WEBVTT
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So problem 24, we're dealing with the limit definition
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of an integral. So a lot of algebra involved
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in solving these guys, I find it easiest just
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to break these up. So break some of the
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algebra into two separate integral. So two X dx
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Minour the integral from 0 to 2 of x cubed
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dx. Let's go do the work of each of
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those separate and then we'll come right back here to
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find our answer. So let's deal with the first
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integral. First the integral from 0 to 2.
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So the integral from 0 to 2 of two x
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. DX by definition is going to be too times
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the limit as N approaches infinity of the sun of
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I equals one to end. Now the width of
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each rectangle, it's just going to be to zero
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over in. So the width of each rectangle is
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too over in the height of each rectangle is going
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to be this function X evaluated so X at two
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over in Um four over in and and so forth
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. You're gonna increment this. So this is gonna
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be too I over in right? So this is
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the limit or it's going to be two times the
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limit as N approaches infinity. And if you look
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at what I have here, this is going to
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be you can replace that i with the formula that
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you know the some Mhm. I go one to
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end of I it's just simply end N Plus 1/2
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. So that's going to take the some of this
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out, the summation sign. So I'm gonna be
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left with two over in and then times two in
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Yeah, N plus one over two in. Yeah
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. So all I did was replace the what you
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see here in N plus 1/2, that replaces the
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I. So in this case, if I look
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at what I have to simplify this guy, this
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is two times the limit as N approaches infinity.
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And so what you see here is that you've got
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um to end over to end and then if you
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look at this, so this is going to be
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two and if I look at this in over N
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plus one, I can write that as one plus
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one over in. Mhm. And so as N
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goes to infinity, this term will go to zero
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. So this just becomes too um times two times
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one plus zero, which is simply four. So
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if I look at where I am in the scope
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of this problem, I have determined that the value
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of this one Is four. Okay, now I
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could have easily gotten that one graphically, you know
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, what is the line, you know, y
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equal to X From 0 to 2. Mhm.
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So when you plug in a two there you get
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four. Oh. Mhm. So right here,
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so if this is too yes and this is for
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the area that one half based sometimes should be four
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. So already got that. So I know that's
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right now. Little bit more work to get the
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integral. Yes, from 0 to 2 of X
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cubed dx. That is going to be the limit
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as in approaches infinity of the some I equal one
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to end with of each rectangle is still too over
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in and then now this is going to be um
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two. Um Yes, I cube so to I
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over in cute. So what is this going to
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give me? Well I know how to evaluate.
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Um let's just go and simplify this is the limit
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in approaches infinity of the sum I equal 12 N
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two cubed is eight. So this is going to
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be what, 16? So to to me so
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16 Over into the 4th. I cubed. Ok
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. And so now we need our formula that the
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sum I equal one to end of I cubed is
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1/4 in squared. Yeah, N plus one squared
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. Yeah. Mhm So let me get the some
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on this one real quick. This is going to
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be the limit. Yeah. Yeah. And approaches
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infinity. Uh So what have we got? 16
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over? End of the fourth And then the sum
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of I cubed. It's just going to be 1/4
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and then you've got in squared N plus one squared
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. So this is going to give me the limit
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and approaches infinity. 16/4 is four. And then
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I can write this as so you got in squared
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. This term will cancel if you write that as
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in square times in squared And then you can write
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this as one plus one over in he squared.
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And so this answer just turns out to be four
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. Go back to your original, So it's 4
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-4. Final answer is zero.