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    nothing to see here
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    What's the equation of the curve ?
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    Oh yeah 11 III, I hear you.
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    (Original post by ihatehannah)
    for 11 III

    you do the area under the line take away the area under the curve, i got the area under the curve to be 5.8333 but i keep getting wrong values for the area under the line, i've integrated y= 0.5x + 3 with respect to x which gave me 0.25x^2, plugged in the values and took away, but keep getting 1.25.
    \displaystyle \int \frac{x}{2} + 3 \, \mathrm{d}x = \int \frac{x}{2} + 3x^0 \, \mathrm{d}x = \frac{x^2}{4} + 3x + c
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    (Original post by ihatehannah)
    for 11 III

    you do the area under the line take away the area under the curve, i got the area under the curve to be 5.8333 but i keep getting wrong values for the area under the line, i've integrated y= 0.5x + 3 with respect to x which gave me 0.25x^2, plugged in the values and took away, but keep getting 1.25.

    http://www.ocr.org.uk/Images/130160-...nced-level.pdf
    You haven't integrated the 3 as well as the 0.5x. Also I'm not sure integration is the best way to find the area under a straight line, using area of trapezium has gotta be easy and less chance for errors.
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    (Original post by Zacken)
    \displaystyle \int \frac{x}{2} + 3 \, \mathrm{d}x = \int \frac{x}{2} + 3x^0 \, \mathrm{d}x = \frac{x^2}{4} + 3x + c
    omg i integrated the first term and differniated the second for some reason
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    (Original post by ihatehannah)
    nothing to see here
    Again... I've already talked to you about this. If you had a question, no matter how silly the mistake, if you've had it answered, have the courtesy to thank the person and not delete the thread, somebody else might find it useful. I find what you're doing very rude and you're putting people off wanting to help you.
 
 
 
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