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Problem

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

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Problem 10 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^1_0 \sqrt{x^3 + 1}\, dx $, $ n = 5 $


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Frank Lin

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Amrita Bhasin

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

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Integrals

Integration

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Video Transcript

Yeah, yeah. So five dot to problem number one. So we're asked to use Riemann sums to estimate this? Uh integral. So our estimate is going to be using five rectangles. So in the interval From 0 to 1, If you divide that into five, everyone is going to be .2. So point to 0.4 0.6 .8. And then one. Now we're also asked to use mid points. So what are the midpoint is going to be? The mid points are going to be .1 0.3 0.5 0.7 and 0.9. So if I were to estimate this with rectangles, I'm going to use the function value at the midpoint. The width of each rectangle is still going to be .2. So my estimate Ah for this integral. So the integral from 0 to 1 Cube root, x cubed plus one dx is going to be estimated by the area, which is the sum of these five rectangles. So what is the width of each rectangle? So the width of each rectangle Is going to be what one? So I got the width is one divided by five. So the width of each rectangle is 1/5. Yeah And then the height of each rectangle is going to be evaluating at the midpoint. So you're going to get .1 Q plus one. And let me just do it sort of this way plus the square root of 0.3 cubed plus one Plus the square root of .5 cubed Plus one plus the square root of .7 cubed plus one Plus the square root of .9 cubed plus one. So all of that turns out to be our estimate. Now there are easier ways of doing this with a calculator than doing all of that out. Let's try that. So let's just create my function is the square root. So the square root no. Of X cubed plus one. And I'm going to store that my function F of X. And so now what I need to do is just create the sequence of all of those mid points. Or I could just on it just sum them all up. I could just say, hey, the area is going to be, What is it? 1/5, so 1 5th Times F of .1. Yeah. Plus F point too. Excuse me, F A 0.1. And then fo 0.3 Plus F 5.5. Yeah, half of mhm. Seven F 0.9. So 1/5 Uh f of one F 2 F three f of five. F seven F. Of nine. So 12345 is the case there. And so what you see there is 1.109 6 7. So this area estimate Is about 1.109. Let me make sure right, that right? 1.109 67 Okay, so that is the estimate of this integral.

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Heather Zimmers

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Video Thumbnail

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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.

Video Thumbnail

40:35

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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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