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Definite Integral Misunderstood

  1. Sep 18, 2007 #1
    Okay, so I went a little ahead in my text and I am having alittle trouble deciphering something. To make it easy, I will provide an example/question.

    If I see the following problem: [tex]\int_2^42xdx[/tex], does it mean

    a.) Evaluate the line y=2x between 2 and 4 to get area=4.

    or

    b.) Antidifferentiate and evaluate the curve [tex]x^2[/tex] from 2 to 4 to get area=12.

    C.) Neither.

    Thank you,
    Casey
     
  2. jcsd
  3. Sep 18, 2007 #2

    Dick

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    If by a) you mean the area under the line y=2x between 2 and 4 then it means both. Except you got the answer to a) wrong. It's a trapezoid. The answer is also 12.
     
  4. Sep 18, 2007 #3
    Okay....let me try this again in different words...I was trying to avoid an extremely wordy explanation, but here goes.

    if you see this [tex]\int_a^b xdx[/tex] ....what do you do?

    Casey
     
  5. Sep 18, 2007 #4

    Dick

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    I find an antiderivative to x, which is x^2/2 and hence conclude the answer is b^2/2-a^2/2. What do YOU do?
     
  6. Sep 18, 2007 #5

    dynamicsolo

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    The Fundamental Theorem of (Integral) Calculus says you would antidifferentiate x to get (1/2)(x^2) and then take the difference (1/2)(4^2) - (1/2)(2^2). It applies to any definite integral of f(x), *provided* the function is continuous on the interval between your integration limits. If the antiderivative (or integral) function of f(x) is F(x), then the definite integral from a to b is F(b)-F(a).

    You'll learn somewhat later what to do if the function f(x) is *not* continuous...
     
  7. Sep 18, 2007 #6
    The answer is 12 for both....okay...I see where my error was.

    Is that coincidence or are a function and its 1st derivative's areas under the same bounds supposed to be equal?! I don't think they are.

    i.e, x from [0,3} =3 and its antiderivative x^2/2 from [0,3] = 9/2

    I am more confused now. Can someone answer post #3 for me:confused:

    Thanks,
    Casey

    Edit: I am too slow! Thanks dynamicsolo.


    SO is this correct to say:

    [tex]\int_a^bxdx=(.5b^2)-(.5a^2)[/tex] =the area under the graph of y=x.<---or is it the area under .5x^2 from [a,b]? AHHH!!!

    Edit again: I saw your post Dick. Thanks. I would stare at it for awhile and then go kick something...because the way I thought you did it was to evaluate .5x^2 from a to b ....however, my text appears to be doing something funny....maybe I am reading it wrong.
     
    Last edited: Sep 18, 2007
  8. Sep 18, 2007 #7

    Dick

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    You are indeed royally confused. If F(x) is an antiderivative of f(x), then F(b)-F(a) is the area under the curve of f(x) between a and b. This is NOT equal to f(b)-f(a). And the area under F(x) is not equal to the area under f(x). It's only the first statement that is true. None of the others.
     
  9. Sep 18, 2007 #8

    dynamicsolo

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    In reply:

    [tex]\int_a^bxdx=(.5b^2)-(.5a^2)[/tex] =the area under the graph of y=x.
     
  10. Sep 18, 2007 #9
    Okay. This is what I thought, so I was not confused by the Definite Integral, I must be confused by my textbook instructions.

    Here is the problem exactly as stated.

    Sketch the region whose signed area is represented by the definite integrand, and evaluate the inegral using an appropriate formula from geometry, where needed.

    [tex]\int_0^3 xdx[/tex]


    Ohhh @#$%.....I get it. I can draw y=x and use .5(bh) to find the area of the triangle=9/2.....OR.....evaluate the antiderivative .5x^2 from [0,3] to get the same answer.

    Casey
     
  11. Sep 18, 2007 #10

    Dick

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    Yessssssssssssssssss.
     
  12. Sep 18, 2007 #11
    NOoooooooooooo.

































    Just kidding. Yes.
     
  13. Sep 18, 2007 #12

    Dick

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    Don't scare me, I'm the sssssnake. I thought I was losing you again.
     
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