Question

How to integrate (6-2x)^2 . (indefinite integral). I'm trying U sub but Idk what I'm doing wrong

Answer 1

what did you use for u ?

Answer 2

6-2x

Answer 3

can you show your sork?
what did you get for du?
what did you get for dx?

Answer 4

U = 6-2x
du = -2dx
\[(1/2) \int\limits_{}^{} -2U^2dx\]
@TuringTest

Answer 5

I'm not seeing where you get the -2 from...
u=6-2x
du=-2dx -> dx=-du/2

Answer 6

It's to get du right ? so I multiply by 1 pretty much 2/2

Answer 7

you have left your integral all mixed up with u integrated with respect to x; you can't do that. I think you are getting confused there.
you need to sub in the values:
u=6-2x
du=-2dx -> dx=-du/2
subbing in for u and dx gives\[\int u^2(-\frac{du}2)=-\frac12\int u^2du\]

Answer 8

Ohhh thank you turing!

Answer 9

Welcome!

Answer 10

We weren't taught how to solve for dx and plug back in but I like that way!

Answer 11

I sort of never saw them do it in class that way either, I never understood why :p

Answer 12

@TuringTest What answer are you getting ? I get \[\frac{-(6-2x)^{3}}{6} +c\] But my TI-89 is giving me\[\frac{4(x-3)^{3}}{3} +C\]

Answer 13

it is the same answer; they appear to have factored out a -2 in your TI-89

Answer 14

Ohh okay! Thanks! :)