Help please =) (1 Viewer)

[LuthienAdrianna]

1) Find the indefinite integral of (x^2)(e(x^3 +1))dx using the substitution u=x^3 +1

...

 

[Trev]

I'm assuming you meant:
int. x²e^x³+1 dx
u=x³+1
du=3x²dx
du/3=x²dx

1/3 * int. e^u du
1/3 * e^u + C
e^x³+1/3 + C

 

[LuthienAdrianna]

I'm assuming you meant:
int. x²e^x³+1 dx
u=x³+1
du=3x²dx
du/3=x²dx

1/3 * int. e^u du
1/3 * e^u + C
e^x³+1/3 + C

I'm lost in all the bold parts... which is practically all of it. ><

 

[KFunk]

With the substitution method you want to change the variable that you're integrating 'with respect to'. i.e. you want du - not dx.

If you let u = x³ + 1

and differentiate it ---> du/dx = 3x²

giving du = 3x² dx ('dx' isn't strictly an algebraic term but you can still move it over like that)

you then want to sub it into the integral.

&int; x²e^x³+1 dx

= &int;e^ux²dx (since u = x³+1)

You then want a 3 in there to make '3x²dx' so you multiply by 1 = 3/3 to give you

= (1/3)&int;e^u(3x²dx)

= (1/3)&int; e^udu ...then integrating as normal...

= (1/3)e^u + c

= (1/3)e^x³+1 + c ...subbing back in the x terms

 

[icycloud]

I'm lost in all the bold parts... which is practically all of it. ><

Edit: Oops beaten by KFunk.

The derivative of u with respect to x is du/dx,
Now, u = x^3 + 1
So, du/dx = 3x^2
Thus, du = 3x^2 dx {you can treat du/dx as a fraction}

Does that help?

 

[LuthienAdrianna]

... I know why I don't really get this... IT'S BECAUSE WE HAVEN'T EVEN DONE IT IN CLASS YET! *doh* My teacher reminded me that we actually left that whole integration using substitution part out and will do it later. I still want to get it, though, so:

You then want a 3 in there to make '3x2dx' so you multiply by 1 = 3/3 to give you

= (1/3)∫eu(3x2dx)

I don't understand that part.

:S