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Simple Question about integral definition

  1. Jul 30, 2013 #1
    1. The problem statement, all variables and given/known data
    so this is my first time learning about integrals , from spivak' calculus
    Actual quote : the integral [itex]\[ \int_{a}^{b} f(x) \, \mathrm{d}x \][/itex]was defined only for a<b we now add the definition
    [itex]\[ \int_{a}^{b} f(x) \, \mathrm{d}x \][/itex]=-[itex]\[ \int_{b}^{a} f(x) \, \mathrm{d}x \][/itex] if a>b "
    isn't he contradicting himself here to write[itex]\[ \int_{a}^{b} f(x) \, \mathrm{d}x \][/itex] a<b is required right?so you can't just write [itex]\[ \int_{a}^{b} f(x) \, \mathrm{d}x \][/itex] when yo usay "if a >b"
    i tried doing problem 7 which involves the function x^3
    we have [itex]\[ \int_{-1}^{1} x^3 \, \mathrm{d}x \][/itex]=[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \][/itex] + [itex]\[ \int_{0}^{1} f(x) \, \mathrm{d}x \][/itex](so far everything is normal) =applying spivak's definition -[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \][/itex] +[itex]\[ \int_{0}^{1} f(x) \, \mathrm{d}x \][/itex] why in the answer books he says this equals 0 ? this dosen't make sense at all since [0;-1] is not an interval?[itex]\[ \int_{-1}^{0} f(x) \, \mathrm{d}x \]/[itex] requires that 0<-1 ..
    Please help i am VERY confused.


    2. Relevant equations
    mentioned above


    3. The attempt at a solution
    mentioned above
     
    Last edited: Jul 30, 2013
  2. jcsd
  3. Jul 30, 2013 #2
    It is an odd function.
     
  4. Jul 30, 2013 #3

    Ray Vickson

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    Homework Helper

    What's the problem? -1 < 0 is certainly true! Why would you think otherwise?
     
  5. Jul 30, 2013 #4
    well i miss typed that, i mean 0<-1 laughs , well i think none is getting me here?
    can someone prove that [itex]\[ \int_{-1}^{1} x^3 \, \mathrm{d}x \][/itex] = 0 ? by splitting the intervals to -1 , 0 and 0 1 , that would help clear these things
     
  6. Jul 30, 2013 #5

    LCKurtz

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    If ##f## is an odd function so ##f(-x) = -f(x)## then for ##a>0##,$$
    \int_{-a}^af(x)\, dx =\int_{-a}^0f(x)\, dx + \int_0^a f(x)\, dx$$Let ##x=-u## in the first integral making it$$
    \int_{a}^0f(-u)\, (-1)du = \int_{a}^0 -f(u)\, (-1)du = \int_a^0f(u)\, du
    =-\int_0^af(u)\, du$$so this integral cancels the second integral, giving ##0##.
     
  7. Jul 30, 2013 #6
    thanks everything is clear now and a little bit offtopic , wow Integral is hard compared to derivatives and limits i might switch to another book spivak became suddenly very hard to me
     
  8. Jul 30, 2013 #7

    Mark44

    Staff: Mentor

    Some LaTeX tips.
    You're putting in way more symbols than you actually need - extra brackets and slashes.
    Use a single pair of LaTeX delimiters for an entire equation, rather than breaking it up into multiple LaTeX expressions.

    Instead of writing this: [itex]\[ \int_{a}^{b} f(x) \, \mathrm{d}x \][/itex], you can write it much more simply this way: ##\int_a^b f(x)~dx##
    Code (Text):
    ##\int_a^b f(x)~dx##
    Or instead of this: [itex]\[ \int_{a}^{b} f(x) \, \mathrm{d}x \][/itex]=-[itex]\[ \int_{b}^{a} f(x) \, \mathrm{d}x \][/itex]
    You can write this:
    ## \int_a^b f(x)~dx = -\int_b^a f(x)~dx##
    Code (Text):
    ## \int_a^b f(x)~dx = -\int_b^a f(x)~dx##
     
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