# Math Help - difficult antiderivative

1. I understood of all that except, how did you get 1/m^3? I understand where you got 1/m from, but how did it change to m^3? Also, at the end of the problem you said to multiply it by 1/m^3. Wouldn't you either leave it how it is or multiply by m^3 on both sides?

Can someone please explain this to me?

Thank you very much

Regards

2. Originally Posted by chocolatelover
I understood of all that except, how did you get 1/m^3? I understand where you got 1/m from, but how did it change to m^3? Also, at the end of the problem you said to multiply it by 1/m^3. Wouldn't you either leave it how it is or multiply by m^3 on both sides?

Thank you very much

Regards
if $u = mx$ then $du = m~dx \implies \frac {1}{m}du = dx$ ........there's your factor of $\frac {1}{m}$

Also, we have to change the $x^2$
if $u = mx$ then $x = \frac {1}{m}u \implies x^2 = \frac {1}{m^2}u^2$ .........there's your factor of $\frac {1}{m^2}$

multiplying these factors together, you get $\frac {1}{m^3}$

when Krizalid was doing the integral, he left off this factor and so he said you should remember to multiply through by it, as his final equation suggested

3. Thank you very much

So this is the final answer, right?

;

4. Originally Posted by chocolatelover
Thank you very much

So this is the final answer, right?

;
yes. but Krizalid suggested to multiply through by the $\frac {1}{m^3}$, that way, you got the answer in the form he had it in on the first page. this is not absolutely necessary though. i think it's fine as it is, it's up to you and your preferences

Page 2 of 2 First 12