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Math Help - application of integrals

  1. #1
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    Talking application of integrals

    let f(x)= (sinx) / x , from 0 which is less than x which is less than or equal to pi
    1, when x = 0

    show that xf(x) = sinx , for 0 which is less than or equal to x which is less than or equal to pi

    i multiplied both sinx/x by x and got sinx for 0<x<(less than or equal to) pi
    and multiplied 1 * x = x which gives me 0 when x=0.
    is that right because i'm unsure if the question also wants it to be continuous

    also the second question says that find the volume of the solid generated by revolving f(x) about the y-axis.
    how am i supposed to calculate that if f(x) is divided into 2 parts
    Last edited by ubhutto; September 22nd 2013 at 11:43 AM.
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  2. #2
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    Re: application of integrals

    Quote Originally Posted by ubhutto View Post
    let f(x)= (sinx) / x , from 0 which is less than x which is less than or equal to pi
    1, when x = 0

    show that xf(x) = sinx , for 0 which is less than or equal to x which is less than or equal to pi

    i multiplied both sinx/x by x and got sinx for 0<x<(less than or equal to) pi
    You multiplied both x and what?

    and multiplied 1 * x = x which gives me 0 when x=0.
    So what? Why are you concerned with x= 0?

    ebr
    is that right because i'm unsure if the question also wants it to be continuous

    also the second question says that find the volume of the solid generated by revolving f(x) about the y-axis.
    how am i supposed to calculate that if f(x) is divided into 2 parts
    If you are taking Caculus, you certainly should know enough algebra to know that if you multiply both sides of f(x)= sin(x)/x by x you will get xf(x)= sin(x). Why do you have any question about that?

    Now, what do you mean by "f(x) is divided into two parts? Normally, I prefer the "disk method" but here it is "x" that would be the radius and it will be difficult to solve for x as a function of y (the inverse function to f) so I think I would use the "shell method". That is, we imagine taking a thin strip (of thickness "dx") at a given x value, between 0 and [/itex]\pi[/tex] (where f(x)= sin(x)/x= 0) and rotate that strip around the y- axis. It describes a cylinder of radius x and height y= sin(x)/x so a cylinder or area \pi x^2(sin(x)/x)= \pi x sin(x). That gives a thin shell of volume \pi x sin(x) dx. Integrate that from x= 0 to \pi.
    Thanks from ubhutto
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