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Integration Question

  1. Jan 24, 2005 #1
    Is [tex] \int^2_1 2x + x + 1 dx= (1)(3) + \frac{1}{2}(1)(3) + 1 [/tex]?
     
  2. jcsd
  3. Jan 24, 2005 #2

    dextercioby

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    Why doncha tell us how u came up with those numbers...??

    Daniel.
     
  4. Jan 24, 2005 #3
    [tex]\int^2_1 2x + x + 1 dx = \int^2_1 3x + 1 dx = \left[\frac{3}{2}x^2 + x \right]^2_1[/tex]
    When you solve that i think u get 5.5

    Though i suspect that original equation would be
    [tex]\int^2_1 2x^2 + x + 1 dx[/tex] or why would they give it to you in that form ... 2x + x ?
     
    Last edited: Jan 24, 2005
  5. Jan 24, 2005 #4
    They were trying to illustrate the basic integrals

    [tex] \int^b_a x dx = \frac{1}{2}(b-a)(b+a) [/tex] and [tex] \int^b_a x^2 dx = \frac {1}{3}(b^3 - a^3) [/tex]

    [tex] \int^b_a c dx = c(b-a) [/tex]

    This is how I got those numbers
     
  6. Jan 24, 2005 #5
    You answer is correct.
     
  7. Jan 24, 2005 #6
    your answer is absolutely correct. (if you sure the first term is 2x instead of 2x^2 as ryoukomaru stated earlier)
     
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