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Substitution homework problem

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫ -2 to 2 (x + 3)(4 - x^2)^1/2 dx by writing it as a sum of 2 integrals and interpreting one of those integrals in terms of an area.

    2. Relevant equations

    None.

    3. The attempt at a solution

    ∫ -2 to 2 (x + 3)(4 - x^2)^1/2 dx
    = ∫ 0 to -2 (x + 3)(4 - x^2)^1/2 + ∫ 0 to -2 (x + 3)(4 - x^2)^1/2

    I then draw a graph which resembled 1/4 of an ellipse (having a major axis on the y-axis), for 0 to 2. The y-intercept = 6 and x intercepts: -3, 2, -2.

    However the answer in the back of my book was 6pi, and I dont understand why
     
  2. jcsd
  3. May 20, 2010 #2
    Re: Substitution

    Hm... looks like you might have to multiply in the (4-x^2)^1/2 into (x+3), then it would be a simple u-substitution in the first half, and probably a trig sub in the second... Very doable but a little long...
     
  4. May 20, 2010 #3

    rock.freak667

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    Re: Substitution

    [tex]\int_{-2} ^{2} (x + 3)(4 - x^2)^{\frac{1}{2}} dx[/tex]

    Try expanding out (x+3)(4-x2)1/2, then split the integral.


    (Hint: (a+b)c = ac+bc)
     
  5. May 20, 2010 #4

    Mark44

    Staff: Mentor

    Re: Substitution

    Not long at all if the OP follows the suggestion of interpreting one of the integrals as the area of a geometric object.
     
  6. May 20, 2010 #5
    Re: Substitution

    Oh wow haha, you're right. It just looks like half of a tilted ellipse. It's not so bad then. XD
     
  7. May 21, 2010 #6

    Mark44

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    Re: Substitution

    In the broadest sense, it's an ellipse, but it's really something else.
     
  8. May 21, 2010 #7

    rock.freak667

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    Re: Substitution

    an ellipse with an eccentricity of zero!
     
  9. May 21, 2010 #8
    Re: Substitution

    Yeah I know it looks like a part of an ellipse, but I dont get why the area interpreted is 6pi? Unless it comes from the circle area formula some how (pi *r^2) and is a circle???
     
  10. May 21, 2010 #9

    Mark44

    Staff: Mentor

    Re: Substitution

    Why are you and thepatient insisting the figure is an ellipse?
     
  11. May 21, 2010 #10
    Re: Substitution

    Ohh no, it just looks like half of a slightly tilted ellipse from -2<x<2, but it isn't an ellipse, it looks like a upside down parabola that was stricken by a line coming from the left and merged onto it, though it's domain is only from -2<x<2.

    What is kind of strange is that by using the formula for the area of an ellipse and dividing by two, using a=2 and b = 6, abpi/2 = 6pi. But I don't think this is the proper way to calculate this since the end of the parabola isn't at y=6, but at 6.96 to the nearest hundredth and when x = -3/4+(1/4)*41^(1/2).

    Either way, integrating it directly gets you 6pi. XD I tried it myself.
     
  12. May 21, 2010 #11

    Cyosis

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    Re: Substitution

    Wrong again. Where did you get the values for a and b from? Secondly I suggest you read post #6 and #7 very carefully and try to understand what they are actually trying to tell you.
     
  13. May 21, 2010 #12

    Mark44

    Staff: Mentor

    Re: Substitution

    No, it's definitely not a parabola. Technically, the figure is a kind of ellipse, but that is irrelevant to this problem. I am assuming that we're talking about splitting up the original definite integral into two separate integrals. The first can be done by an ordinary substitution, and the other by using the suggestion given in the problem.

    My guess is that TsAmE has not yet been exposed to the integration technique of trig substitution. However, by recognizing that this integral --
    [tex]\int_{-2}^2 \sqrt{4 - x^2} dx[/tex]

    -- represents the area of a well-known geometric figure, TsAmE should be able to arrive at the value of the definite integral.
     
  14. May 21, 2010 #13
    Re: Substitution


    Sorry, it's just that I went to graphing directly using the function given and got: g-1.jpg That's why I said tilted parabola. But I see that after you split the integral you just get a simple u-sub and the area of a simple shape.


    OP ignore everything I say, hopefully I'm not confusing you. XD I completely understand what mark is saying.
     
  15. May 22, 2010 #14
    Re: Substitution

    Yeah I havent done trig substitution yet. I know how to substitute, but how does that get you 6 pi? Since your lower and upper limits are -2 and 2 respectively? If they contained radians, then that would make sense.
     
  16. May 22, 2010 #15

    Mark44

    Staff: Mentor

    Re: Substitution

    This integral represents the area of a common geometric figure (not an ellipse or parabola). What's the figure, and what's its area?
     
  17. May 22, 2010 #16
    Re: Substitution

    Not sure, but if I say y = (4 - x^2)^1/2, then I get x^2 + y^4 = 4, which is a circle with radius (2)^1/2
     
  18. May 22, 2010 #17

    Mark44

    Staff: Mentor

    Re: Substitution

    Almost. The equation y = (4 - x^2)^(1/2) represents only part of a circle. The equation x^2 + y^2 = 4 represents an entire circle.
     
  19. May 22, 2010 #18
    Re: Substitution

    Makes sense. Since it is 1/4 of a circle, the area equal:

    1/4 * pi r^2
    = 1/4 * pi * 2
    = 1.57 :confused:
     
  20. May 22, 2010 #19

    rock.freak667

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    Re: Substitution

    (4-x2)1/2 between 0 and 2 give 1/4 of the circle, so between -2 and 2, will be twice the area.
     
  21. May 22, 2010 #20
    Re: Substitution

    True that, but then:

    Area = 1/2 * pi r^2
    = 1/2 * pi * 2
    = pi

    but not 6pi?
     
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