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Evaluate the given integral.$$\int_{0}^{1} \frac{2 x-1}{\left(2 x^{2}-2 x+4\right)^{1 / 2}} d x$$

$$0$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Baylor University

University of Nottingham

Idaho State University

Lectures

05:53

In mathematics, an indefinite integral is an integral whose integrand is not known in terms of elementary functions. An indefinite integral is usually encountered when integrating functions that are not elementary functions themselves.

40:35

In mathematics, integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. Given a function of a real variable (often called "the integrand"), an antiderivative is a function whose derivative is the given function. The area under a real-valued function of a real variable is the integral of the function, provided it is defined on a closed interval around a given point. It is a basic result of calculus that an antiderivative always exists, and is equal to the original function evaluated at the upper limit of integration.

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Our task is to find the integral from 0 to 1 of two. X minus one over two X squared minus two X plus four, uh, to the one half power t X. So if I didn't my new substitution correctly, I would want to use two x squared minus two x plus forward in for the U. And so the derivative of that would be four X minus two. That quantity would be D X and what I would have my students do First of all, factor out of two out of that and also divide that piece over. So if I factor out of two, I'm left for two X minus one. Do you is equal to DX. So now, as I go back to that integral, I'm just doing substitution real quick. First of all, I believe that two x minus one alone. Um, now, I would also instead of writing mhm over, I would take that piece to the negative one half power. Um and this piece is equally you because I said look that way and then replaced d x with that one over. There is a multiplication, by the way 22 x minus one D. U. And the reason why I do that is to show students that these two X minus one will cancel. And you just need to change the bounds if I plug in zero for all of these exes, because now it's in terms of you. I get four. Well, zero minus zero is for If I plug in one, I get to minus two plus four. Well, at this point, I don't really need to do anything, because if I start and stop at the same value, my area must be zero, and that's your final answer now. I would also go through the work just because I would double check that I have the right answer. But it is kind of a waste of time when you add 12 negative one half of one half. And if you multiply by the reciprocal, it's too well. That will cancel with this one half one half times to cancel and your balance from 4 to 4 and then when you plug in the bounds, it's really the square root of four, which is two minus the square root of for which is two and two minutes to is still zero. No matter what you do, you're going to get an answer of zero

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