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The graph of $ g $ consists of two straight lines and a semicircle. Use it to evaluate each integral.

(a) $ \displaystyle \int^2_0 g(x) \, dx $(b) $ \displaystyle \int^6_2 g(x) \, dx $(c) $ \displaystyle \int^7_0 g(x) \, dx $

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01:09

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Campbell University

Oregon State University

University of Nottingham

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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The graph of $g$ consists …

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The graph of g consists of…

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The graph of $f$ consists …

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Think About It The graph o…

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The graph of is shown. Eva…

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The graph of $ g $ is show…

given that we want to find the integral of these three, we could use geometric shapes. The first one is the area under the curve from 0 to 2, and that forms a triangle. It has a height of four on a base of two and finding the area Z base times height, which is eight but then divided by two. Therefore, that first section or in a girl has an area of four, and it is positive. The next section is, ah, semi circle below the X axis from 2 to 6. The radius there is too, and the area would be pie R squared then, because it's a half a circle, we could divide that by two. That would end up giving us four pi over two or two pi. However, like we said, because it's below the X axis of negative, so that would be negative. Two pi. The last one there is adding up the um area from 0 to 6, which are the combination of the previous two answers that we've gotten. But then, at the end, we have a triangle with an area of positive one half. So if we combine the four that's also the same thing as 8/2, plus the one half. So finally we could be right the entire integral from 0 to 7, as dine has and then minus two pi there.

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