WEBVTT
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So what we want to do here, since we
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want to find to regions with equal area, is
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we want to split our region that's enclosed between the
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red and green graph and half. And what's nice
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is our parabola is symmetric, so we can split
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it smack dab right through the center along the Y
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axis right there. And just consider one of these
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pieces to find half the area of this region.
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So let's consider the right hand side. So let's
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computer integral with respect to why so our right hand
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function will call it f of why. And so
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we're gonna convert our green y equals X squared to
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a function in terms of why instead of X.
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So we just saw for why they're so f of
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why is equal to the square root of why Andi
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, we're choosing the positive square root because we're in
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the first quadrant. So when we plug in positive
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why values will come out with positive X values as
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a result, and then our left hand function is
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gonna be our blue function that we're using to split
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our region in half. And that is tthe e
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y axis, which is X equals zero Our.
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In other words, we call that G of why
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equals zero Now for our bounds, we have our
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lower bound that's the X axis, and then our
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upper bound is going to be our red y equals
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four. So with the X axis, that's why
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equals zero. And then why equals four is our
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upper bound, So to create are integral for 1/2
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of the area of this region that's equal to the
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integral from 0 to 4 of our left hand function
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, or are excuse me a right hand function minus
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our left hand function, So f of y minus
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ci of why? Which is the square root of
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Why? Why? To the 1/2 minus zero?
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We don't even need to write that. We just
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have d y. So evaluating this we have 2/3
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. Why, to the three halves evaluated from 0
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to 4 and plugging in our bounds, we have
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2/3 of four to the three halves minus Well,
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when we plugged in zero, we just get zero
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. So we don't need to write that 2/3 times
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wth e um square root of four cute, which
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is the square root of 64 which we know is
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eight. So this is 2/3 times eight, which
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is 16 over three. So that is the value
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of half of the area of our region. Great
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. So what we want to do now is find
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that be value where we can find a line.
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Why equals B that will split our region, Um
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, in half as well, such that both of
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the areas are equivalent. So we're going to do
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very similar process here. This is still f of
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why equals thes square root of why on the left
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hand side, we call this G of y that's
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equal to the negative square root of why and our
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bounds here for our bottom half of the area,
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while the lower bound is again the X axis.
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And so that's going to be that. Why equals
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zero. Our upper bound is gonna be our blue
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line. Why equals B? We don't know what
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B is yet. We're going to find that in
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this process. So to construct the integral for this
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, we'll have that half of the area which we
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now know is actually 16 over three. That's equal
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to the integral fromthe lower bound zero to the upper
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bound B of our f of why minus g of
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why? Which is the square root of why minus
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the negative square root of why So why did the
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1/2 minus and minus Why to the 1/2? Do
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you? Why? So that's equal to the integral
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from zero to be of two. Why to the
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1/2? Do you? Why, which is equal
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Thio? 4/3. Why? To the three halves
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evaluated from zero to be and then we plug in
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our bounds. We get 4/3 be to the three
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halves minus. When we plug in zero, we
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just get zero. So we don't need to worry
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about that. And that's equal. Thio again 16
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over three And now we just sell for B.
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So let's multiply both sides by three over four goodbye
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, goodbye and goodbye and four goes into 16 4
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times. So we have four equals B to the
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three halves and we want Thio isolate Be so using
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properties of exponents those go away So we have that
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be is equal to four to the two thirds.
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Now we can check our work Thio verify that that
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is actually the correct be constant for our line y
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equals beat. So let's look at and consider,
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um d top half of this enclosed region and using
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our be that we found four to the 2/3.
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Let's calculate the integral and verify that this area in
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here can be calculated Using B to B equals four
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to the 2/3 and in fact, results in six
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over 13. So our, um, our bounds
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for our actor region upper half the lower bound is
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gonna be y equals b, which is why equals
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we know now, four to the 2/3 and our
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our upper bound is going to be our red y
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equals four. So when we want to check,
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we want to verify that 16 over three, which
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is our 1/2 of our area, is equal to
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the integral from four to the 2/3 24 off our
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f of y minus g of way. So that
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is why to the 1/2 minus minus. Why to
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the 1/2 do you? Why, which is the
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integral from four to the 2/3 to the 4 to
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4 of two? Why? To the 1/2 do
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you want, and that's equal to 4/3. Why
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? To the three halves evaluated from four to the
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2/3 24 and so plugging in our upper and lower
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bounds. We have 4/3 times four to the three
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halves, minus 4/3 times four to the 2/3 23
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house. And then we want to simplify this 4/3
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times four to the three halves. Well, we
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already know that this guy is the square root of
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64 which is eight. So eight times for his
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32 over three. And then here the buy properties
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of exponents. These guys cancel each other out,
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and we have minus four times four over three.
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So minus 16 over three. And yes, indeed
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, that's equal to 16 over three. So we've
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got that. The line that divides our region in
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half is why equals four to the 2/3. That
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is the B value that satisfies this problem.