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Prove that $ \displaystyle \int^b_a x^2 \, dx = \frac{b^3 - a^3}{3} $.

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12:13

Frank Lin

Calculus 1 / AB

Chapter 5

Integrals

Section 2

The Definite Integral

Integration

Missouri State University

Campbell University

University of Michigan - Ann Arbor

Idaho State University

Lectures

05:53

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.

40:35

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.

04:58

\begin{equation}\text …

06:33

Prove that $ \displaystyle…

05:15

Prove that the integral on…

I think so. Problem number 28 they're asking us to prove the definite integral from A to B of X squared is be cubed minus a cube. Okay, with the limit definition of the derivative. So this is quite involved algebraic lee to do this. It looks pretty simple. It will be simple when you get to the fundamental theorem and have shortcuts other than the limit definition of an integral, but it can be done and that's what we're going to do right now. So the integral from A to B yeah, of X squared dx is going to be equal to the limit as N approaches infinity of some of the area of all the in trying rectangles. So this is I equals one to end. Now the width of each rectangle is going to be b minus A over in. So that is the width of the rectangle. The height of the rectangle is going to be determined by where it falls on the parabola X squared. So we're going to say we started the limited lower and limit of integration of a. So A plus and then it's going to be, I'm gonna move by B minus a ever in and then to be minus ever inch, I'm going to implement that so it's going to be b minus A over in times I all of that squared. Mhm Yeah, yeah and sorry about that. I mean back up it is all yeah, this quantity squared. So you plug in a it's the lower limit of integration and then you increments one, B minus every in two, B minus every end increment all of that. And then the function is squaring uh that entire quantity. So now we have to square that and work with the algebra that would come up with. So this is the limit as in approaches infinity of the some. Okay, I equal one to end, You have B minus A over in and I'm just going to use all the b minus hes with parentheses here for clarity. And now when I square this I get a squared. Yeah. Plus and then I get to a b minus A I over in plus. And then you square that last term b minus a squared I squared over N squared. Yeah. So now we just need to continue working with the um algebraic simplification that happens now. What I want to do next is I want to take out the summation. So I'm gonna I've got a summation of this term here is the sum of I and here is the sum of ice squared. So what I'm using here is that three different properties. The some from my equal one to end of a squared is just going to be a squared times in the sum Yeah, I equal one to end of I is going to be in n plus 1/2 and then finally the some I equal one to end of I squared yeah Is going to be 16 and N plus one, two n Plus one. So those are all the summation properties that I would need. So go to the next line, let's just change our colors here. This is going to be the limit as in approaches infinity and the sum is going to come out now. So I'm going to have b minus a over in now to take care of the A squared then that some is going to be a squared in uh huh Plus to a B minus a over in And then the sum of I is just going to be in in plus one over to Mhm Yeah. Plus and then that last term is going to be b minus a squared and then I'm going to have in n plus one two n plus one over six in squared. So sorry about breaking that on the line, but that is where we are. So let's just rewrite that so they can make sure that we've got everything we need. So this is the limit in approaches infinity of B minus a over in times A squared N plus um B minus A and N Plus one over to in plus b minus a squared times in and plus one two n plus one Over six in squared. Yeah. And now we would just continue our simplification. So let's go ahead and distribute this B minus A into the parentheses. So this is the limit N approaches infinity. So in the first term you're going to have a squared b minus a and the ends cancel. Okay? Yeah. Plus In the next term I'm going to have um this end will cancel this in, will cancel the two and the two and then I'm going to be left with in that case um A times b minus a squared, Yeah, a times b minus a squared. And when I look at this value of the end right here, this in um it's gonna be dividing the N plus one. The easier way to write that, it's just gonna be one plus one over in. So that's a um B minus a squared one plus one over N. Plus. Now you look at the next expression and what you see here is I'm going to see this in, we'll cancel with this in right here. Okay? And then if you look at it, you're going to have b minus a cubed, so you're going to have plus B minus a cubed over six. And now when I look at this in dividing these two terms, the easiest way to write this is just going to be one plus one over in part me and two plus one over N. Now what we can do is take the limit as N approaches infinity as N approaches infinity. This term We'll go to zero. This term will go to zero and this term will go to zero so we can remove the limit and this is going to be a squared B minus a. Yeah plus A Yeah mhm B minus a squared plus. And then you're left with two mm. Yeah, yeah B minus a cubed over six. Sure now continuing on, it looks like there is a common b minus A in all of this. So this is b minus A. And then I'm left with a squared plus a d minus a. Yes plus one third, B minus a squared. Now we're way beyond the calculus, we're just into the algebraic simplification here. So this is B minus a. Then you have a squared plus a B mhm minus a squared, Mhm Plus one third. B squared minus to a B plus a squared. Yeah. Right, mm. This is B minus a. One nice thing here. Is that a square? Money's a squared is zero. And so this leaves me with a B yeah plus one third, B squared minus two thirds A B plus one third A squared, mm hmm. Yeah. Yeah, yeah. Okay, this is B minus a. And then let's look at the terms that we have first so I can write this as what one third B squared. And if you look at the abbey terms, you think of this as that's three thirds minus two thirds is going to be plus one third plus one third A b plus one third A squared. That tells you can factor out of 1/3. So this is one third, B minus a Yes, B squared plus a B plus a squared. You've done a lot of work to this effect and you have to go back to when you first learned a factor, the difference of cubes. So be cubed minus a cubed is b minus a B squared plus a B plus a squared that matches this identically. So your final answer here is be cubed minus a cubed over three. So a classic problem where calculus ended about a couple of steps into this and it was just all algebraic simplification to get to the final answer that the integral from A to B of x squared is be cubed minus a cubed over three.

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