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anonymous

  • 4 years ago

Find the area of the region bounded by the curves f(x)=(3x+1)^2 g(x)=2^0.4x i checked my work a lot of times but I couldn't get 8.3522

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  1. jessicap
    • 4 years ago
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    Ok, I'm going to try this one.

  2. anonymous
    • 4 years ago
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    oh thanks. May be I better show you what i have done so you can find my mistakes

  3. jessicap
    • 4 years ago
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    Sounds good. I'm looking at the graphs right now.

  4. anonymous
    • 4 years ago
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    |dw:1327278455526:dw|

  5. anonymous
    • 4 years ago
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    is that what you get?

  6. anonymous
    • 4 years ago
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    \[\int\limits_{-1.259}^{0.62} 2^{0.4x}-[(3x+1)^{2}-7]dx \]

  7. anonymous
    • 4 years ago
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    \[\left[ (2^{0.4x}/0.4\ln2 - 3x ^{3}-3x ^{2}+6x \right] \]\[\int\limits_{-1.259}^{0.62} 2^{0.4x}-9x ^{2}-6x+6 dx\]\[\int\limits_{-1.259}^{0.62}2^{0.4x}-[9x ^{2}+6x+1-7]dx\]

  8. anonymous
    • 4 years ago
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    \[\left[ 2^{0.4x}/0.4\ln2 - 3x ^{3}-3x ^{2}+6x \right]\]

  9. jessicap
    • 4 years ago
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    Sorry, it's taking me some time to check the algebra/arithmetic. :) But so far your approach looks good.

  10. anonymous
    • 4 years ago
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    it's okay :) I can't find what's wrong here :(

  11. anonymous
    • 4 years ago
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    oh btw did I get the interval right? [-1.259, 0.62]

  12. jessicap
    • 4 years ago
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    Did you get those values by setting the equations equal to themselves and solving for x?

  13. anonymous
    • 4 years ago
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    No. I got it from calculator. :)

  14. jessicap
    • 4 years ago
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    Well, I have to admit that I was having difficulty solving the equation, but what I did was put it into wolframalpha.com\[9x ^{2}+6x+1=2^{0.4x}\] It gave me x=0 and 1.246 E -8. This is the link: http://www.wolframalpha.com/input/?i=9x%5E2%2B6x%2B1%3D2%5E%28.4x%29

  15. anonymous
    • 4 years ago
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    well i just look at the intersections of f(x) and g(x) on calculator. I'm confused now.

  16. jessicap
    • 4 years ago
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    Ok, never mind what I said if that confuses you. When I look at my calculator I get [-0.64, 0].

  17. anonymous
    • 4 years ago
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    oh my bad. I typed a wrong equation f(x)=(3x+1)^2 -7 sorry sorry sorry

  18. jessicap
    • 4 years ago
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    No problem! Glad you found your mistake. Let's try it with these limits.

  19. anonymous
    • 4 years ago
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    :) I'm so sorry.

  20. jessicap
    • 4 years ago
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    Don't be! I just started this too and have made errors that are much worse :) It has helped me to work through this with you.

  21. anonymous
    • 4 years ago
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    hehe this problem takes me like more than one hour to find the answer that my teacher stated.

  22. anonymous
    • 4 years ago
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    my answer is 9.913 do you get the same one?

  23. jessicap
    • 4 years ago
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    Almost have it . . .

  24. jessicap
    • 4 years ago
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    This is my equation:\[\int\limits_{-0.64bj }^{0}2^{0.4x}-(9x+6x+1) dx\] My answer is way off, I got 3.018

  25. anonymous
    • 4 years ago
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    you didn't change the equation, did you? cuz it's supposed to be 9x+6x+1-7

  26. jessicap
    • 4 years ago
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    My bad it is supposed to be\[9x ^{2}\], but where did the -7 come from? I didn't see it in the original equation.

  27. jessicap
    • 4 years ago
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    I mean question.

  28. anonymous
    • 4 years ago
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    i just said that I typed a wrong equation f(x)=(3x+1)^2-7 :)

  29. jessicap
    • 4 years ago
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    Ok now I get it so it is\[\int\limits\limits_{-0.64}^{0}2^{0.4x}-( 9x-3x+1-7) dx \]which equals\[\int\limits_{-0.64}^{0}2^{0.4x}-9x+3x+6dx\]

  30. jessicap
    • 4 years ago
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    Is this right?

  31. anonymous
    • 4 years ago
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    yup but the interval is different cuz you changed the equation. You should check the intersections with your calculator too.

  32. jessicap
    • 4 years ago
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    Oh yes

  33. anonymous
    • 4 years ago
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    hey wait it is \[2^{0.4x}-9x ^{2}-6x+6\]

  34. jessicap
    • 4 years ago
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    You are right, I meant 6x. So are you saying that the minus sign is already distributed?

  35. anonymous
    • 4 years ago
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    Yes

  36. jessicap
    • 4 years ago
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    So I should take it out when I plug it in to my calculator.

  37. anonymous
    • 4 years ago
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    Yup. Did you find the interval?

  38. jessicap
    • 4 years ago
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    Here we go: [-1.277,0.63] . However, the first number is not exact.

  39. anonymous
    • 4 years ago
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    I got [-1.259, 0.62]

  40. jessicap
    • 4 years ago
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    hmmm this is why I think an equation is better if only I could solve it correctly.

  41. jessicap
    • 4 years ago
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    But I think we are in the ballpark. This was tricky with the x in the exponent.

  42. anonymous
    • 4 years ago
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    you're trying to find the interval by solving the equations? I can't find it by hand, you know.

  43. jessicap
    • 4 years ago
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    Ok, well I was estimating at first; I get the same answer that you did using the calc function.

  44. anonymous
    • 4 years ago
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    yes then we just need to integrate that equation. I started over and get 9.913

  45. jessicap
    • 4 years ago
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    The integration is fine. I used all of the digits for the intersection and got 6.702.

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