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

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Homework Help Overview

The problem involves finding the integral of the function 3x²e^(2x³) with respect to x, using substitution techniques. The discussion centers around the integration process and the application of substitution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the substitution method, specifically using u = 2x³ and the corresponding differential du. There are questions about the integration process and the manipulation of differentials, particularly why du/2 = 3x²dx is used. Some participants express confusion about the integration sign and the substitution steps.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the substitution method. Some guidance has been provided regarding the relationship between du and dx, but there is no explicit consensus on the final outcome. Participants are encouraged to seek further clarification and practice with similar problems.

Contextual Notes

There is mention of confusion stemming from the teaching style of the original poster's instructor, which may be affecting their understanding of the substitution method. Participants are also discussing the format for representing mathematical expressions in the forum.

staples82
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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
 
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[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?
 


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


staples82 said:
I'm wondering why you did du/2=3x^2dx


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.

staples82 said:
also how did you get the math text?

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


Ok, so then that's 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
 


staples82 said:
Ok, so then that's 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

Post back if you need more help.
 

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