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anonymous

  • one year ago

Estimate the area under the curve f(x) = x^2 + 1 from x = 0 to x = 6 by using three circumscribed (over the curve) rectangles. Answer to the nearest integer.

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  1. anonymous
    • one year ago
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    So the interval would be from [0,2] [2,4] [4,6]?

  2. anonymous
    • one year ago
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    yes

  3. anonymous
    • one year ago
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    do you know which endpoints you're using?

  4. anonymous
    • one year ago
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    Would the endpoints not be 1 to 6?

  5. anonymous
    • one year ago
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    you only have three intervals. I meant endpoints in terms of right or left of the interval

  6. anonymous
    • one year ago
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    can you draw the rectangles? |dw:1441160832388:dw|

  7. anonymous
    • one year ago
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    |dw:1441160946061:dw| Something like this?

  8. anonymous
    • one year ago
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    yes. so the width of all the rectangles will be 2. The heights will the y-values at 2, 4, and 6. Make sense?

  9. anonymous
    • one year ago
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    Ok, so how would this help me setup my integral?

  10. anonymous
    • one year ago
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    This is the integral\[\int\limits_{0}^{6}(x^2+1) dx\] But that's not what you're being asked to find/evaluate. You're being asked for an estimate of the area, and you use the rectangle you drew to find it. For example, the area of the first rectangle is 2f(2) because 2 is its width and f(2) is its height. So you really just need to find the areas of the 3 rectangles and add them up

  11. anonymous
    • one year ago
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    So (4+8+12)=24?

  12. anonymous
    • one year ago
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    \[A=2f(2)+2f(4)+2f(6)\] \[A=2[f(2)+f(4)+f(6)]\] You have to plug in 2, 4, and 8 into \(f(x)=x^2+1\) to get f(2), f(4), and f(6) |dw:1441161661208:dw|

  13. anonymous
    • one year ago
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    Why 8?

  14. anonymous
    • one year ago
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    sorry, typo. it's supposed to be 6

  15. anonymous
    • one year ago
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    118, final answer?

  16. anonymous
    • one year ago
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    yes

  17. anonymous
    • one year ago
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    Appreciate it.

  18. anonymous
    • one year ago
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    no problem

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