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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.$$f(x)=x^{2},[0,1]$$

(a) $1 / 3$(b) $\frac{1}{\sqrt{3}}$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 7

Substitution and Properties of Definite Integrals

Integrals

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.

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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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Okay, So an average value problem just follows this pattern of one over you subtract your upper bound, minus your lower bound, and then you do the integral from the lower bound to the upper bound, and they give you in the direction that it's 0 to 1 of the function D x, Um, so this is really set up in a nice way where you can just go straight ahead like one minus zero. Still, 11 divided by one is just one multiplied by one. So I can change the value, but and then the anti derivative. You add one to your exponents and you multiply by the reciprocal of that experiment, and that goes from 0 to 1. Um, so as you solve this, you plug one in for X or one cube distill one times one third is one third and then minus zero. Cubed is still zero and zero times. Anything is still zero. So the average value is one third of this problem. And that answer makes sense, Um, in the context of the question. But what we want to do now. So that's your answer to part A. What we now want to do is take the original function, which is X squared, and set it equal to one third. Well, to solve a square problem, you have to square root. Uh, and I would just square root. Each piece is just kind of about a habit. And you do get to answers, plus or minus one over root three, but negative Route three. You know, negative one over. Root three is not in that interval that they gave you from 0 to 1. So we ignore the negative answer here, and we only want the positive answer here. Actually was positive one of the three. So I'll circle that as you're correct. Answer to be. Or you might have a teacher that wants you to rationalize the denominator. So Route 3/3 is another good answer. Uh, and that's an X value that is bigger than zero smaller than one. So that makes perfect sense.

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