General Integral rules

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  • #1
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Homework Statement



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

Homework 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

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)
 

Answers and Replies

  • #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]

?
 
  • #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]?
 
  • #4
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oh no, sorry the answer is: (5/2)*e^(x^2)+C... (at work, brb)
 
  • #5
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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
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(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:
  • #8
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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.
 
  • #9
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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:
  • #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
 
  • #11
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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:
  • #12
malawi_glenn
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That looks good :-)
 
  • #13
233
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Great :)
 

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