Integration by Substitution - help

[Dinoduck94]

I'm trying to understand 'Integration by Substitution' and there's one bit that I just can't wrap my head around.

If I use the following question as an example:

$$\int x ~ e^{-x^2} dx$$
STEP 1
$$u = -x^2$$$$\dfrac{du}{dy} = -2x$$
STEP 2
$$\dfrac{1}{2} \int e^u +C$$
STEP 3
$$\dfrac{1}{2} * -e^{-x^2} +C$$
Answer:
$$\int x ~ e^{-x^2} dx = - \dfrac{e^{-x^2}}{2} + C$$

The material that I have, explains everything except where the $$\dfrac{1}{2}$$ comes from, in Step 2 - and where the other 'x' term went (was it differentiated to 1?); and I'm just getting more confused the more I search online.

Can someone help with my understanding, please?

Thanks

 

[skeeter]

[MATH]u=-x^2[/MATH]
[MATH]du = -2x \, dx[/MATH]
[MATH]\int x \cdot e^{-x^2} \, dx = \int e^{-x^2} \cdot x \, dx = {\color{red}-\dfrac{1}{2}} \int e^{-x^2} \cdot {\color{red}-2} x \, dx = -\dfrac{1}{2} \int e^u \, du[/MATH]
see why the [MATH]\color{red}-\dfrac{1}{2}[/MATH] is necessary?

 

[Dinoduck94]

See why the [MATH]\color{red}-\dfrac{1}{2}[/MATH] is necessary?

Yes, I think I do, thank you.
Multiplying the value we have for [MATH]du[/MATH] by [MATH]- \dfrac{1}{2} [/MATH] makes it match the [MATH]x[/MATH] term in the function.

I still don't understand where that [MATH]x[/MATH] goes from that point? If you integrate [MATH]x[/MATH] you get [MATH] \dfrac{x^2}{2} [/MATH], but we don't see that here. I assume that I am misunderstanding; Is it differentiated instead, into [MATH]1[/MATH]?

Thank you for your help.

 

[skeeter]

[MATH]u = -x^2 \implies \dfrac{du}{dx} = -2x \implies du = -2x \, dx[/MATH]
[MATH]-\dfrac{1}{2} \int e^{-x^2} \cdot \left({\color{blue}-2x \, dx} \right)[/MATH]
[MATH]-\dfrac{1}{2} \int e^u \, \color{blue} du [/MATH]
the “x” is contained in the “du”