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integral from 1 to 2 of 1/(2x -1) dx

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Basically is x^2 - x and just substitute 1 and 2 into the function and add them.
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?
Yes, when you apply a \(u\)-substitution, you will change the bounds according to the choice of \(u\).

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Acess denied is that what you get ln(2x-1)/2 for your solution?
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.
thank you.
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)
why dont you sub u=2x-1 back in?
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...
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?
Yes, you only need to do the u-substitution and change the bounds once. Then the definite integral proceeds as normal
You're welcome! :)

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