Problem

Find the exact area of the region under the graph…

03:53

Problem 29 Easy Difficulty

(a) Express the area under the curve $ y = x^5 $ from 0 to 2 as a limit.

(b) Use a computer algebra system to find the sum in your expression from part (a).

(c) Evaluate the limit in part (a).

Answer

$\lim _{n \rightarrow \infty} \frac{2}{n} \sum_{i=1}^{n} \left(\frac{2 i}{n}\right)^{5}$

Watch More Solved Questions in Chapter 5

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Video Transcript

So the problem 29 they give us a function f. of X. is X to the 5th. We know that that curve lies above the X. Axis. So we're asked to find, Hey express as a limit the um area under the curve from 0-2. So to express this as a limit and a some. So what's going to happen is you're gonna divide this up into rectangles. So as you divide it up into rectangles, you're trying to approach an infinite number of rectangles. So in is the number of rectangles. So we're gonna let that go to and finish. You're gonna have the limit as in approaches infinity. And now we're going to have to sum up the areas of all the rectangles. So I'm gonna pick another variable over my indexing. I equal. Okay. And then I've got in rectangle so I'm gonna count them from one to end. Okay. And now the height of those rectangles. Okay. If I use the right some then as I go from 0 to 2, what is the distance from 0 to 2? That's two minus zero divided by in is to over in. So each of these rectangles has to be a width of two over in to make this work. So the width of the rectangle is too over in. Okay. And then the height of the rectangle is to be determined by this function X to the 5th. So the first rectangle, if I take the right side, that is going to be when at the first rectangle. So it's going to be i times what to over in to the fifth power. So that first value is one times two over, in, the next rectangle is at two times two are in three and then all the way up to end. So this is the sum that needs to happen. So this is a limit infinite some uh value of the area under that curve. So now what they said to do is to use uh computer algebra system to evaluate um this some okay so let's just go to that and figure that out. So it's going to be the sum. So we're going to say the some Mhm. From I equal 1 2 n. of two over N times I time is too over in Raised to the 5th power and you see what I come up with there. So this is 16 N plus one squared. So this is equal equal to 16 In plus one squared. And then I had a quadratic term too two in -1. So to n squared plus two in minus one Divided by and that was three into the 4th, so three in to the fourth. So this is just, pardon me, the the this is just the thumb part. Okay, so what you're seeing here is that is just the sum part. Okay, so now what we want to do is want to take that some and we want to take the um the limit of that. So now I need to figure out what is the limit as N approaches infinity? Okay. Of 16 N plus one squared. Mhm two in squared. Okay Plus two in -1 over. I'm gonna write this as three times in squared times in square to help me with this limit, that's three into the fourth. So this is going to be the limit as in approaches infinity 16/3. Now that first term is going to be N plus one over in squared Ok, And then the next one is going to be mhm to N squared? Yeah Plus two in -1 over in squared. And now you look as n goes to infinity, then this term is going to go to zero. This term is going to go to zero. This term is just going to go to two. This term will go to zero. This term will go to one. So this limit is going to evaluate to The value of 32 over three. So that is the exact area of the curve from 0 to 2 of extra the fifth. And we found that by writing this as a sum and a limit. We used our computer algebra system to figure out what the value of the sum was. And then we just used our own Strengthen and recognition of algebra and limits to figure out that the limit was 32/3.

Robert D.

Florida State University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 1

Areas and Distances

Problem

Find the exact area of the region under the graph…

Problem 29 Easy Difficulty

(a) Express the area under the curve $ y = x^5 $ from 0 to 2 as a limit.

(b) Use a computer algebra system to find the sum in your expression from part (a).

(c) Evaluate the limit in part (a).

Answer

$\lim _{n \rightarrow \infty} \frac{2}{n} \sum_{i=1}^{n} \left(\frac{2 i}{n}\right)^{5}$

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Catherine R.

Heather Z.

Caleb E.

Samuel H.

Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

Recommended Videos

Watch More Solved Questions in Chapter 5

Video Transcript

Calculus: Early Transcendentals

Grace H.

Anna Marie V.

Kayleah T.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview