- #1

Mattara

- 348

- 1

[tex](4x-3)^2[/tex]

This was quite straightforward. Or so I though.

It can easily be turned into

[tex]16x^2 - 24x + 9[/tex]

and then integrated to

[tex] \frac {16x^3}{3} - 12x^2 + 9x[/tex]

choosing C = 0

Then I started to think. Couldn't this be integrated using the chain rule backwards (with lack of better wording)?

[tex](4x-3)^2[/tex]

then becomes according to my integration

[tex] \frac {(4x-3)^3}{12}[/tex]

If I differentiate the above, I get the initial function. What I'm having a hard time understanding is that the different approaches yields different results. I plotted both of them in my graph calculator and they indeed give different results.

Any hints on what I did wrong is greatly appreciated. Thank you for your time. Have a nice day.