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Problem

Use the Midpoint Rule with the given value of $ n…

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Problem 9 Medium Difficulty

Use the Midpoint Rule with the given value of $ n $ to approximate the integral. Round the answer to four decimal places.

$ \displaystyle \int^8_0 \sin \sqrt{x}\, dx $, $ n = 4 $


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Frank Lin

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Amrita Bhasin

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

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Integrals

Integration

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Top Calculus 1 / AB Educators
Anna Marie Vagnozzi

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Integrals - Intro

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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40:35

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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Watch More Solved Questions in Chapter 5

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Problem 9
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Video Transcript

So if we want Thio, use the midpoint formula for this, Um, all we really need to do is first to find what is our Delta X? So remember, Delta X is supposed to be B minus a over and, um, so are being This case is a r a. Is zero and the number of victims we want US force that's going to be, too. And then we need to find our set of partition. So we could either just use, like, this formula here, um, to get them. Because remember, each X I is going to be x I minus one plus Delta X So we could go about doing it that way, or we can just create our first set. I normally like to create the single like the set of partitioning points, and then I would just find the midpoint of all of those and I can just plug them indirectly. That's the way I'm going to kind of do this. So, first, if we were department partition, this normally. So we start from zero a, and they're just gonna keep adding Delta acts like we had here until we hit B. So it would be 0246 and eight. And then we have this new partitioning set, which is going to be for the midpoint. So then we just average these. So the average of zero and two is one. The average of two and four is three or in 656 and 87 And so these are going to be the numbers that we would end up plugging into here. So let's go ahead and actually, um, set this up So our Delta X in this case, it's just going to be, um too. So was kind of rape this out. So be to times and then our f is what we have on the inside here, So it's going to be a sign of Route one plus sign of Route three plus sign of Route five plus sign of Route seven that we could just go ahead and plug these into a calculator on. Actually, heart went ahead and found these earlier, so this would be to and then 0.84147 plus 0.98703 and then plus 0.78675 and then, obviously for 70.47577 And then we just need to multiply all these together or add them all together than multiply. So that would give us eso approximately 6.18 e o. Actually, how many do we need to round 24 decimals? So that would be good enough. So this is what we would approximate. So this is approximately equal to the integral from zero a. A sign of the square root of X dx.

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Top Calculus 1 / AB Educators
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Video Thumbnail

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Video Thumbnail

40:35

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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