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Evaluate the integral.

$ \displaystyle \int^{2}_{1} \biggl( \frac{x}{2} - \frac{2}{x}\biggr) \,dx $

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00:55

Frank Lin

00:44

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Missouri State University

Harvey Mudd College

University of Nottingham

Boston College

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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were asked to the integral from 1 to 2 of X over two, but divided by two is the same thing is multiplied by one half. So that's what I would leave that as -2 over X. Whatever works for you. Um But I would do this because then when I'm looking at the anti derivative and I add one to the exponents. You know, I'm looking at The exponents now to the second power and then multiply by the reciprocal will half times half is 1/4. Now looking at the next one, I just haven't memorized and a lot of people know that when it's one over X, the anti directive is natural log but you can work it out if you tried your power rule and add one to the expo you'd be divided by zero. So that's an issue. And we're doing our bounds from 1-2. Well now we can plug in our upper bounds, Well two squared is four and 1 4th of four is 1 -2. Natural log of two and then minus your lower bound, which is one plugged in for each of these. Well one squared is once we're looking at 1/4 -2 Natural Log of one. And I expect my sister know that natural log of one is zero, so two times zero is just zero, so that's gone. Um and then from here you just might want to reduce like 1 -1 4th is three. Force when you combine them, combining this term with this term and just leave this alone and you're fine. There's nothing left to do. This is a perfect answer.

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