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anonymous

  • one year ago

Use left and right endpoints and the given number of rectangles to find two approximations of the area of the region between the graph of the function and the x-axis over the given interval. g(x) = 2x2 − x − 1, [3, 5], 4 rectangles

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  1. jim_thompson5910
    • one year ago
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    how far did you get?

  2. anonymous
    • one year ago
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    I got it wrong :/

  3. jim_thompson5910
    • one year ago
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    can you show me what steps you have so I can see where you went wrong

  4. anonymous
    • one year ago
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  5. jim_thompson5910
    • one year ago
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    how did you get the `5+6+7+8` ?

  6. anonymous
    • one year ago
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    that's what it told me to do in the problem before

  7. anonymous
    • one year ago
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    it's a sum

  8. jim_thompson5910
    • one year ago
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    maybe you were thinking of f(3) + f(3.5) + ... ??

  9. anonymous
    • one year ago
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    should it be?

  10. jim_thompson5910
    • one year ago
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    let's draw out the graph

  11. jim_thompson5910
    • one year ago
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    |dw:1442286116197:dw|

  12. jim_thompson5910
    • one year ago
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    here is f(x) |dw:1442286125598:dw|

  13. jim_thompson5910
    • one year ago
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    |dw:1442286156489:dw|

  14. jim_thompson5910
    • one year ago
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    divide the interval from 3 to 5 into four pieces |dw:1442286172862:dw|

  15. jim_thompson5910
    • one year ago
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    the left rectangle approximation will look like this |dw:1442286205016:dw|

  16. jim_thompson5910
    • one year ago
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    agreed so far?

  17. anonymous
    • one year ago
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    yes!

  18. jim_thompson5910
    • one year ago
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    ok so you have the delta x part correct that is delta x = (b-a)/n delta x = (5-3)/4 delta x = 2/4 delta x = 1/2 delta x = 0.5 this is the width of each rectangle |dw:1442286351913:dw|

  19. anonymous
    • one year ago
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    yes!

  20. jim_thompson5910
    • one year ago
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    you can factor out the delta x terms which is what you have on your paper the heights of each rectangle is based on f(x) the first rectangle has a height of f(3) |dw:1442286417246:dw|