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Determine the area of the region between the given curves.$$f(x)=x^{2}-4, y=0, x=-2 \text { and } x=2$$

$32 / 3$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 8

Applications of the Definite Integral

Integrals

Campbell University

Harvey Mudd College

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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Determine the area of the …

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Find the area of the regio…

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Sketch the region enclosed…

I feel like they actually give you more information that you need in this problem because if you look at the graph of Y equals x squared minus four And you look at the equation y equals zero. It actually intersects at x equals negative two and X equals positive too. So you could find those bounds without them telling you. So it was just generous. They gave you those values. The other thing is you want to make sure that because we're finding area, we want the positive area. So what you actually want to do is the upper function of Y called zero minus the lower function of X squared minus four. So what we're actually going to do is the integral from negative to mhm positive for minus X squared dx. And there's a lot of different ways to evaluate it. I'm just going to show you one way where you add one to your exponents, multiply by the reciprocal of your new exponents. Mhm. And then you're ready to plug in your bounds. So starting with plugging in your upper bound of 24 times two is eight And then two cubed is 8/3 And then minus plugging in negative too, so you negative eight. Um and then be careful does not have to, the third power is still negative eight, so it's going to change the sign this. Yeah. Um So if you start to distribute this which is something I should have done in, you know, different problems. Um I think what I really could do is I could just find one of these values and then double it. Um Trying to think of a better way of explaining it but anyway what we could do as well as get the same denominator, so 24/3 is the same as uh eight. And no matter how you do this problem, Whether you do 24 -8 is 16 and double that to give you 32/3, or just continue to add All of the numerator and denominator, first of all, the numerator and the denominator stays the same, so the correct answer is 32/3.

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