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Math Help - Integral

  1. #1
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    Integral

    If x\sin (\pi x) = \int\limits_1^{{x^2}} {f(t)dt} where f is a continuous function find f(4) and f'(4)

    use Fundamental therom of Calculus

    so I know that I'm going to let
     u = x^2

    and then differentiate both sides but how do i defferentiate

     x\sin (\pi x)

    and it's derivate how do i do that

    ---------------------------------------------------------------------------------------------------

    Another question with this F.T.C stuff

    Use F.T.C find a function and a number c such that

    6 + \int\limits_c^x {\frac{{f(t)}}{{{t^2}}}} dt = 2\sqrt x for x>0

    So I know i gotta differentiate both sides but how would i find a number C when F.T.C would just eliminate it since derivative of a number is 0
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  2. #2
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    Quote Originally Posted by treetheta View Post
    If x\sin (\pi x) = \int\limits_1^{{x^2}} {f(t)dt} where f is a continuous function find f(4) and f'(4)

    use Fundamental therom of Calculus

    so I know that I'm going to let
     u = x^2

    and then differentiate both sides but how do i defferentiate

     x\sin (\pi x)

    and it's derivate how do i do that
    x\sin(\pi x) = \int_1^{x^2} f(t) \, dt

    derivative ...

    x \cdot \pi \cos(\pi x) + \sin(\pi x) = f(x^2) \cdot 2x

    let x = 2 ...

    2 \cdot \pi \cos(2\pi) + \sin(2\pi) = f(4) \cdot 4

    f(4) = \frac{\pi}{2}


    take the derivative again ...

    \frac{d}{dx}[x \cdot \pi \cos(\pi x) + \sin(\pi x) = f(x^2) \cdot 2x]

    -x \pi^2 \sin(\pi x) + \pi \cos(\pi x) + \pi \cos(\pi x) = 2 \cdot f(x^2) + 4x^2 \cdot f'(x^2)

    substitute x = 2 again, and determine f'(4)
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  3. #3
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    lol I don't know why I thought the derivative would be complicated I forgot pi is just a number
    Thanks (:

    can you help me out with the other one?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by treetheta View Post
    Another question with this F.T.C stuff

    Use F.T.C find a function and a number c such that

    6 + \int\limits_c^x {\frac{{f(t)}}{{{t^2}}}} dt = 2\sqrt x for x>0

    So I know i gotta differentiate both sides but how would i find a number C when F.T.C would just eliminate it since derivative of a number is 0
    Take the derivative:

    \frac{f(x)}{x^2} = \frac{1}{\sqrt{x}}

    f(x) = \frac{x^2}{\sqrt{x}} = x^{3/2}

    Plug f(x)=x^{3/2} back into original equation to find c.

    Spoiler:
    6+\int_c^x\frac{t^{3/2}}{t^2}\,dt = 2\sqrt{x}

    6+2\sqrt{x}-2\sqrt{c} = 2\sqrt{x} \implies 6=2\sqrt{c} \implies \boxed{c = 9}
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  5. #5
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    Oh snap, wow, who new it would work out so nicely, wow thank you so much
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  6. #6
    Super Member redsoxfan325's Avatar
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    Your welcome.
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  7. #7
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    Use F.T.C find a function and a number c such that

    6 + \int\limits_c^x {\frac{{f(t)}}{{{t^2}}}} dt = 2\sqrt x for x>0

    So I know i gotta differentiate both sides but how would i find a number C when F.T.C would just eliminate it since derivative of a number is 0
    6 + \int_c^x \frac{f(t)}{t^2} \, dt = 2\sqrt{x}

    derivative ...

    \frac{f(x)}{x^2} = \frac{1}{\sqrt{x}}

    f(x) = x^{\frac{3}{2}}

    go back to the original integral equation and sub in f(t) ...

    6 + \int_c^x t^{-\frac{1}{2}} \, dt = 2\sqrt{x}

    6 + [2\sqrt{t}]_c^x = 2\sqrt{x}

    6 + [2\sqrt{x} - 2\sqrt{c}] = 2 \sqrt{x}

    c = 9
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  8. #8
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    How would I go about doing

    \mathop {\lim }\limits_{x \to  - \infty } \sqrt {x + {x^2}}  - x

    I multiplied by the conjugate, to get

    \mathop {\lim }\limits_{x \to  - \infty } \frac{x}{{\sqrt {x + {x^2}}  + x}}

    just don't know how to continue I'm sure I could factor out x, but I don't know how to do that, could someone show me or would L'hospital work
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  9. #9
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by treetheta View Post
    How would I go about doing

    \mathop {\lim }\limits_{x \to  - \infty } \sqrt {x + {x^2}}  - x

    I multiplied by the conjugate, to get

    \mathop {\lim }\limits_{x \to  - \infty } \frac{x}{{\sqrt {x + {x^2}}  + x}}

    just don't know how to continue I'm sure I could factor out x, but I don't know how to do that, could someone show me or would L'hospital work
    Well, \sqrt{x+x^2}\to\infty as x\to-\infty, and -x\to\infty as x\to-\infty, so it makes sense that \lim_{x\to-\infty}\sqrt{x+x^2}-x = \infty

    Maple confirms that \infty is the correct answer.

    ---------------------------------------

    However, that raises a question of my own: Where did I go wrong below:

    \lim_{x\to-\infty}\sqrt{x+x^2}-x = \lim_{x\to-\infty}\left((\sqrt{x+x^2}-x)\cdot\frac{\sqrt{x+x^2}+x}{\sqrt{x+x^2}+x}\right  ) =\lim_{x\to-\infty}\frac{x}{\sqrt{x+x^2}+x} = \lim_{x\to-\infty}\frac{x}{\sqrt{x^2(1/x+1)}+x} = \lim_{x\to-\infty}\frac{x}{x\sqrt{1/x+1}+x} = \lim_{x\to-\infty}\frac{x}{x(\sqrt{1/x+1}+1)} = \lim_{x\to-\infty}\frac{1}{\sqrt{1/x+1}+1}

    Now we can take the limit as x\to-\infty to get \frac{1}{\sqrt{0+1}+1} = \frac{1}{2}

    I feel like I made a really fundamental error and am just not seeing it. I know that if x\to+\infty, the limit is \frac{1}{2}, so where is my reasoning faulty?
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  10. #10
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    Hmm yes, I got this answer as well but however in the choices i have to pick from it's not there...
    however -1/2 is so i convinced myself that maybe -infnity had something to do with it

    oh btw, redsoxfan325 you are godlike math genius (:

    could you help me out with this one as well:

    \int {\frac{1}{{x\sqrt {{x^6} - 4} }}}

    I really don't know where to go with this I attempted to trigsubstiution but then that got really messy, I don't think parts will work, and I dont think theres anything i can normally substitute ):
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  11. #11
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by treetheta View Post
    Hmm yes, I got this answer as well but however in the choices i have to pick from it's not there...
    however -1/2 is so i convinced myself that maybe -infnity had something to do with it

    oh btw, redsoxfan325 you are godlike math genius (:

    could you help me out with this one as well:

    \int {\frac{1}{{x\sqrt {{x^6} - 4} }}}

    I really don't know where to go with this I attempted to trigsubstiution but then that got really messy, I don't think parts will work, and I dont think theres anything i can normally substitute ):
    \int\frac{dx}{x\sqrt{x^6-4}} = \int\frac{dx}{x\sqrt{x^6(1-\frac{4}{x^6})}} = \int\frac{dx}{x^4\sqrt{1-\left(\frac{2}{x^3}\right)^2}}

    Let u=\frac{2}{x^3}. Thus du = -\frac{6}{x^4}

    So our integral now looks like: -\frac{1}{6}\int\frac{du}{\sqrt{1-u^2}} = -\frac{1}{6}\sin^{-1}(u) = \boxed{-\frac{1}{6}\sin^{-1}\left(\frac{2}{x^3}\right)}

    Thanks for the compliment!
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  12. #12
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    you make everything look so easy,

    is there a trig formula for
    -\frac{1}{6}\int\frac{du}{\sqrt{1-u^2}} = -\frac{1}{6}\sin^{-1}(u)

    to equal eachother?

    I think this will be the last one I shall ask you becuase i'm starting to feel bad ><

    \int {\frac{{{x^5} - 1}}{{{x^4} + {x^2}}}}
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  13. #13
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by treetheta View Post
    you make everything look so easy,

    is there a trig formula for
    -\frac{1}{6}\int\frac{du}{\sqrt{1-u^2}} = -\frac{1}{6}\sin^{-1}(u)

    to equal eachother?

    I think this will be the last one I shall ask you becuase i'm starting to feel bad ><

    \int {\frac{{{x^5} - 1}}{{{x^4} + {x^2}}}}
    First, the inverse sine integral should be found in any table of integrals. Second, break the integral up into two parts:

    \int\frac{x^5-1}{x^4+x^2} = \int\frac{x^5}{x^4+x^2}\,dx - \int\frac{1}{x^4+x^2}\,dx

    First integral:

    \int\frac{x^5}{x^4+x^2}\,dx = \int\frac{x^3}{x^2+1}\,dx

    Let x=\sqrt{u-1}. Thus dx=\frac{1}{2\sqrt{u-1}}

    The integral is now \int\frac{(u-1)^{3/2}}{u}\cdot\frac{1}{2\sqrt{u-1}}\,du = \frac{1}{2}\int\frac{u-1}{u}\,du = \frac{1}{2}\int 1-\frac{1}{u}\,du = \frac{u}{2}-\frac{1}{2}\ln|u| = \frac{x^2+1}{2}-\frac{1}{2}\ln|x^2+1|

    Second integral:

    \int\frac{1}{x^4+x^2}\,dx = \int\frac{1}{x^2(x^2+1)}\,dx

    \frac{1}{x^2(x^2+1)} = \frac{Ax+B}{x^2}+\frac{Cx+D}{x^2+1} \implies (A+C)x^3+(B+D)x^2+Ax+B = 1

    Thus, A=0, B=1, C=0, D=-1

    So \int\frac{1}{x^2(x^2+1)} = \int\frac{1}{x^2}\,dx-\int\frac{1}{x^2+1}\,dx = -\frac{1}{x}-\tan^{-1}(x)

    Subtracting the first and second integrals gives \boxed{\frac{x^2+1}{2}-\frac{1}{2}\ln|x^2+1| + \frac{1}{x} + \tan^{-1}(x) + C}

    Sorry it took so long. That one was tough.
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  14. #14
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    Don't worry about time man, you can get anything I would have never thought to sperate it good call there, can you suggest any practice type questions (: I got an exam tmrw been scrambling all over the place try to do as much as possible
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  15. #15
    Super Member redsoxfan325's Avatar
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    1.) \int\frac{1}{x^4+1}\,dx (Good luck, it's a hard one!)

    2.) \int-\frac{x^4}{(1+x^2)^{3/2}}\,dx

    3.) \int\sqrt{1+e^x}\,dx

    4.) \int\sin(\ln(x))\,dx

    5.) \int\frac{1}{x\ln(x)\ln(\ln(x))}\,dx (This is an easy u-substitution...if you know what to use for u!)

    6.) \int\frac{x^3}{x+\sqrt{x}}\,dx

    I think that's enough to keep you busy for a while, as most of these aren't easy.
    Last edited by redsoxfan325; April 29th 2009 at 11:05 PM. Reason: Added another integral
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