Find integral of 3x^2 e^2x^3 for x

[staples82]

Homework Statement

Use substitution to find: (3x^2)*(e^2x^3)*(dx)

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

Homework Equations

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

 

[rock.freak667]

\int 3x^2 e^{2x^3}dx


u=2x^3 \rightarrow \frac{du}{dx}=6x^2 \rightarrow du=6x^2 dx

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

You understand up to here right?

Now you can write the integral as this

\int e^{2x^3} (3x^2 dx)

Now do you see what you can put at 2x^3 and (3x^2 dx) as?

 

[staples82]

I'm wondering why you did du/2=3x^2dx, also how did you get the math text?

 

[rock.freak667]

I'm wondering why you did du/2=3x^2dx

Well you see, when you did the substitution of u=3x^2, 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 \frac{du}{dx}=6x^2, if we multiply by the 'dx' we'll see that du=6x^2 dx. But in the integrand there is no '6x^2', only 3x^2. BUT 6x^2=2(3x^2), so that is why we divided by 2.

also how did you get the math text?

just use [*tex]What you want to represent mathematically[/tex*] without the *

 

[staples82]

Ok, so then that's why e^u/2 is the answer, I think I have to look at a few more problems, see if i can clear it up

 

[rock.freak667]

Ok, so then that's why e^u/2 is the answer, I think I have to look at a few more problems, see if i can clear it up

Post back if you need more help.