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Integration Questions

  1. Jan 25, 2005 #1
    Hello all

    Here are a few problems I encountered:

    Find the area bounded by the parabola [tex] y = \frac{1}{2}x^2 + 1 [/tex] and the straight line [tex] y = 3 + x [/tex] Ok, so I first set the two equations equal to each other:

    [tex] 3 + x = \frac{1}{2}x^2 + 1 [/tex] After solving for x, do I just use those two values for the upper and lower limits, and evaluate:

    [tex] \int^b_a \frac{1}{2}x^2 + x - 2 [/tex]?

    Also how would you evalutate the following integrals:

    [tex] \int^b_a (x+1)^a dx [/tex]
    [tex] \int^b_a sin \alpha x {} dx [/tex]
    [tex] \int^b_a cos \alpha x {} dx [/tex]

    For the first one would I use a geometric progression? I know that for [tex] \int^b_a x^a dx [/tex] you divide up the interval using the following points:
    [tex] a, aq, aq^2, . . . , aq^n^-1, aq^n = b [/tex] So that means for [tex] a+1 [/tex] we have [tex] a+1, (a+1)q, (a+1)q^2, . . . (a+1)q^n-1, (a+1)q^n [/tex]?

    For the other integrals, do I just make use of the identity [tex] 2\sin u sin v = \cos(u-v) - \cos(u+v) [/tex]?

    Thanks :smile:
     
    Last edited: Jan 25, 2005
  2. jcsd
  3. Jan 25, 2005 #2

    dextercioby

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    1.Why did you set equal those 2 equations involving 2 parabolas??Shouln't have it been a parabola & a line??
    2.The first integral is realized through an elementary substitution...Do you see it?
    3.The second one is accomplished through a substitution as well.
    4.The third one is similar to the second one... Meaning you need to do the same substitution...
    Daniel.
     
  4. Jan 25, 2005 #3
    yes but I am not supposed to use substitutions. I have to actually use the limiting process! This is the beginning of the integrals chapter.
     
  5. Jan 25, 2005 #4

    dextercioby

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    Okay,u should have said that from the beginning...What about the first exercise??Why 2 parabolas??

    Daniel.

    P.S.U edited your message... :tongue2: Solve for "x" and then determine what curve is on "top" of the other,compute the 2 areas and then subtract them (if they're both positive/negative) or add them if one of them is negative.
     
  6. Jan 25, 2005 #5
    Thanks. Any help or hints for the last few problems are appreciated.

    Also if we have to evaluate [tex] \int^1_0 (1-x)^n dx [/tex] where n is an integer do I just expand the bracket? Would it be [tex] 1 - nx + x^n +... [/tex]?

    Thanks
     
  7. Jan 25, 2005 #6

    dextercioby

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    Without substitution??Okay,the u'd have to use the binomial formula:
    [tex] (1-x)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{n-k} x^{n-k} [/tex]

    Daniel.
     
  8. Jan 25, 2005 #7
    ok so we know [tex] k = 0, n = 1 [/tex]. So it would be [tex] C^0_1 (-1)^1 x^1 = -x [/tex] So is is [tex] \int^1_0 -x dx = \frac {-1}{2}(b^2 - a^2) = \frac {-1}{2}[/tex]?

    Thanks
     
    Last edited: Jan 25, 2005
  9. Jan 25, 2005 #8
    I think it's [tex] \frac {-1}{2} [/tex] because I just plugged in the values.

    Is this right?
     
  10. Jan 25, 2005 #9

    dextercioby

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    Are u asking whether
    [tex] \int_{0}^{1} -x \ dx [/tex]

    is equal to [tex] -\frac{1}{2} [/tex]
    ??If so,then the answer is YES.

    Daniel.
     
  11. Jan 25, 2005 #10
    thank you. is that the answer to [tex] \int^1_0 (1-x)^n [/tex] after simplifying?
     
  12. Jan 26, 2005 #11

    dextercioby

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    No,the answer is:

    [tex]\int_{0}^{1} (1-x)^{n} dx=\frac{1}{n+1} [/tex]

    Daniel.
     
  13. Jan 26, 2005 #12
    how did you get this? Did you substitute the values for k and n?

    Thanks
     
  14. Jan 26, 2005 #13

    dextercioby

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    Nope,i made a substitution.

    Daniel.
     
  15. Jan 26, 2005 #14
    I used the binomial formula and got [tex] _-x [/tex]
     
  16. Jan 26, 2005 #15

    dextercioby

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    What?Please explain your result...

    Daniel.
     
  17. Jan 26, 2005 #16
    Because as you said using the binomial forumula [tex] (1-x)^{n}=\sum_{k=0}^{n} C_{n}^{k} (-1)^{n-k} x^{n-k} [/tex] shouldn't I just substitute [tex] k = 0, n = 1 [/tex] and this equals [tex] 1(-1^{1-0})(x^{1-0}) = - x [/tex]

    Thanks for your help
     
  18. Jan 26, 2005 #17

    dextercioby

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    No,"k" is a summation index.And "n" is natural,arbitrary and finite...

    Daniel.
     
  19. Jan 26, 2005 #18
    Then how would you use the binomial formula?

    Thanks
     
  20. Jan 26, 2005 #19

    dextercioby

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    By writing explitely each term from the sum and then applying the fact that
    [tex] \int_{a}^{b} [f(x)+g(x)] \ dx=\int_{a}^{b} f(x) \ dx +\int_{a}^{b} g(x) \ dx [/tex]

    The property of addition for the definite integrals...

    Daniel.
     
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