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anonymous
 3 years ago
integral from 1 to 2 of 1/(2x 1) dx
anonymous
 3 years ago
integral from 1 to 2 of 1/(2x 1) dx

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Basically is x^2  x and just substitute 1 and 2 into the function and add them.

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

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

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

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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.

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1so... 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
 3 years ago
Best ResponseYou've already chosen the best response.0why dont you sub u=2x1 back in?

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1Well, 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...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?

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

AccessDenied
 3 years ago
Best ResponseYou've already chosen the best response.1You're welcome! :)
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