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Math Help - definite integrals

  1. #1
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    definite integrals

    hi

    have been given question : write down the definite integral that will be given of area under the curve y=x^3 cos(1/6x) between x=5/2pie and x=3pie.

    unfortunately i do not even know how to begin integration of y=x^3 cos(1/6x). can some one give me some help please.

    wayne
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  2. #2
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    Integration by parts thrice! Some different approaches to a very similar examples here:

    http://www.sosmath.com/CBB/viewtopic...176529#176529]
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  3. #3
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    thanks for the link. have to admit i have no idea about integration by parts as i only started on calculus on monday,so your link is very helpful, will read carefully and post my attempts later this evening.

    would like to thank you again for you time and patients i really appreciate your help.

    wayne
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  4. #4
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    Quote Originally Posted by smartcar29 View Post
    hi

    have been given question : write down the definite integral that will be given of area under the curve y=x^3 cos(1/6x) between x=5/2pie and x=3pie.

    unfortunately i do not even know how to begin integration of y=x^3 cos(1/6x). can some one give me some help please.

    wayne
    Are you asked to do that integration? The question you post here only asked you to write the integral, not to actually do the integration.
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  5. #5
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    hi

    i guess i have misunderstood the question, i guess i just plug in the limits to find the area between the two limits. i thought i needed to integrate y=x^3 cos(1/6x).

    thanks

    wayne
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  6. #6
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    We need to know the wording of the problem. If it refers to "the curve y = x^3 cos(1/6x)..." then no, plugging the limits into that will just give you the y values of those two points on the curve (or the difference between them). HallsofIvy is pointing out, I think, that the exercise may just be to use appropriate notation to state the quantity:

    A = \int_{\frac{5}{2}\pi}^{3\pi} x^3 \cos(\frac{1}{6}x)\ dx

    On the other hand, if it said "the integral y = x^3 cos(1/6x) gives the area under a curve", then plug away...

    What is the wording?
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  7. #7
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    the exact question reads

    write down a definite integral that will give the value of the area under the curve y=x^3 cos(1/6x) between x=5/2 pie and 3pie.

    the expression x^3 cos(1/6x) takes no negative values for 5/2pie<x<3pie.
    you are not asked to evaluate the integral by hand.

    part 2 of question use mathCAD to find the area described in part 1

    and thats all she wrote.
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  8. #8
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    Yes, so HoI is right, of course. For now, you just write down,

    A = \int_{\frac{5}{2}\pi}^{3\pi} x^3 \cos(\frac{1}{6}x)\ dx

    Then use the machine to get

    \int_{\frac{5}{2}\pi}^{3\pi} x^3 \cos(\frac{1}{6}x)\ dx = [..........]_{\frac{5}{2}\pi}^{3\pi}

    ... and so on.
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