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Finding the area beneath a curve help

  1. Jun 23, 2014 #1
    1. Hi, I have been asked to calculate the area beneath a curve. The curve is
    y = 2x^2 + 7x + 24 and the values of x to find the area within are 5 and 2.



    3. My attempt stands as follows: A = 5,2∫ (2x^2+7x+24)dx
    [ 2x^3/2 + 7x^2/2 + 24x ]
    From there I am unsure on what to do as i am unsure on calculus as a whole.



    Thanks
     
  2. jcsd
  3. Jun 23, 2014 #2

    HallsofIvy

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    I presume that the "5,2" in front of the integral indicates that you are integrating "from x= 2 to x= 5". Do you know what that means? Yes, [itex]2x^3/3+ 7x^2/2+ 24x+ C[/itex] (C can be any constant) is the general anti-derivative. Now, evaluate that between x= 2 and x= 5. That is find the value of [itex]2x^3/3+ 7x^2/2+ 24x+ C[/itex] at x= 5, at x= 2 and subtract the first from the second.

    If F'(x)= f then [itex]\int f(x)dx= F(x)+ C[/itex] and [itex]\int_a^b f(x)dx= F(b)- F(a)[/itex]
     
  4. Jun 23, 2014 #3

    Student100

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    Look at the way you calculated the integral.

    Also when calculating a definite integral what makes it different than an indefinite one? Why can you disreguard the arbitary constant?

    He actually got the indefinite wrong, but yes this is what you should have gotten.
     
    Last edited: Jun 23, 2014
  5. Jun 23, 2014 #4
    Ok I got (250/2 + 175/2 +24 ) - (16/2 + 28/2 + 24) = 236.5 - 46 = 190.5 units of area?


    Does that sound correct?

    Thanks,

    Ps, would y = -x^2 + 1 use the same method?
     
  6. Jun 23, 2014 #5

    Student100

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    No, look at your orginal calculation, then look at HallsofIvy post. Do you see a difference?
     
  7. Jun 23, 2014 #6

    He has misplaced a 3 for a 2, but I didn't use that in my calculation so 190.5 should be correct?
     
  8. Jun 23, 2014 #7

    Student100

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    His was right, yours was wrong. You raised it to the 3rd power.
     
  9. Jun 23, 2014 #8
    Ah I see what you mean.
     
  10. Jun 23, 2014 #9
    So I got 151.497 units of area this time using the correct indefinite
     
  11. Jun 23, 2014 #10

    Student100

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    Can you post your work? That's incorrect.

    Edit: Looking at your previous post you probably forgot the x after the 24 again.
     
  12. Jun 23, 2014 #11

    2x^3/3 + 7x^2/2 + 24x

    = [ 250/3 + 175 /2 +24] - [ 16/3 + 28/2 + 24 ]

    = 194.83 - 43.33 = 151.497?
     
  13. Jun 23, 2014 #12

    Student100

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    24x

    Look at your steps very close.
     
  14. Jun 23, 2014 #13
    I'm unsure on calculus, I understand as much that you multiply by the limits of x and then subtract it but with the x I do not know
     
  15. Jun 23, 2014 #14

    Student100

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    We're past the calculus, you just got to look at your arithmetic, you are adding 24 instead of 24x or 24(5) and 24(2).
     
  16. Jun 23, 2014 #15

    Sorry I see now.

    As a final answer I got 223.499 now, I am really hoping that is correct
     
  17. Jun 23, 2014 #16

    Student100

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    That's right, you might want to practice keeping the answers in fraction forum, however. Just keep practicing.
     
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