1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    Re: Integration

    Post back if you need more help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?