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Show that the area between the demand and supply functions is $\int_{0}^{x_{e}}(D(x)-S(x)) d x.$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 8

Applications of the Definite Integral

Integrals

Missouri State University

Oregon 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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All right, let's start by drawing a picture. So we've got x equals zero here. And we've got the function Why you calls E to the negative acts, which looks something like this. And as X goes off to infinity, uh Function approaches the horizontal lesson to it. Why is zero this value right here? When X equals zero, you plug that in and get one. So this Y value right here is one and this goes on forever. And ah let me drop smoother than that. It just keeps getting closer and closer to the X axis. So uh the integral From 0 to Infinity of. Eat the negative X squared dx. That is this area in blue. Well, a different way to interpret this area instead of in terms of X. Is to look at it in terms of why in this region in blue Uh x goes from 0 to infinity. But why ah Goes all over the interval from 0 to 1. And if we have Y equals E. To the negative X squared negative X squared. Then we have Ellen, Y equals negative X squared X squared equals negative L N Y X equals the square root of negative Ellen. Why? And since these values of X we're getting are in the first quadrant, they're all positive. We want the positive square root here. So this curve given by Y equals E to the negative X squared is also given by X equals these positive square root of negative Ellen. Why? So the other way to interpret this area is by looking at the integral in terms of why? Why? Like we said, it goes from 0 to 1. We are integrating with respect to Why in this area here is the area under the function the square root negative Ellen. Why? Which is the other integral given in the problem. Therefore, by considering the area of this region in the plane, we get that the two integral roles are equal

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