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Evaluate the integral.

$ \displaystyle \int^{2}_{1} \biggl( \frac{1}{x^2} - \frac{4}{x^3} \biggr) \,dx $

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01:08

Frank Lin

00:28

Amrita Bhasin

Calculus 1 / AB

Chapter 5

Integrals

Section 4

Indefinite Integrals and the Net Change Theorem

Integration

Missouri State University

Campbell University

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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We're looking at the integral from 1-2 of this function one over x squared Uh -4 over x cubed dx. And before doing the anti direct of what I would do is rewrite this is X to the name of second power minus four X. To the negative third power dx. And now we can follow our rules where we add one to the exponents. Now we multiply by the reciprocal of that new experiment. Just remember that when you add to a negative uh A lot of students make that mistake where adding 12 negative three, they write negative four, but negative three plus one is negative two. And when you divide negative forward by negative two, you get a positive too From 1- two. Um And it might be worth your while to remember that this is equal to negative one over X plus two over X squared From 1- two. But sometimes I don't even do that because I don't know, it just makes sense in my head to leave this alone. Uh When you plug into and for X. Squared, you get to force or one half. Um So that's actually zero. But then when you have to plug in, your lower balance would be negative one plus two. Um so you get one there but remember it's a negation of that value, so it should be negative one as your final answer. So let's circle that and move on. Okay.

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