin(\frac{1}{8}\pi)} = 1.003451067

    x = \frac{1}{6}\pi
    y = \sqrt {1+sin(\frac{1}{6}\pi)} = 1.004558806

    x = \frac{1}{4}\pi
    y = \sqrt {1+sin(\frac{1}{4}\pi)} = 1.00683035

    x = \frac{1}{2}\pi
    y = \sqrt {1+sin(\frac{1}{2}\pi)} = 1.013613404


    Markscheme Values
    Spoiler:
    Show
    x = 0

y = 1

    x = \frac{1}{8}\pi
    y = 1.175875602

    x = \frac{1}{6}\pi
    y = 1.306562965

    x = \frac{1}{4}\pi
    y = 1.387039845

    x = \frac{1}{2}\pi
    y = 1.414213562


    Maybe i am inputting them into the calculator wrong, but i have tried to work out whats going on but i cant seem to figure it out.

    Any help would be appreciated.
    Radians, as has been said

    Also

    You are using the incorrect x values

    Are those x values shown on the MS?
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    (Original post by TenOfThem)
    Radians, as has been said

    Also

    You are using the incorrect x values

    Are those x values shown on the MS?
    Yes, they are the correct x values.

    And thank you for the radians comment, now it makes sense and adds correctly

    Markscheme - HERE
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    (Original post by _Morsey_)
    Yes, they are the correct x values.

    And thank you for the radians comment, now it makes sense and adds correctly

    Markscheme - HERE
    Begging your pardon, but TenOfThem is correct and THIS IS THE MARKSCHEME
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    (Original post by _Morsey_)
    Yes, they are the correct x values.

    And thank you for the radians comment, now it makes sense and adds correctly

    Markscheme - HERE
    No, they are not

    That does not seem to be the correct MS ... what Q were you doing
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    (Original post by joostan)
    Begging your pardon, but TenOfThem is correct and THIS IS THE MARKSCHEME
    Thanks for the MS

    At least they are using the correct values
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    (Original post by TenOfThem)
    Thanks for the MS

    At least they are using the correct values
    Lol, the OP posted the paper that the q was from
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    (Original post by joostan)
    Lol, the OP posted the paper that the q was from
    oh aye
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    Oops, i was doing the Jan 08 paper at the time, but this question was at the back of my mind.

    Nevertheless, thanks
 
 
 
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