Problem

(a) Find an approximation to the integral $ \disp…

15:15

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Problem 25 Easy Difficulty

Use the form of the definition of the integral given in Theorem 4 to evaluate the integral.

$ \displaystyle \int^1_0 (x^3 - 3x^2) \, dx $

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Video Transcript

Yeah, yeah. So this question we're looking at the definite integral from 0 to 1 X cubed minus three X grade and were asked to use the limit definition of an indefinite integral, which is sort of the long form of doing this. But to make sure that we understand the concept that the integration is just really um the summation of triangle areas. Excuse me, rectangle areas. Easiest way to do this is to break this up, I fine into two separate integral and handle each one so that it breaks the algebra up into pieces so I can write this as three times the integral from 0 to 1 of x squared dx. Now we're going to go do some work. We are going to come back here for our final answer once we figure out the value for each of these intervals using our limit notation. So the integral from 0 to 1 of x cubed DX is equal to by definition the limit as in approaches infinity of the area of These in rectangles. So from my equal one to end, the width of each rectangle is going to be the limits of integration divided by N. So one over in so one over in. And then if you look at it it's going to be um the height of the correct angle is determined by X cubed. So it's going to be um I know you're gonna have I over in cubed so you're gonna have one over N two over end three over in and so forth to come up with that. So this is the limit From I equals one To end one over in. And now we can jews are formula for what are the um The values for the summation. So in this case we're going to end up with this is going to be won over into the 4th and then the sum of I cubed and then if I go ahead and apply that formula, this is going to be one over into the fourth and the sum from one to end of I cubed, It's just going to be 1 4th and then that is in squared. Uh huh, times N plus one squared. Yeah. Mhm. And then if I look at that, so sorry about this, a little bit sloppy notation that what has to go in there is still the limit as N approaches infinity. So let me get to a new line and write this more precisely. It's the limit As in approaches infinity of one over into the 4th times 1 4th in squared And plus one squared. Well that is the limit as N approaches infinity A 1/4. And then we can just write this. Well you've got into the fourth and n squared. So that would tell me that this would cancel and just leave me with an N squared. Let's move that inside of that binomial term. And that is just going to be one plus one over in Squared. And so as in approaches infinity this term goes to zero. That final answer is 1/4. So what that tells me is that this first integral Evaluates to 1/4. Now we just need to evaluate what is the value of the integral? three Integral from 0 to 1 X squared dx. Well that is three times the limit in approaches infinity of the sum I equal one to end. Um The width of each um rectangle is still one over in and then you are left with when you square that term you're going to have I over in squared. So this is three times the limit as N approaches infinity of one over n cubed I squared. Um Sorry, I left off this some there. Let me just erase that. Yeah, Some I equal 1, 2 N one over n cubed I squared. And then we use our formula for what the sum of I square it is going to be. So this is three times the limit. Yeah, N approaches infinity of one over n cubed and then the sum of I squared is just 16 comes in Times in plus one times two in plus one. And so now if you look at it 3/6 is one half. So this is one half and the limit in approaches infinity. And if you look at it, this end will cancel and that makes this an end squared. And then I can write this as one plus one over in and two plus one over N as n goes to infinity. This term goes to zero. This Kirm goes to zero. So your value Ah it's simply going to be one to go back up here, 1 4th -1, so that is 1/4.1 is 4/4, answer is -3 fourths. Mhm. Mhm.

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Problem

(a) Find an approximation to the integral $ \disp…

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James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Samuel Hannah

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Samuel Hannah

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Kristen Karbon

Samuel Hannah

James Stewart

Calculus: Early Transcendentals