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Homework Help: Find integral of 3x^2 e^2x^3 for x

  1. Jul 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Use substitution to find: (3x^2)*(e^2x^3)*(dx)

    I don't know how to do the integration sign...
    2. Relevant equations



    3. The attempt at a solution
    u=2x^3
    du=6x^2

    this is where I got confused, my teacher isn't that good and I'm trying to understand how to substitute...

    I know that its supposed to be e^u*du=e^u+c

    the answer was (e^2x^3)/2+C
     
  2. jcsd
  3. Jul 5, 2008 #2

    rock.freak667

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    Re: Integration

    [tex]\int 3x^2 e^{2x^3}dx[/tex]


    [itex]u=2x^3 \rightarrow \frac{du}{dx}=6x^2 \rightarrow du=6x^2 dx[/itex]

    which is the same as [itex]\frac{du}{2}=3x^2 dx[/itex]

    You understand up to here right?

    Now you can write the integral as this

    [tex]\int e^{2x^3} (3x^2 dx)[/tex]

    Now do you see what you can put at [itex]2x^3[/itex] and [itex](3x^2 dx)[/itex] as?
     
  4. Jul 5, 2008 #3
    Re: Integration

    I'm wondering why you did du/2=3x^2dx, also how did you get the math text?
     
  5. Jul 5, 2008 #4

    rock.freak667

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    Re: Integration


    Well you see, when you did the substitution of [itex]u=3x^2[/itex], a 'u' will appear to be integrated and you can't integrate 'u' with respect to x. So what we need to get something to replace 'dx' with and [itex]\frac{du}{dx}=6x^2[/itex], if we multiply by the 'dx' we'll see that [itex]du=6x^2 dx[/itex]. But in the integrand there is no '[itex]6x^2[/itex]', only [itex]3x^2[/itex]. BUT [itex]6x^2=2(3x^2)[/itex], so that is why we divided by 2.

    just use [*tex]What you want to represent mathematically[/tex*] without the *
     
  6. Jul 5, 2008 #5
    Re: Integration

    Ok, so then thats why [tex]e^u/2[/tex] is the answer, I think I have to look at a few more problems, see if i can clear it up
     
  7. Jul 5, 2008 #6

    rock.freak667

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    Re: Integration

    Post back if you need more help.
     
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