1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?