anonymous
  • anonymous
\[\int\limits_{}^{}2x(2x-1)^{1/2}dx\] I know I need to solve by substitution so u= 2x -1 du/dx = 2 how do I solve for x?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
solve \[u=2x-1\] for x as in elementary algebra
anonymous
  • anonymous
\[x=\frac{u+1}{2}\]
anonymous
  • anonymous
but why maybe I'm brain dead today Im just not getting it :L

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TuringTest
  • TuringTest
\[u=2x-1\implies 2x=u+1\]
amistre64
  • amistre64
no need to solve for "x"; just solve for "2x"
anonymous
  • anonymous
good point!
anonymous
  • anonymous
oh lol wow I'm dumb ha
anonymous
  • anonymous
Or put x=2x and solve for z.
amistre64
  • amistre64
not that solveing for x is bad :)
anonymous
  • anonymous
Thanks
anonymous
  • anonymous
so u= 2x -1 du/dx = 2 u = 2x - 1 (u + 1)/2 = x \[\int\limits_{}^{} u^{1/2}(1 + u)/2\] = 1/2 (u^(1/2) + u^(3/2))du = 1/2(2u^(3/2)/3 + 2u^(5/2)/5) + c
anonymous
  • anonymous
not subing but thanks for the help :)
TuringTest
  • TuringTest
that is a sub now just sub back in 2x-1 for u and you have the final answer

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