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General Integral rules

  1. Apr 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve (e^(x^2))` , use this result to solve §5x*e^(x^2)dx

    2. Relevant equations

    * is multiply
    ` is derive
    1.§e^xdx=e^x+C
    2.§e^kxdx=1/k*e^kx+C
    3.§a^xdx=1/ina*a^x+C (a is a number)

    4.§k*f(x)dx=k*§f(x)dx

    3. The attempt at a solution

    (I'm alittle confused as to how you 'use' the previous one to solve the next.
    The previous example simply states this when solving: [in((x^2)+4)]`
    [in((x^2)+4)]`= (in u)`= (1/u)*u`= 1/((x^2)+4)*2x=(2x)/((x^2)+4)
    Then you're going to use this to solve §(3x)/((x^2)+4)dx
    gives: §(3x)/((x^2)+4)dx = 3/2§(2x)/((x^2)+4)dx= 3/2*in((x^2)+4)+C.
    It's not written, so I suppose they are using it by simply putting 3/2 on the outside, using rule 4)

    So my try went:

    (e^(x^2))`=(e`u)`*u`(um, using the corerule as it is named in norwegian) = e^(x^2)*2x
    , which was correct.
    Then you use it..
    §5x*e^(x^2)dx=(ADSFSDF PRESSING SHIFT FOR A LONG TIME CAN*T WRITE LOL F:: WINDOWS LOL)
     
  2. jcsd
  3. Apr 16, 2008 #2

    malawi_glenn

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    Why not learn LaTeX if you hate pressing shift? :-)

    what is:

    in

    integral?

    corerule is called "chain rule" in english (I am swedish)

    So you got [itex] \frac{d}{dx}(e^{x^2}) = 2xe^{x^2} [/itex], which is correct!

    Now what is your result if you use this forumula:
    [tex] \int F(x) dx = f(x) + C [/tex]

    if:

    [tex]\frac{dF}{dx} = f [/tex]

    ?
     
  4. Apr 16, 2008 #3

    HallsofIvy

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    Hopefully you know that if F'= f, then [itex]F= \int f dx + C[/itex]. Since
    [tex](e^{x^2})'= 2xe^{x^2}[/tex], what is
    [tex]\int 2x e^{x^2}dx[/tex]?

    and, from that, what is
    [tex]\int x e^{x^2}dx[/itex]?
     
  5. Apr 16, 2008 #4
    oh no, sorry the answer is: (5/2)*e^(x^2)+C... (at work, brb)
     
  6. Apr 16, 2008 #5
    Where do I find LaTeX?

    Yeah, I meant chain rule. in is natural logarithm, e^inp=p.

    My try went:
    §5x*e^(x^2)dx=
    5/2§2x*e^(x^2)dx (cause here is where you 'use it', right? =
    5/2*2*(1/(1+1))x^(1+1)*(e^u)`*u` (chain rule) =
    5/2*x^2*e^(x^2)*(1/(2+1))x^(2+1) =
    5/2x^2*e^(x^2)*(1/3)x^3+C.
     
  7. Apr 16, 2008 #6

    malawi_glenn

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  8. Apr 17, 2008 #7
    (e^u)`*u` I'm making a fault here? no chain rule in integral? e^(x^2)=e^(x^2)? , but why isn't chain rule being used? (like in §e^(3x-2)dx=(1/3)e^(3x-2)+C.) it looks like it should, heh. And how do the x^2 dissapear..
     
    Last edited: Apr 17, 2008
  9. Apr 17, 2008 #8
    no, it's rule 2? e^(x^2)= 1/x^2*e^(x^2), it would make the x^2 dissapear anyway.. and would make sense at the least.
     
  10. Apr 17, 2008 #9
    so.. :

    testing..
    [tex]\int 5x e^{x^2}dx = 5/2 \int 2x e^{x^2}dx = 5/2[/tex]\times2\times \frac{1}{1+1}x^1+1
     
    Last edited: Apr 17, 2008
  11. Apr 17, 2008 #10

    malawi_glenn

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    pace

    are you to solve

    [tex]\int 5 x e^{x^2}dx[/tex]

    ?

    Then use the derivative of:
    [tex] e^{x^2} [/tex]

    And use the:
    [tex] \int F(x) dx = f(x) + C [/tex]

    where:
    [tex]\frac{dF}{dx} = f [/tex]

    That's all you have to do
     
  12. Apr 17, 2008 #11
    yes, i am. Yeah, I figured.

    Like this right?:

    [tex]\int 5 x e^{x^2}dx[/tex] = 5/2§2xe^(x^2) = 5/2§ (and as in opposite of the above e^(x^2)*2x) = (e^(x^2)) , which gives simply 5/2(e^(x^2))+C.
     
    Last edited: Apr 17, 2008
  13. Apr 17, 2008 #12

    malawi_glenn

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    That looks good :-)
     
  14. Apr 18, 2008 #13
    Great :)
     
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