Question

If integral from 0,2 (2x^3-kx^2+2x)dx=12, then k must be?
I've tried working it out several times but I still don' get the right answer.

Answer 1

\[\int\limits_{}^{?}\]

Answer 2

\[ int_{0}{2} \]

Answer 3

\[\int\limits_{0}^{2} (2x^3-kx^2+2k) dx =12\]

Answer 4

\[\int\limits_{0}^{2} 2x^3 - kx^2 + 2k = 12 \]

Answer 5

= 2x^4/4 - kx^3/3 +2kx from 2 to 0 , = 12

Answer 6

2(2^4)/4 - k(2^3)/3 + 2(k)(2) - [ 0 + 0 + 0 ] = 12

Answer 7

did you get that so far?

Answer 8

yeah... instead of adding one to the power I was just finding the derivative..

Answer 9

well we are integrating, correct

Answer 10

give me a medal

Answer 11

yeah :) thanks... let me see if i get it now

Answer 12

Got it :)

Answer 13

:)

Answer 14

hey can u help me in another one real quick? sorry

Answer 15

ok

Answer 16

\[\int\limits_{}^{} (\sec^2x)(\tan ^2x)dx=\]

Answer 17

u = tan x

Answer 18

du = sec^2 x

Answer 19

so you do u sub?

Answer 20

yep

Answer 21

ok thanks

Answer 22

so i got integral u^2 du