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(a) determine a definite integral that will determine the area of the region shown and (b) evaluate this integral.

(a) $\int_{-2}^{2}\left(4-x^{2}\right) d x$(b) $32 / 3$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Oregon State University

Harvey Mudd College

Baylor University

University of Nottingham

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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(a) determine a definite i…

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(a) Set up an integral for…

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Construct and evaluate def…

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So as you look at this problem, I think it's pretty evident that we are doing, um, pretty Not a great problem, but we're looking at Why equals four minus X squared, and they even give you the bounds that we care about. X equals negative two and X equals positive, too. And we want that area in here. Well, since we want an area under the curve, then the math equation that we want to set up is an integral from your lower bound X equals negative two to your upper bound X equals positive, too, of the function we care about, which is that four minus X squared D X. This is actually your first answer to part a. All right, So from here, we're actually ready to find the anti derivative, which is adding one to your expletive, or you can think of it as four Xs derivative is for I don't know if that helps anybody, but I like to add one to the X one divide by your new explanation. But you can't think this as the derivative of one third X cube is X squared and then do your bounce from negative to to to. So as I'm ready to plug in my bounds, it's going to be four times two. I figure you can do that in your head. Four times two is eight two. Cubed is eight and then minus. If I plug in negative two, I get negative. 84 times negative two negative two. Cubed is also a negative eight thirds. What's kind of nice about this? I don't know how many students would recognize it. Um, but each one of these are the same. If I change 8 to 24 3rd because 24 divided by three is eight and subtract those, I get 16 3rd. And over here it's going to be the same thing. Except that both the opposite has negative 16 3rd that when I subtract the negative turns into adding, so 16 plus 16 is 32 thirds should be your correct answer. So this is it for part B and part A was up here, so my two answers are circled in green

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