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ShironlomBest ResponseYou've already chosen the best response.0
Basically is x^2  x and just substitute 1 and 2 into the function and add them.
 one year ago

GalaxySBest ResponseYou've already chosen the best response.0
my answer was ln(2x1)/2 then i substituted 1 and 3, dont you have to change the bounds, because you are using usubstitution?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
Yes, when you apply a \(u\)substitution, you will change the bounds according to the choice of \(u\).
 one year ago

GalaxySBest ResponseYou've already chosen the best response.0
Acess denied is that what you get ln(2x1)/2 for your solution?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
Well, when you change the bounds of the integral, you don't need to backsubstitute. The bounds are changed to accommodate the variable change to u. If you backsubstitute, you'd just use the original bounds for x. Your antiderivative of 1/(2x  1) being ln(2x  1)/2 is correct, though.
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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)
 one year ago

GalaxySBest ResponseYou've already chosen the best response.0
why dont you sub u=2x1 back in?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
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 ufunction. 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 backsubstitute...
 one year ago

GalaxySBest ResponseYou've already chosen the best response.0
I see. So when doing usubstitution, and we change the bounds, DO NOT CHANGE BACK! Leave the bounds the same and sub in for u. Yes?
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
Yes, you only need to do the usubstitution and change the bounds once. Then the definite integral proceeds as normal
 one year ago

AccessDeniedBest ResponseYou've already chosen the best response.1
You're welcome! :)
 one year ago
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