1. ## Integration!

The problem is:

∫(x^2)e^[-(x^3)/4]

The attempt at a solution:

Is it equals to (-4/3)e^[-(x^3)/4]???

I am really confuse here. Any help would be appreciated

2. Originally Posted by HeheZz
The problem is:

∫(x^2)e^[-(x^3)/4]

The attempt at a solution:

Is it equals to (-4/3)e^[-(x^3)/4]???

I am really confuse here. Any help would be appreciated
Yes, that is the correct answer (don't forget the "+C")! What is confusing you with this problem (or the substitution process in general)?

3. ∫(x^2)e^[-(x^3)/4]

Juz confusing that we didnt take (x^2) into account when integrating? Juz need to integrate e^[-(x^3)/4] so it becomes:

[x^2] * [-4/3(x^2)]e^[-(x^3)/4]

and so the (x^2) just cancel out itself to become (-4/3)e^[-(x^3)/4]

4. Originally Posted by HeheZz
∫(x^2)e^[-(x^3)/4]

Juz confusing that we didnt take (x^2) into account when integrating? Juz need to integrate e^[-(x^3)/4] so it becomes:

[x^2] * [-4/3(x^2)]e^[-(x^3)/4]

and so the (x^2) just cancel out itself to become (-4/3)e^[-(x^3)/4]
You actually do take the $\displaystyle x^2$ term into account!

When you make the substitution $\displaystyle u={\color{blue}-\tfrac{1}{4}x^3}$, we see that $\displaystyle \,du=-\tfrac{3}{4}{\color{red}x^2\,dx}$

At this stage, let's go back to the original integral:

$\displaystyle \int {\color{red}x^2}e^{-x^3/4}{\color{red}\,dx}$.

Observed that I highlighted the part that appeared in our $\displaystyle \,du$ term! So we see that $\displaystyle \,du=-\tfrac{3}{4}{\color{red}x^2\,dx}\implies -\tfrac{4}{3}\,du={\color{red}x^2\,dx}$.

Therefore, $\displaystyle \int {\color{red}x^2}e^{{\color{blue}-x^3/4}}{\color{red}\,dx}=\int e^u(-\tfrac{4}{3}\,du)=-\tfrac{4}{3}\int e^u\,du$.

Then integrating results in $\displaystyle -\tfrac{4}{3}e^u+C$ and back substitution gives us the desired result $\displaystyle \int x^2e^{-x^3/4}\,dx=-\tfrac{4}{3}e^{-x^3/4}+C$.

Does this clarify what's going on here?

5. Ahhh!! Thank You very much. I kind of forget these stuff ><. Thanks again for explaining it to me