anonymous
  • anonymous
integral from 1 to 2 of 1/(2x -1) dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Basically is x^2 - x and just substitute 1 and 2 into the function and add them.
anonymous
  • anonymous
my answer was ln(2x-1)/2 then i substituted 1 and 3, dont you have to change the bounds, because you are using u-substitution?
AccessDenied
  • AccessDenied
Yes, when you apply a \(u\)-substitution, you will change the bounds according to the choice of \(u\).

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anonymous
  • anonymous
Acess denied is that what you get ln(2x-1)/2 for your solution?
AccessDenied
  • AccessDenied
Well, when you change the bounds of the integral, you don't need to back-substitute. The bounds are changed to accommodate the variable change to u. If you back-substitute, you'd just use the original bounds for x. Your antiderivative of 1/(2x - 1) being ln(2x - 1)/2 is correct, though.
anonymous
  • anonymous
thank you.
AccessDenied
  • AccessDenied
so... integral from 1 to 2 of 1/(2x - 1) dx u = 2x - 1 du = 2 dx --> 1/2 du = dx = integral from u(1) to u(2) of 1/u * 1/2 du = 1/2 integral from 1 to 3 of 1/u du = 1/2 * ln(u) from 1 to 3 ### here we use the bounds 1 and 3 for u. = 1/2 * ln(3) - 1/2* ln(1) = 1/2 * ln(3)
anonymous
  • anonymous
why dont you sub u=2x-1 back in?
AccessDenied
  • AccessDenied
Well, if we substitute it back in, we'd basically be changing the variable again. We need to change the bounds around to do that. Notice, however, that our function x of u: u = 2x - 1 u + 1 = 2x x = (u + 1)/2 Is the inverse of the u-function. We change these bounds: x(1) = (1 + 1)/2 = 1 x(3) = (3 + 1)/2 = 2 You would just get your original boundaries back. There would be no point in that boundary change if we just back-substitute...
anonymous
  • anonymous
I see. So when doing u-substitution, and we change the bounds, DO NOT CHANGE BACK! Leave the bounds the same and sub in for u. Yes?
AccessDenied
  • AccessDenied
Yes, you only need to do the u-substitution and change the bounds once. Then the definite integral proceeds as normal
anonymous
  • anonymous
thanks.
AccessDenied
  • AccessDenied
You're welcome! :)

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