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Antiderivative problem

  1. Jan 8, 2006 #1
    Problem states:

    Find and antiderivative [tex]F[/tex] of [tex]f(x)=x^2\sin(x^2)[/tex] such that [tex]F(1)=0[/tex].

    I thought at first that I would use [tex]u=x^2[/tex] to simplify the matter but I don't seem to get anywhere. I can't seem to figure out a way to start it off either...I was considering the Squeeze Theorem because they are functions, essentially.

    Any advice/help?
  2. jcsd
  3. Jan 8, 2006 #2
    How do you use the squeeze theorem to find an antiderivative? Try integration by parts. If you haven't learned integration by parts yet, then are you sure it isn't [tex] x\sin(x^2)[/tex]
  4. Jan 8, 2006 #3
    Nope...sorry, I haven't learned integration by parts yet...that will be covered in the second Calculus class that I will be taking in this Winter session at my college.

    And yes, I am sure it isn't [tex]x\sin(x^2)[/tex]...that would be a completely superficial problem that I should not have any problems with =P.

    I was thinking about this problem some more (been stuck on it since last night and it's almost the night of the next day haha) and I thought that I could possibly get an antiderivative that had [tex]\frac{1}{2}[/tex] in it but then I realized that no matter what I came up with...the product rule would ruin my plan. If only that [tex]x^2[/tex] was a constant haha...

    About an hour ago, I thought about a method that I could use by adding in a few functions that wouldn't change the value of [tex]f(x)[/tex] and then I came up with a plan all of a sudden and it seemed to work in my head. I worked it over a few times in my head and realized that it may be the answer...I would have to use the law that states [tex]\sin^2(x)+\cos^2(x)=1[/tex].

    Sadly, that didn't produce the answer either because I ended up with:

    This didn't get me anywhere because the product rule caught me none the less...the [tex]x^2[/tex] was the problem in that case.

    While typing all of this up, I decided to look in the back of the book to see what the answer is so that I could get some insight as to how to go about this problem....and I was a bit dumbfounded upon seeing it...I guess I forgot that an antiderivative can just be the general notation of an integral and the integrand...you don't necessarily need the actual antiderivative function...*sigh*...mmmm the answer was like this:

  5. Jan 9, 2006 #4


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    That happened to me in a differential equations course once- I reduced the problem to an integral and spent hours trying to do the integral. The answer in the back of the book gave the solution in terms of the integral!:rolleyes:
  6. Jan 9, 2006 #5


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    This is probably a properties of definite integrals question, cause there is no way you can integrate this that I know of without using IBP.
    Last edited by a moderator: Jan 9, 2006
  7. Jan 10, 2006 #6


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    [tex] \int x^{2}\sin x^{2}dx= -\frac{1}{2}\left( \cos x^{2}\right) x+\frac{1}{4}\sqrt{2}\sqrt{\pi }\func{FresnelC}\left( \frac{\sqrt{2}}{\sqrt{\pi }}x\right) + C [/tex]

    Well, set x=1 in the RHS and equal it to 0.

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