WEBVTT
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the proble y equals X squared over two divides the
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disc X square plus y square less than equal today
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into two parts. Let's find the area of these
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parts s o N red. We have our circle
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. This is the boundary of the disc, and
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Red Export plus y squared equals eight. So it's
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a circle with a radius radically. And then it's
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intersected by our problem at thes two points highlighted agreed
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. So there's a blue problem like equals X squared
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over two. So a one let's do Nobody won
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the area inside the circle So inside of the red
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but a bluff above the blue problem and then a
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two is everything else in a circle Or if you
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want everything in the circle but below the problem.
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So let's find a one first. So here will
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find a one. Then use the fact that a
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one plus a two Well, if you add a
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one and a two together, you're getting the area
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of the entire circle so we know the area of
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a circle. Oh pi r squared. We're here
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. We have our squared equals eight. So this
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equals a pie and then we have softer A too
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Wei have this so really, all of our work
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is going to be on finding a one. And
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then once we find it, we find a to
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using this formula circled and then we're done so is
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focused on a one first. So here, this
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looks like in a roll with respect to the X
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from this point of intersection and green to the other
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one. And then we have top minus bottom.
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So this is the area between two curves. So
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notice that the talkers, which is part of the
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circle X squared plus y squared, equals eight.
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If you saw for why you get to parts of
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the circle the plus radical, which is the top
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half. And then on the bottom half of the
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circle down here, below the X axis, you
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have the negative radical. So that's our tankers.
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So we're doing top man in his bottom, and
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then the bottom curve is down here on blue.
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That's X squared over two. Now we need the
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points of intersection. So let's go on to the
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next page to do that. Well, we're used
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The fact that why was explored over too, and
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then we'LL plug it into this equation over here,
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so I'm gonna put equals a because at the point
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of intersection, run the on the circle, not
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inside the disc. And so the radius of that
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circle isthe root. Wait, So this is the
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equation of the circle now playing in excluding him too
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, for why we obtained this equation. So let's
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go ahead and simplify this Get X to the force
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over four plus X squared minus a equals zero.
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And then let's go ahead and multiply both sides by
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four. And this is a contract IQ, but
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not an expert in X square. And this one
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will factor as X squared plus c and then X
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squared minus four. Now, if you see this
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first term, X squared plus eight is always positive
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. It's always bigger than zero because X squared is
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bigger than zero. So, really, this term
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here is bigger than our equal to eight, which
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is bigger than zero. So this first term is
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number zero and then we set this term zero,
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and we get X equals plus or minus two.
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So these are the X values of the points of
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intersection. So we have negative tune into top man
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in his bottom And now, to make this a
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little easier I'LL go ahead and basically using the cemetery
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on the original area A want so recall a one
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was of this form So a one is the collection
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these two regions right here and basically by cemetery instead
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of finding it from negative to to to we'LL just
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find it from zero to and then we'll supply by
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two so we can just do this instead, Not
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a necessary step. Might make it a little easier
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. And now, because you could see this first
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, Integral has a radical. We should use the
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tricks up here for the second one. We could
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just use the hour rule. So let's just go
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ahead and split this into two animals. So writing
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that all we have minus and then we cancel the
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twos Explorer dx, Sierra two and we would just
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evaluate these separately, put them together, and then
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we'LL get our anyone so looking at this easier integral
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. Here. You can just do Tyrol. You're
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excused over three zero to, and then we get
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eight over three after plugging in the end points so
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that just leaves us with this remaining in a girl
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over here and we'LL do it tricks up for this
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Let's go to the next page. This was our
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inaugural. So here this is a trick sub Let's
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do X equals radically signed data And since this is
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a definite no rule, we might be best off
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here If we go out and simple by this radical
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we'LL see why shortly it could help us avoid writing
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the triangle. So here this is R X,
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go ahead and take the X And now they will
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see why this was a good choice because now we
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could try to find the new limits of immigration in
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terms of data. So plugging in X equals zero
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into this equation and also using the fact that when
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you do a tricks of involving sign, you have
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to make sure it is between negative power to in
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parenting. So let's use these fast tto find data
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. So on the left side for X, we
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have zero and the right side to room two Scient
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Ada, this means sign has to be zero,
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and sense data is in this interval Over here.
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We must have data equals zero is the only solution
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into sensible now for the upper limit. Tune said
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the left hand side X equal to two and on
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the right two room to sign Data Sol for sign
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. And this is a point in the circle and
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this interval, the point that we want this state
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a equals power for So that's our new upper limit
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lower and let's do one more step here we have
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eight minus X squared inside the radical that becomes radical
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a minus and then X squared is eight signed square
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. We could fatten throughout a radical one minus sign
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square. And then if we like we can write
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this radical is to room two use with Agron identity
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and then finally just cancelled a square with the square
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root. Okay, so that takes care of the
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Radicals. Let's go ahead and plug this all into
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our our new interval. In terms of data,
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we have to our limits change. Now we have
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zero pi over four. We just simplify the radical
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that was to room two cosign data and then the
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ex. I was too rude to cause and data
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self here we have radical eight times radical eight.
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So we have eight and then we multiply by this
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two on the outside, we could probably sixteen and
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then we have co signed Square. So trying to
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remember what to do in this case and then for
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this trick in a rural there's no science present.
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So in this case, the best way to go
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is one plus coastline to date over. So let's
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pull that too out. We have sixteen over to
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since eight zero power before one plus co sign to
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data, and this we can integrate. We have
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data plus sign to Theatre Over to and points Sierra
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Power before Dan. Let's go to the next patient
144
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, plugging those end points and simplify so plugging in
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pi over four for data. But and then when
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we plug in zero, we have zero plus sign
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zero over, too, so we could ignore the
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second term sign. A pie or two is one
149
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. So let's go ahead and multiply. That ate
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through and we get right over here and then we
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have to buy and then times in an eight times
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a half, it's to my place for so remember
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that this was just the first integral We also had
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to subtract eight over three from this from the previous
155
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integral that we got using the powerful. So let
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me take a step back. A one was equal
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to This is the animal that we just completed using
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the tricks up. Originally, we had this other
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expression that we used. So this number corresponds to
160
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this first in the room. So we have to
161
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pi plus four. And then we're subtracting the previous
162
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answer the quadratic and we had eight over three for
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this one. So put these fractions together. Forest
164
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well over three. So we subtract eh for over
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three. So this is a one and then recall
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for a tu es tu is everything else in the
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circle. The formula that we had for a two
168
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was a pi minus a one. A pi was
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the area of the whole circle. A one is
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over here on the right, So plugging this in
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we have a pie minus to Piper's for over three
172
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. So we have six by minus four over three
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and there's a final answer