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    (Original post by kiiten)
    Then i subbed in 3 and 1 separately then took them away to get 25? :dontknow:

    So 14 x 3 - 1/3 × 3^2 + 9 × 3^-1 for the first term (3)
    for 3 it should be  \displaystyle 14(3) - \frac{3^3}{3} + 9 \times 3^{-1}
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    (Original post by DylanJ42)
    for 3 it should be  \displaystyle 14(3) - \frac{3^3}{3} + 9 \times 3^{-1}
    Thanks, careless mistake :sigh:

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    (Original post by kiiten)
    Thanks, careless mistake :sigh:

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    dw we all make them
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    (Original post by DylanJ42)
    dw we all make them
    Wait, that was a typo. I still get 36 then 11 when i sub in 1? (Answer: 25) :s

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    (Original post by kiiten)
    Wait, that was a typo. I still get 36 then 11 when i sub in 1? (Answer: 25) :s

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    What seems to be the problem? Sorry I wasn't part of the conversation so I don't know what's going on!

    Also: it would be good if you could post your working out, that way we may be able to pinpoint where you're making mistakes!
    Furthermore, I sometimes find it helpful that when a question is getting out of hand and messy, start it again on a new piece of paper

    This is where I think we've gotten up to so far:

    So you need to find \displaystyle \int^{3}_{1} 14-x^{2}-9x^{-2} \ \mathrm{d}x

    Your integration is correct, we now have \displaystyle \int_1^3 14-x^{2}-9x^{-2} \ \mathrm{d}x = \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right]_1^3

    First you substitute in the limits (3 and 1), then you subtract what you get for the lower limit from what you get with the upper limit, so you have

    =\ \left[14(3)-\dfrac{3^{3}}{3}+\dfrac{9}{3}\ \right] - \left[14-\dfrac{1}{3}+9\right]

    Which is the area enclosed between x= 1 and x= 3 :

    When x=3\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] = 36

    This is correct, however:

    When x=1\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] \not= 11

    As \left[ 14 + 9 - \dfrac{1}{3} \right] = \dfrac{68}{3}
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    (Original post by edothero)
    What seems to be the problem? Sorry I wasn't part of the conversation so I don't know what's going on!

    Also: it would be good if you could post your working out, that way we may be able to pinpoint where you're making mistakes!
    Furthermore, I sometimes find it helpful that when a question is getting out of hand and messy, start it again on a new piece of paper

    This is where I think we've gotten up to so far:

    So you need to find \displaystyle \int^{3}_{1} 14-x^{2}-9x^{-2} \ \mathrm{d}x

    Your integration is correct, we now have \displaystyle \int_1^3 14-x^{2}-9x^{-2} \ \mathrm{d}x = \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right]_1^3

    First you substitute in the limits (3 and 1), then you subtract what you get for the lower limit from what you get with the upper limit, so you have

    =\ \left[14(3)-\dfrac{3^{3}}{3}+\dfrac{9}{3}\ \right] - \left[14-\dfrac{1}{3}+9\right]

    Which is the area enclosed between x= 1 and x= 3 :

    When x=3\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] = 36

    This is correct, however:

    When x=1\ \ \rightarrow\ \ \left[ 14x - \dfrac{1}{3}x^{3} + 9x^{-1} \right] \not= 11

    As \left[ 14 + 9 - \dfrac{1}{3} \right] = \dfrac{68}{3}
    Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
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    (Original post by kiiten)
    Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
    Sorry I'm really busy right now. Look at the graph. Area under the curve from x= 1 to 3 is a rectangle + curve.
    Zacken can help you here. Sorry again.
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    (Original post by kiiten)
    Yeah, i must have typed it wrong into my calculator. So i got 40/3 and the area has to = 5 1/3 so you have to minus (4x2). But where does the 2 come from?
    As edothehero said, the area you've found using integration is the area under the curve. But you want the shaded area so you want to subtract the area of that rectangle.
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    (Original post by Zacken)
    As edothehero said, the area you've found using integration is the area under the curve. But you want the shaded area so you want to subtract the area of that rectangle.
    As i said before you have to minus 8 (4x2) from 40/3 to get 5 1/3. But, where does the 2 come from?

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    (Original post by kiiten)
    As i said before you have to minus 8 (4x2) from 40/3 to get 5 1/3. But, where does the 2 come from?

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    What's the area of the rectangle? Width * height, no...?
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    (Original post by Zacken)
    What's the area of the rectangle? Width * height, no...?
    Height is 4 and i dont know about the width - maybe 3? The question says that the shaded area has to equal 5 1/3 so i did 40/3 - 5 1/3 to get 8. Thats why i thought the width might be 2?
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    (Original post by kiiten)
    Height is 4 and i dont know about the width - maybe 3? The question says that the shaded area has to equal 5 1/3 so i did 40/3 - 5 1/3 to get 8. Thats why i thought the width might be 2?
    You're integrating from 1 to 3, the rectangle has width 3-1 = 2.
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    (Original post by Zacken)
    You're integrating from 1 to 3, the rectangle has width 3-1 = 2.
    Oh, i must have got confused from the diagram because the curve goes inwards at the top so i thought it would be less. But, i see what you mean. Thanks
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    (Original post by kiiten)
    Oh, i must have got confused from the diagram because the curve goes inwards at the top so i thought it would be less. But, i see what you mean. Thanks
    The curve is, the rectangle isn't.
 
 
 
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