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Homework Help: Space between these functions of x

  1. Jan 14, 2009 #1
    i need to find the are trapped between the graphs of
    y=-x3
    y=arctg[tex]\sqrt{x}[/tex]
    x=2
    x=1

    y=arctg[tex]\sqrt{x}[/tex] will be my top function so the area will be

    [tex]\int[/tex]arctg[tex]\sqrt{x}[/tex]-(-x3)dx from 1 to 2

    in order to to this i can split the integral up and get
    [tex]\int[/tex]arctg[tex]\sqrt{x}[/tex]dx -[tex]\int[/tex]-x3dx both 1-2

    |[tex]\int[/tex]-x3dx| (1-2) is 3.75 with a simple table integral

    but i dont know how to integrate the arctg

    what i think could work is...

    arctg[tex]\sqrt{x}[/tex]=t
    tg(t)=[tex]\sqrt{x}[/tex]

    [tex]\int[/tex]tdx while dt=dx/1+x2

    [tex]\int[/tex]t(1+x4)dt
    =[tex]\int[/tex]t(1+tg4t)dt
    =[tex]\int[/tex]tdt+[tex]\int[/tex](tg4t)dt

    another option
    [tex]\int[/tex]t(1+x4)dt
    =1/2[tex]\int[/tex](1+tg4t)dt2
    =0.5[tex]\int[/tex]dt2+0.5[tex]\int[/tex](tg4t)dt2

    sort of feels like im getting lost in it around here,,, any ideas??
     
  2. jcsd
  3. Jan 14, 2009 #2
    how about this,,

    lets call t2=p

    [tex]\int[/tex](tg2p)dp=[tex]\int[/tex][tex]\frac{sin^2p}{cos^2p}[/tex]

    =[tex]\int[/tex](1/cos2p)dp-[tex]\int[/tex]dp
    =tgp-p
    =tg(t2)-t2
    =tg[{arctg[tex]\sqrt{x}[/tex]}2]-(arctg[tex]\sqrt{x}[/tex])2
    =x-(arctg[tex]\sqrt{x}[/tex])2

    is this correct
     
    Last edited by a moderator: Jan 15, 2009
  4. Jan 14, 2009 #3

    Dick

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    To integrate arctg(sqrt(x)), set u=sqrt(x). So du=dx/(2*sqrt(x))=dx/(2u). Or 2udu=dx. The integral in terms of u becomes arctg(u)*2udu=arctg(u)*d(u^2). Now integrate by parts.
     
    Last edited: Jan 14, 2009
  5. Jan 14, 2009 #4
    on my table of integrals i dont have an integral for arctg, how do i do this? do i need to do something to the effect of

    arctg u =v
    tgv=u and then carry on from there (similar to what i posted? ) or is there a abetter way?
     
  6. Jan 14, 2009 #5

    Dick

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    To integrate arctg(u)*d(u^2) you don't need to know how to integrate arctg(u). You only need to know how to differentiate it. a*db=d(a*b)-b*da. That's how integration by parts works, right?
     
  7. Jan 14, 2009 #6
    cant see it, could you please show me.
     
  8. Jan 14, 2009 #7

    Dick

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    The integral of arctg(u)*d(u^2)=arctg(u)*u^2-the integral of u^2*d(arctg(u)). That's what integration by parts means, yes? What d(arctg(u))? Differentiate arctg(u).
     
  9. Jan 14, 2009 #8
    ive read up a bit on this and found an equation

    [tex]\int[/tex]udv=u*v-[tex]\int[/tex]vdu

    is this what you mean??

    so in my case
    arctgU is my u function
    and u^2 is my v function
    where u is the root of x

    so

    [tex]\int[/tex](arctg(u))d(u2)=(arctg(u))*u2-[tex]\int[/tex]u2d(arctg(u))

    now d(arctg(u))=[tex]\frac{1}{1+u^2}[/tex]*d(u^2)

    so [tex]\int[/tex]u2d(arctg(u))=[tex]\int[/tex]U2/(1+u2)d(u2)

    correct up till here??

    =[tex]\int[/tex]1-[tex]\frac{1}{1+u^2}[/tex]du2====u2-ln|1+u^2|

    which is x-ln|a+x|
     
  10. Jan 14, 2009 #9

    Dick

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    You are getting there. But d(arctg(u))=du/(1+u^2) (NOT d(u^2)). So the integral you are left with is u^2*du/(1+u^2). You'll find there aren't any logs in the problem.
     
  11. Jan 14, 2009 #10
    how did you come to d(arctg(u))=du/(1+u^2)
    is (in general) d(t) not equal to the differential of t multiplied by dx???

    in this case d(arctg(u))= differential of (arctg(u)) * d(u^2)

    how do i know which d() to multiply by?
     
  12. Jan 14, 2009 #11

    Dick

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    Exactly as you said. d(f(x))=f'(x)*dx. d(arctg(x))=(d/dx arctg(x))*dx. d/dx(arctg(x))=1/(1+x^2). So d(arctg(x))=dx/(1+x^2). I don't know where you are getting the d(u^2) stuff.
     
  13. Jan 15, 2009 #12
    i see my problem, what is the differential of arctg(x^0.5) is it not the 1/(1+x^0.5) multiplied by 1/2(x^0.5) ??
     
  14. Jan 15, 2009 #13

    Gib Z

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    Well, it appears you are using the chain rule, which is good, but you are forgetting to square something. Look at what the derivative of arctan x is again.
     
  15. Jan 15, 2009 #14

    Dick

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    No! The differential of arctg(sqrt(x)) is 1/(1+sqrt(x)^2) times 1//(2*sqrt(x)). Like Gib said, you forgot the square in the derivative of arctg. That's EXACTLY the same thing as saying the differential of arctg(u) is 1/(1+u^2) times du where u=sqrt(x). Look at it!
     
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