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(a) determine the average value of the given function over the given interval, and (b) find the $x$ -coordinate at which the function assumes its average value.$$f(x)=m x+d,[a, b]$$

(a) $\frac{m(a+b)+2 d}{2}$(b) $1 / 2(a+b)$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

Oregon State University

Harvey Mudd College

Baylor University

University of Nottingham

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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(a) determine the average …

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Determine the average valu…

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Suppose that the average v…

Okay, so if we're doing the average value, they don't even give you the bounce. It doesn't really matter because it's a linear function. What's new is that you have to subtract your balance from each other or dividing by that value of this equation m x plus D DX. Um, so we could probably just jump right into this. We have one over B minus A. So that's part of your average value. The anti derivative of this would be adding one to your exponent and divide by your new exponents, plus d x. And that's from a to B. Um, so what you need to do is plug in your balance, so leave this one over B minus a alone. But if I plug in those bounds, I would have, um m times B squared over two plus d times B minus. Now it's the quantity. So I'm gonna write to minuses. Mm. Times a squared over two minus D times A. I believe what the what a better answer would be would be to, um, I guess, factor. Because each one of these shares, um, similar things. So if you were to group pieces together, um, you know you can factor out a B. I should have done this to begin with. So we have MB over to plus D minus. You can factor out an a here, which is m a over to. Yeah, um, I'll make that a plus, indeed. But even that's not exactly the same. I guess what I was trying to do is factor out of B minus A by not seeing how to do that. Yes. I feel like I did this stuff wrong. What I should have done is rewrote this and switch these two around. So I'm looking at one over B minus a. And if I group those two together, I get factor out in em over to it's a B squared, minus a squared. And then the next one, I can factor out a G. I could then have a B minus a mhm. Uh, but even then, what I would have to do is a difference of squares right here. This is kind of complicated. I don't like this problem a minus a. So we could have a common term there to factor out and cancel with this peace. So if I'm doing this correct, correct what I end up with is m times B plus a all over to plus D I'm gonna stop right there. Uh huh. Because now what you need to do is you have to set this equation equal to the original problem. MX pleasant t and sold for ex Excuse me. Um, now I kind of have an issue here. Um, I guess I could subtract d over. That's fine. So what I would want to do is multiply that to over. So I have m B plus m A. After I distribute equals two X. And if I want to solve for X, I need to divide everything by two. Um, well, all of these events were cancelled, so that's kind of nice. So we're left with B plus a all over to. Here we go. Um, so it took me a while, but I did get to there, so hopefully you enjoyed watching this video A B

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