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(a) determine a definite integral that will determine the area of the region shown and (b) evaluate this integral.

(a) $\int_{1}^{2} \frac{4}{x^{1 / 2}} d x$(b) $8(\sqrt{2}-1)$

Calculus 1 / AB

Chapter 5

Integration and its Applications

Section 6

The Definite Integral

Integrals

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University of Michigan - Ann Arbor

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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 a definite i…

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so you don't really need the visual to be able to do this graph, but it's nice they give it to you that the function is four over X to the one half and they give you the X values are X equals wand and X equals two, and this is the area that we want. So to put this into a math equation, it's the same as saying the integral from 1 to 2, no infinite number of rectangles. And I would rewrite that function with a negative because if we're in the denominator, we have a negative exponents D s. And it makes doing the integral that much easier because all you have to do is add one to your extra negative. One half plus one is positive one half and multiply by the reciprocal as you look at the that four in front times are reciprocal. So four times two gives me eight own 1 to 2, and now we're ready to just plug in the bounds and remind yourself that the to the one half power means the square root looking at eight times the square root of two minus eight. Well, the square to one is still 11 to any power. Still, Um, and you can leave your answer like that. Um, actually, I'll just get rid of that. And this is okay. Or what you could have done is look at this previous problem and factor out that h uh, eight. So you have eight times a quantity of route two minus one. Uh, hopefully that made sense of everything I said. I guess I should reiterate where I put all of these values lower bound here, upper bound here and then the function rewritten with rational exponents, and then work it out. And either one of these circle and green are correct answers.

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