Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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 \(u\).

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
thanks.
You're welcome! :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question