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Evaluate the integral by interpreting it in terms…

02:46

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Problem 34 Medium Difficulty

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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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

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Integration

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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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Area Under Curves - Overview

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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Video Transcript

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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Video Thumbnail

40:35

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