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Integrating (x^4+x^2)^0.5 form 3^0.5 to -3^0.5

  1. Sep 29, 2014 #1
    3. The attempt at a solution
    Let x=tan u
    dx=sec^2(u)*du

    When x=3^0.5,u=pi/3
    x=-3^0.5,u=-pi/3
    S sec^2(u)d(sec u)
    =1/3[2^3-2^3]=0
     
  2. jcsd
  3. Sep 29, 2014 #2
    But it is wrong
     
  4. Sep 29, 2014 #3

    SteamKing

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    It's not clear why you chose this substitution, x = tan u, for this integral.

    In any event, you have not shown what indefinite integral you obtained after making the substitution, so plugging in the limits to evaluate the definite integral is rather pointless.

    Please show all of your work.
     
  5. Sep 29, 2014 #4
    S(x^4+x^2)^0.5 dx
    =S(x^2(x^2+1))^0.5
    Let x=tan u
    dx=sec^2(u)du
    When x=(3)^0.5,u=pi/3
    x=-(3)^0.5,u=-pi/3
    =S(tan^2(u)(tan^2(u)+1)^0.5*(sec^2(u))du
    =Ssec^3(u)tan(u)du
    =Ssec^2(u)d(sec u)
    =1/3*sec^3(u)
    By substitution
    Ans=0
     
  6. Sep 29, 2014 #5

    pasmith

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    Use the fact that the integrand is even to obtain [tex]
    \int_{-\sqrt{3}}^{\sqrt{3}} \sqrt{x^4 + x^2}\,dx = 2 \int_0^{\sqrt{3}} \sqrt{x^4 + x^2}\,dx
    [/tex] which will avoid lurking sign errors (such as that if [itex]-\frac13\pi \leq u < 0[/itex] then [itex]\sqrt{\tan^2 u} = |\tan u| = -\tan u[/itex].) Also, why not consider the easier [itex]u = x^2[/itex] instead of a trig substitution?
     
  7. Sep 29, 2014 #6
    But why i can't calculate in this why ,I just dun know what is going wrong
     
  8. Sep 29, 2014 #7

    Mark44

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    For this integral --
    $$ \int \sqrt{x^4 + x^2}dx$$
    -- the simplest thing to do is to factor x2 from the two terms in the radical, bringing out a factor of x, not x2 as you show.
    ## \sqrt{x^4 + x^2} \neq x^2 \sqrt{x^2 + 1}##
    After you fix the mistake above, then use an ordinary substitution. You could use a trig substitution, but it's almost always better to use a simpler method than a more complicated one.

    Edit: Because of the symmetric limits of integration, you need to bring out a factor of |x|, not x as I said earlier.
     
    Last edited: Sep 30, 2014
  9. Sep 29, 2014 #8

    Ray Vickson

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    Such problems are always tricky, and you are well advised to ALWAYS look first at the "symmetry" of the problem, which in this case means that because your ##f(x) = \sqrt{x^2 + x^4}## is even, ##\int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx##; more generally, for ##a,b > 0## it is best to write ##\int_{-a}^b f(x) \, dx = \int_0^a f(x) \, dx + \int_0^b f(x) \, dx##. Sometimes you can get correct answers by substituting in the limits at ##-a## and ##b##, but in other cases you get completely wrong answers, usually because formal algebraic manipulation of the antiderivative formula can give results that are incorrect over part of the ##x##-region (due to errors such as saying ##x = \sqrt{x^2}##, for example). Your antiderivative is ##(1/3)\sec^3 u = (1/3) (x^2+1)^{3/2}= F(x)##, and the derivative of ##F(x)## is ##F'(x) = x\sqrt{1+x^2}##. This is equal to ##f(x)## only when ##x \geq 0##. In fact, ##f(x) = |x| \sqrt{1+x^2}##, not ##x \sqrt{1+x^2}##. Since ##f(x)## is an even function its true antiderivative ##G(x)## should be an odd function. So, the correct antiderivative is
    [tex] \int f(x)\,dx = (1/3) \text{sign}(x) (x^2+1)^{3/2} = (1/3) \text{sign}(u) \sec^3(u), [/tex]
    where
    [tex] \text{sign}(w) = \begin{cases} 1, & w > 0\\
    -1, & w < 0
    \end{cases}
    [/tex]

    Note added in edit: of course the Fundamental Theorem of Calculus applies, so the only issue is whether or not one's antiderivative formula ##F(x)## applies over the whole relevant range of ##x##. That is what one needs to be careful about.
     
    Last edited: Sep 29, 2014
  10. Sep 30, 2014 #9
    if i substitute x^2=u
    when x=(3)^0.5 , u=3
    when x= - (3)^0.5,u=3
    the integral will be from 3 to 3 which is equal to 0
     
  11. Sep 30, 2014 #10
    it is [(x^2)*(x^2+1)]^0.5
    sorry that im not expressing this clear
     
  12. Sep 30, 2014 #11

    Ray Vickson

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    This is nonsense. I already wrote out for you a complete explanation, but it seems you paid no attention at all to what I told you. Don't you get it? Carelessly substituting ##x^2 = u## without thinking can fail!. It is OK for ##x > 0##, but for ##x < 0## it gives the "wrong" ##du##, because when ##x < 0## and it increases by a little bit ##dx > 0##, then ##u## decreases and so we ought to have ##du < 0##. In other words, you need to use [tex] du = \text{sign}(x) 2 dx,[/tex]
    NOT ##du = 2 dx##. I keep making the same point over and over again: the antiderivative ##F(x)## is not just ##(1/3)(u+1)^{3/2} = (1/3)(x^2+1)^{3/2}##; it should be ##F(x) = \text{sign}(x)(1/3)(x^2+1)^{3/2}##. In terms of ##u## this would be ##\text{sign}(x)(1/3)(u+1)^{3/2}##, which looks a bit weird because it does not involve just ##u## alone.
     
  13. Sep 30, 2014 #12

    Mark44

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    I misread what you wrote. BTW, I like the other suggestions made in this thread better than my own. I corrected my earlier post to say that you can bring out a factor of |x|, rather than what I wrote about bringing out a factor of x.
     
  14. Sep 30, 2014 #13
    Simply: [tex]\int\sqrt{x^4+x^2}dx=\int\sqrt{x^2(x^2+1)}dx=\int x \sqrt{x^2+1}dx[/tex]
    then by the simple substitution [tex]u = x^2+1[/tex] the integral transforms to:
    [tex]\frac{1}{2}\int\sqrt{u}du = \frac{1}{2} \frac{u^\frac{3}{2}}{\frac{3}{2}}=\frac{u^\frac{3}{2}}{3}[/tex], the constant of integration was omitted because the original integral is definite.

    Substituting back the value of u we conclude that: [tex]\int\sqrt{x^4+x^2}dx=\frac{1}{3}(1+x^2)^\frac{3}{2}[/tex].
    The evaluation should be no problem at all.
    About the factor |x|, I believe that it is not necessary and that you can only leave it as x. It is true that the square root of a positive number gives two answers, one positive and one negative both with the same magnitude. But we have to keep in mind that we are integrating a FUNCTION [tex]y(x)=\sqrt{x^4+x^2}[/tex] And as you may see from the lack of negative sign of the root, the function is taking into account the positive part of [tex]y^2 =x^4+x^2[/tex] Therefore no absolute value is needed. Although, I have to admit that Vickson's answer is much more general.
     
  15. Sep 30, 2014 #14

    Mark44

    Staff: Mentor

    This is oversimplified in light of the limits of integration of the original problem, and I admit to making the same mistake myself. Because the limits of integration are positive and negative, you have to incorporate the absolute value. Ray's posts explain this in detail.
    No. The square root of a positive number is a function, which means it produces one result, not two. It is true that any positive number has two square roots, but, for example, ##\sqrt{9} = 3##, not ##\pm 3##.
     
  16. Oct 1, 2014 #15

    Ray Vickson

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    Unfortunately, this is incorrect. The antiderivative of ##f(x) = \sqrt{x^4+x^2}## is NOT ##(1/3)(1+x^2)^{3/2}##; it is (up to a constant) ##F(x) = \text{sign}(x) (1/3) (1 + x^2)^{3/2}##. That last one is the form you must use if you want the fundamental theorem of calculus to give you the right answer from ##\int_{-a}^a f(x) \, dx = F(a) - F(-a)##. You made essentially the same error as the OP: the transformation ## u = x^2 + 1## is not monotone, and so leads to trouble when you integrate through ##x = 0##. As I said before, in integrating you need to use ##du = \text{sign}(x) 2 x dx##. More simply put: the antiderivative of an even function must be an odd function (plus a constant), but your claimed antiderivative is even.
     
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