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ggrree

  • 2 years ago

Find the are of the region enclosed between the graphs y = 3x and y =3x^2 - 6x I know the answer, and I think I know how to solve it, but I'm having trouble computing the actual answer. Help would be appreciated.

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  1. Chlorophyll
    • 2 years ago
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    = 13.5

  2. nikvist
    • 2 years ago
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    \[y=3x\quad,\quad y=3x^2-6x\]\[3x^2-6x=3x\]\[3x^2-9x=0\quad;\quad x_1=0\quad,\quad x_2=3\]\[S=\int\limits_0^3(9x-3x^2)dx\]

  3. ggrree
    • 2 years ago
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    is that all there is to it, nikvist ? A portion of the region lies below the x axis, so I think that your answer is incorrect.

  4. ggrree
    • 2 years ago
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    that's correct, chlorophyll! Could you please show your work? I am having trouble getting there.

  5. nikvist
    • 2 years ago
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    check my work again. I think my work is correct

  6. Chlorophyll
    • 2 years ago
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    Just exactly as the nikvist's progress!

  7. Chlorophyll
    • 2 years ago
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    Take derivative then plug x = 3 in!

  8. Chlorophyll
    • 2 years ago
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    In mean Integral!

  9. ggrree
    • 2 years ago
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    I am not understanding how he got \[\int\limits_{0}^{3} 9x - 3x^2 dx\]. Why are the signs reversed here? shouldn't it at the very least be \[\int\limits_{0}^{3} 3x^2 - 9x dx\] ? since that is what he wrote the line before.

  10. ggrree
    • 2 years ago
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    I computed and got a negative value, which doesn't make sense :(

  11. Chlorophyll
    • 2 years ago
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    since when you sketch the graph, y = 3x is above y = 3x^2 - 6x

  12. Chlorophyll
    • 2 years ago
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    That's why you subtract: 3x - ( 3x^2 -6x)

  13. ggrree
    • 2 years ago
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    |dw:1332618620433:dw|

  14. Chlorophyll
    • 2 years ago
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    Yep, then we calculate from the intersection part!

  15. Chlorophyll
    • 2 years ago
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    Make sense?

  16. Chlorophyll
    • 2 years ago
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    Next time, pls don't be hurry in make your comment that the solver is incorrect!

  17. ggrree
    • 2 years ago
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    of course I don't mean to be disrespectful at all( I am very grateful that you are helping me :) ) but there are things that I don't understand. When I did this problem I looked at the graph and saw that a part was under the graph, so I did: \[region = \int\limits_{0}^{3} 3x dx - \int\limits_{2}^{3} 3x^2 +3xdx - \int\limits_{0}^{2} 3x^2-3xdx\]

  18. Chlorophyll
    • 2 years ago
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    We just calculate the portion from the intersection down! The closed limited area, otherwise the graphs go unlimited!

  19. Chlorophyll
    • 2 years ago
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    Since the line is above the curve in the interval [ 0, 3]

  20. Chlorophyll
    • 2 years ago
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    That's why you subtract: 3x - ( 3x^2 -6x) = -3x^2 + 9x

  21. Chlorophyll
    • 2 years ago
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    Are you with me?

  22. ggrree
    • 2 years ago
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    "We just calculate the portion from the intersection down! The closed limited area, otherwise the graphs go unlimited!" This I don't understand... I understand that in general, if you want to find the area of a region enclosed by two curves f(x) and g(x) which intersect at points A and B, and f(x) > g(x) you must do \[\int\limits_{A}^{B} f(x) - g(x) dx\] but here a part of the first graph is below the x axis, so wouldn't the integral there be negative?

  23. Chlorophyll
    • 2 years ago
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    In case you have the straight line cut the third degree one, it's possible to have more than 1 limited portion!

  24. Chlorophyll
    • 2 years ago
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    In here, since quadratic just curve 1 time, so we're luckily have only 1 limited portion to calculate :)

  25. Chlorophyll
    • 2 years ago
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    I'm trying to explain as person somewhat gain more practice than the newbie!

  26. Chlorophyll
    • 2 years ago
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    This topic isn't as tough as you thought at all!

  27. ggrree
    • 2 years ago
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    ok, so the formula I used is correct? (for things that only intersect twice to make one area)

  28. Chlorophyll
    • 2 years ago
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    Technically the formula is applied for all cases! It just different at how many closed area to be calculated!

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