anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits_{}^{?}\]
anonymous
  • anonymous
\[ int_{0}{2} \]
anonymous
  • anonymous
\[\int\limits_{0}^{2} (2x^3-kx^2+2k) dx =12\]

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More answers

anonymous
  • anonymous
\[\int\limits_{0}^{2} 2x^3 - kx^2 + 2k = 12 \]
anonymous
  • anonymous
= 2x^4/4 - kx^3/3 +2kx from 2 to 0 , = 12
anonymous
  • anonymous
2(2^4)/4 - k(2^3)/3 + 2(k)(2) - [ 0 + 0 + 0 ] = 12
anonymous
  • anonymous
did you get that so far?
anonymous
  • anonymous
yeah... instead of adding one to the power I was just finding the derivative..
anonymous
  • anonymous
well we are integrating, correct
anonymous
  • anonymous
give me a medal
anonymous
  • anonymous
yeah :) thanks... let me see if i get it now
anonymous
  • anonymous
Got it :)
anonymous
  • anonymous
:)
anonymous
  • anonymous
hey can u help me in another one real quick? sorry
anonymous
  • anonymous
ok
anonymous
  • anonymous
\[\int\limits_{}^{} (\sec^2x)(\tan ^2x)dx=\]
anonymous
  • anonymous
u = tan x
anonymous
  • anonymous
du = sec^2 x
anonymous
  • anonymous
so you do u sub?
anonymous
  • anonymous
yep
anonymous
  • anonymous
ok thanks
anonymous
  • anonymous
so i got integral u^2 du

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