WEBVTT
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we want to find the area of the enclosed region
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. Ah, we have the equations y equals 1/4
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X squared. That's the wide problem. We have
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two x squared, which is the narrow problem we
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have X plus y equals three, which is the
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question of a line s. So let's label that
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X plus y equals three and we're only interested in
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X greater than zero. So we need to figure
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out what our limits of integration will be. We
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have this first point of the second point, and
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this third point, let's label them A B and
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C A is the intersection of our two problems.
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To X squared equals 1/4 X squared. So this
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tells us that X is equal to zero. That's
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the order. Looking at the second point of intersection
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we have to solve for two x squared are narrow
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. Problem equals the equation of her line. So
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our line is why equals three minus x. So
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this is just a quick erotic. So solving it
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gives us two solutions we have one of them is
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X equals one and the other is X equals negative
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three over two. But we only care about ex
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being positive, so we retake X equals one.
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Next, we're looking at the third point of Intersection
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, which is going to be the intersection of our
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wide problem. 1/4 X squared and our line y
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equals three minus X. Again, it's a quadratic
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so we can solve it of by using the quadratic
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formula whereby factoring and what we get is we have
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the solutions X e quote negative six where X equals
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two. But since excess positive, we take X
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equals two so we can label these points along the
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X axis. This is X equals zero X equals
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one and X equals two. So we're going to
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integrate to determine this area. Let's open up a
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new page. Now here we have area equals into
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grow work. First, we're going from 0 to
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1. So we look at our top function,
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which is to two x squared over here, and
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our bottom function is 1/4 X squared. So we're
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going to subtract those Ah, we have bracket to
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X squared minus 1/4 X squared DX plus integral.
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We're going from 1 to 2. And now let's
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take a look at what is the top function.
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The top function is the lines. So that's going
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to be three minus X. And then the bottom
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is the wide problem. 1/4 X squared. So
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we have three minus x on. We're subtracting the
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wide problem minus 1/4 X squared DX. This first
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sent a girl we can do in one piece because
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two X squared minus 1/4 X squared is just 7/4
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X squared. Ah, so we integrate the anti
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derivative of 7/4. X squared is seven over 12
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x to the three. This is going from 0
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to 1 plus. Now we take the anti derivative
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of this second integral. We take three X minus
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half X squared. Mmm. And then here we
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have minus one over 12 X to the three.
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This is going from 1 to 2 and then after
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plugging in our numbers, So here we have a
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seven over 12 minus zero. Next we plug in
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to hear we have a six minus two squared is
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four former toos, too, to Cuba's 88 over
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12 is three over four. Now we plug in
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13 minus 1/2 minus one over 12 and then simplifying
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everything gives us the solution. Three over two