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Determine the area of the indicated region.Region bounded between $f(x)=2 x, g(x)=-2 x+8$ and the $x$ -axis.

8

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

Missouri State University

Harvey Mudd College

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.

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 bounded …

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so there's actually a couple of ways of doing this problem. But what most people would probably do is first of all, graph these two equations. So we have one function that I think it's f is defined as two X and then another function change colors is, uh, starts up at eight. It was down like this, uh, so we'll call that G is negative two x plus e. So if you examine the graph, then you'll be able to figure out that they intersect each other at the ordered pair to four. And if you don't believe you set the two functions equal to each other, F and G equals, so you can add to exit the left side. To get four X equals eight, divided by two next equals two, and then you can figure out the Y value by plugging into either of the functions. But the faster one is two times two equals four. The reason why I do that is now we can set up an integral and find the area under the curve. By the way, it's it's mashed with the X axis, so that's why I'm doing this area of the integral from 0 to 2 of the black curtain, the FX curve that I have. So that'll be two x dx and then add to it the integral from 22 X equals four. I guess I don't really show you the word. If we set that equal to zero and add two X to the right sign divided by two, you would see that that has a zero X intercepted exit was four of that function. Uh, if you look at the graph, the answer is kind of obviously like area of a triangle, one half base times height or the basis for the hiatus. For, uh, the answer should be two Well, half of four is two times four is eight. Anyway, let's actually work it out. The integral of two X is X squared from 0 to 2. Now this interval will be negative X squared plus eight x, and that's from 2 to 4. So, as I plug in, my upper bound two squared is four minus my lower bound zero square to zero and then add to it plugging in eight. So negative 64 plus eight times eight or 64 minus negative. Four plus uh, 16 and I don't know, getting four minus 12. I must have done something wrong. I get the right answer, but I don't know what I did, because I should get it positively. I guess the easy way out of this is to say that we should get a positive answer. So the answer is the absolute value of that. I really don't know what I didn't right. 0 to 2. Oh, that's what I did here on this upper Bound is supposed to be four. So I mean, negative 16 plus 32. This is very wrong. Sorry about that. So we have four plus 16 minus 12 scenario. That gives me eight 20 minutes sort of equity. Sorry about that.

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