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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

Campbell University

Baylor University

Idaho State University

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.

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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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Okay. All right. Uh What I like about this problem is it shifted up one and we have a per apple. I don't really care how quickly it goes up. But we have this equation why was two X squared plus one? And we only care about the bounds from Mexico's negative as they give us that To x equals two. So to find the area under the curve. Uh And that's why I've grafted to give you that visual. We just have to do the integral from negative 1-2 of that functioning gave us. Yes. So as we go through the process of adding one to the exponent, multiply by the reciprocal of your new exponents. From negative 1-2, we can now plug in our upper bound so plugging into and from both those values two cubed is eight Times two would be 16/3 plus two. And then you have to subtract off plugging in your lower bound well negative one to the third power stays negative so it's gonna be negative two thirds minus one. So what I would probably do is change all of my whole numbers to be a denominator of three. So 6:30 is the same thing as to The same thing with this is 3/3 is the same thing as one. So now I can start doing the arithmetic. 16 plus six would be 20 to thirties -2 -3 is negative 5/3. And when you subtract a negative it's going to turn into adding the positive. So we're looking at a correct answer was right over here, 27/3 which is equal to nine the correct answer.

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