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