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Use the properties about odd and even functions to evaluate the given integral.$$\left.\int_{-3}^{0} 2 x+3^{-} d x \text { (Hint let } u=2 x+3 .\right)$$

$$9 / 2$$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Campbell University

Baylor University

University of Nottingham

Boston College

Lectures

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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.

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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Use the properties about o…

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Evaluate the integral usin…

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Write the integral as the …

Okay, So there's a overall theme in this problem that they want you to do. Is this being an even function? Um, so being even means a symmetric with the Y Axis graph looks something like this. I don't know exactly where Negative three and three are. As far as this goes, it doesn't really matter. I'll just to say this area to the left is equal to the area to the right. So the premise is that you can find again. I don't know where three or negative three are because it doesn't matter is just to do the integral from 0 to 3 and then double um because the area on the left side is equal to the area on the right side, eso is a very rough sketch as to what we're doing, What the theme is in this problem, and you still need to do the problem where you do. The integral of four is four X and then the integral of this anti derivatives one third X cube. The reason why they ask you to do that is because plugging in zero is a lot easier than plugging in negative three. So plug in three first you get 12. Well, that's nice. Three cubed. This 27 divide by three is nine and then plug into zero. Here, you get zero plugging zero. Here, you get zero. You're getting at zero there. Eso The theme in this problem is now it's easier to do. 12 minus nine is three. And just double that answer to get a final answer of six instead of plugging in negative three earlier here and here, which you could have done. Andi, I really would have been that bad. It would have been just negative. 12. Um, yeah, we've been negative. 12 minus nine, minus negative line. So you would have had three minus negative threes. What you would have had you would still get six, so it doesn't really matter.

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