Integral/Antiderivative Problem

[HWilliams44]

Hi! So, I was working on the antiderivative problem and I'm not sure I did it correctly. Can anyone check my work? I've attached an image of my work below. Thanks!

 

[pka]

Hi! So, I was working on the antiderivative problem and I'm not sure I did it correctly. Can anyone check my work? I've attached an image of my work below.

I wish we could make it a criminal offence to use u-substitution.

$\displaystyle \displaystyle\int {{x^{ - 4}}{e^{{x^{ - 3}}}}dx = } \tfrac{{ - 1}}{3}{e^{{x^{ - 3}}}} + C$

Now differentiate the RHS to see how it works.

 

[HWilliams44]

Wait, how did you get that? I'm confused....

 

[pka]

Wait, how did you get that? I'm confused....

Did you do as I asked and differentiate $\displaystyle \tfrac{{ - 1}}{3}{e^{{x^{ - 3}}}}$?

If you can then you will see clearly how I got it.

 

[HWilliams44]

Oh, okay! Sorry; I didn't understand what RHS meant.... So, the answer would be (-1/3)e^x^(-3) times (1/5)x^(-5) + C, right?[FONT=MathJax_Math][/FONT][FONT=MathJax_Math][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Main][/FONT][FONT=MathJax_Math][/FONT]

 

[DrPhil]

Hi! So, I was working on the antiderivative problem and I'm not sure I did it correctly. Can anyone check my work? I've attached an image of my work below. Thanks!

Your biggest blunder in this one is that you tried to integrate a function containing both $\displaystyle x$ and $\displaystyle u$. Once you start to use a substitution, EVERY instance of $\displaystyle x$must be replaced by a function of $\displaystyle u$. If you can't do that in a manner that leads to a simpler integrand, then that substitution is not appropriate.

[I don't agree with pka that all u-substitutions are criminal. Rather, in studying his response, I would recommend letting $\displaystyle \displaystyle u = e^{x^{-3}}$. Then when you differentiate $\displaystyle u$, you hope to get cancellations.]