# Integration!

#### 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 #### Chris L T521

MHF Hall of Fame
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)?

#### 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]

#### Chris L T521

MHF Hall of Fame
∫(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?

• HeheZz

#### HeheZz

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