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Determine the area of the region between the given curves.$$f(x)=-2 x^{2}-1, y=0, x=-1 \text { and } x=2$$

9

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 8

Applications of the Definite Integral

Integrals

Harvey Mudd College

University of Michigan - Ann Arbor

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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Determine the area of the …

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Find the area of the regio…

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All right. Um because we are, I just want to show you to grab a quick because if we just did this graph white was negative two X squared minus one. I don't really care about how steep that parabola is. If we just found the integral of that we would get a negative answer. So as you look at the interval from x equals negative one. So they have to tell you the bounds two x equals two. Uh what we're really doing is taking the upper function of Y0 and subtracting off the lower functions negative X squared minus one The X. Which is why it's the same thing as doing the integral from negative 1-2 of positive two X squared plus one dx. But no matter how you think of the problem, the integral is going to be the same. You know, calculus where you add one to the expert and multiply by the reciprocal of that exponents and then think about the derivative of what we just wrote down. You see equal that same thing with your bounds from they have wanted to. Well now you can start plugging in your upper mounts well to Cuba's eight times that to be 16/3. Us too. And then you have to subtract off plugging in your lower bounds and negative one to the third power still negative and plugging in naval on him for that. X. Now if I were doing this problem I would change my whole numbers to have a denominator three. So six thirds of the same thing as two and three thirds is the same thing as one. Um And just make sure that you simplify correctly, you know that your denominator stays the same native to minus three being negative five. So then as you add those pieces together because subtracting a negative the same thing as adding a positive, we get 27/3 which would even nine, which is the correct answer.

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