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Problem

The graph of a function $ f $ is given. Estimate …

06:15

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Problem 4 Easy Difficulty

(a) Find the Riemann sum for $ f(x) = 1/x $, $ 1 \le x \le 2 $, with four terms, taking the sample points to be right endpoints. (Give your answer correct to six decimal places.) Explain what the Riemann sum represents with the aid of a sketch.

(b) Repeat part (a) with midpoints as the sample points.


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Mutahar Mehkri
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Frank Lin

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 2

The Definite Integral

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Video Transcript

All right. We've got a question here where we're asked to find the renewed some of the f of X Equal to 1 over X When the X is anywhere from 1, two. Alright. With four terms, we're told that our and taking a sample points and to be the right endpoints give your And to correct to successful places Explain the Riemann sum represents with the aid of a sketch and repeat Part A with mid points as the sample points. Okay? So first of all, We'll start off by Running out our series agreements um equal to Series and -1, which we know it would be Actually 4 terms. So just before And then you have a I equals one. We're doing right endpoints part a Functional I multiplied by the change. Yeah. All right. So the change of X Can be calculated by taking a or B minus a over N. This situation RB is to our a one, You have 1/4 And then your endpoint would just be um It will just be 1 was Your change in X. Which would then make that 54 And then you just keep going and adding a 1/4 each time. Finally exit 4 would just be a once again we're We're taking that one plus 1/4 here because we do have it we're asked to find on the right end points. All right. So our agreement some at the right end point He's going to be your series or equals one function of X. I multiplied by the change in X. Which is the same thing as one or four. That'll be one of the 4 multiplied by Function of 5 4. What's the function of six or four? What's the function of 7 or 4? Option two? And we know that our function that's secrets of war. Okay so you just plug that in for each value. You just plug in 546474. So on and so forth, calculate those values, plug these entire values in, Add them all up At all these up, Multiplied by 1 4th. And you will get an answer that comes out to be a point 63 452 Alright. Port B of the question is asking us to just repeat But with the mid points now And the way to do this is take serious. So now we look at em You could have equal to four. I was one But instead of the function of I Scuse me except I you're going to have a Um we're gonna substitute for midpoint points. That's what that We're horizontal line there indicates all of that multiplied by the changing X. All right. So how would we then find our mid points For X. Sub 1? Midpoint We write the midpoint between one possible four. So then we would add these two up. So one plus 5/4 Is the same thing as 9 or 4. And then you divide it by 2 And you will get no no. Keep in mind of it. All right. Does that make sense? Let me just clarify all the times you take the The mid point by taking a left endpoint 1, You're right and .54. You add those two together, That's equal to 9 or 4 and then you divide by two so you can get it Right and then we're just gonna continue doing that for each midpoint. Which would then be your next mid .54-6 or 4. The midpoint ends up being 11/8. The mid .3 will be from between 6 or 47 04. And that would be 34. And then finally, the midpoint except poor will be from 7/4 to to Comes out 15 over All right. So then when we calculate for our agreement, some of the mid points we have em is equal to the series for by someone Equals to 1. Excuse Me. The change in X We know is 1 of the war. And then you would take your function X and high. Uh point. All right. And all of that will come out to be 1/4. Function of 9/8 was the function of 11/8. What's function of 13/8? What's The Fortune of 15? 8? And it will come out to be 26-9-1 2 to 0. All right. That will be our final answer there. Hope that clarifies the question. So much water.

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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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