Integration!

May 2010
6
0
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
May 2008
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2,046
Chicago, IL
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)?
 
May 2010
6
0
∫(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
May 2008
2,844
2,046
Chicago, IL
∫(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?
 
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May 2010
6
0
Ahhh!! Thank You very much. I kind of forget these stuff ><. Thanks again for explaining it to me :D