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What I want to do in this video[br]is find the area of this region

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that I'm shading in yellow.

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And what might seem challenging[br]is that throughout this region,

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I have the same lower function.

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Or I guess the[br]lower boundary is y

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is equal to x squared[br]over 4 minus 1.

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But I have a different[br]upper boundary.

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And the way that[br]we can tackle this

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is by dividing this[br]area into two sections,

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or dividing this region into two[br]regions, the region on the left

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and the region on[br]the right, where

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for this first region,[br]which I'll do--

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I'll color even more in[br]yellow-- for this first region,

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over that entire interval in x.

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And it looks like x is[br]going between 0 and 1.

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y equals-- when x is equal to[br]1, this function is equal to 1.

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When x is equal to 1, this[br]function is also equal to 1.

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So this is the point 1 comma 1.

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That's where they intersect.

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So for this section, this[br]subregion right over here,

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y equals square root of x is the[br]upper function the entire time.

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And then we can have[br]a-- we can set up

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a different-- we can[br]separately tackle figuring out

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the area of this region.

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From x is equal to 1[br]to x is equal to 2,

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where y equals 2 minus[br]x, is the upper function.

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So let's do it.

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So let's first think[br]about this first region.

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Well, that's going to be[br]the definite integral from x

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is equal to 0 to[br]x is equal to 1.

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And our upper function is square[br]root of x, so square root of x.

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And then from that, we want to[br]subtract our lower function--

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square root of x minus x[br]squared over 4 minus 1.

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And then of course,[br]we have our dx.

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So this right over here, this is[br]describing the area in yellow.

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And you could imagine it, that[br]this part right over here,

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the difference between[br]these two functions

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is essentially this height.

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Let me do it in a[br]different color.

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And then you[br]multiply it times dx.

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You get a little[br]rectangle with width dx.

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And then you do that for each x.

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Each x you get a[br]different rectangle.

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And then you sum them all up.

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And you take the limit as[br]your change in x approaches 0.

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So as you get ultra,[br]ultra thin rectangles,

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and you have an[br]infinite number of them.

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And that's our definition,[br]or the Riemann definition

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of what a definite integral is.

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And so this is the area[br]of the left region.

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And by the exact same[br]logic, we could figure out

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the area of the right region.

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The right region--[br]and then we could just

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sum the two things together.

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The right region, we're[br]going from x is equal to 0

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to x-- sorry, x is equal to[br]1 to x is equal to 2, 1 to 2.

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The upper function is 2 minus x.

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And from that, we're going to[br]subtract the lower function,

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which is x squared[br]over 4 minus 1.

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And now we just[br]have to evaluate.

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So let's first simplify[br]this right over here.

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This is equal to the[br]definite integral

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from 0 to 1 of square root of x[br]minus x squared over 4 plus 1,

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dx-- I'm going to write[br]it all in one color now--

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plus the definite integral[br]from 1 to 2 of 2 minus x,

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minus x squared over 4.

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Then subtracting a negative is[br]a positive 3-- or a positive 1.

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We could just add it to this 2.

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And so this 2 just becomes a 3.

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I said 2 minus[br]negative 1 is 3, dx.

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And now we just have to[br]take the antiderivative

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and evaluate it at 1 and 0.

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So the antiderivative[br]of this is-- well,

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this is x to the 1/2.

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Increment it by 1.

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Increment the power by[br]1, you get x to the 3/2,

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and then multiply[br]by the reciprocal

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of the new exponent-- so[br]it's 2/3 x to the 3/2.

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Minus-- the antiderivative[br]of x squared over 4

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is x to the third, divided by 3,[br]divided by 4, so divided by 12,

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plus x.

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That's the antiderivative of 1.

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We're going to[br]evaluate it at 1 and 0.

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And then here the[br]antiderivative is

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going to be 3x minus x[br]squared over 2 minus x

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to the third over 12.

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Once again, evaluate it[br]at-- or not once again.

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Now we're going to[br]evaluate at 2 and 1.

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So over here, you evaluate[br]all of this stuff at 1.

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You get 2/3 minus 1/12 plus 1.

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And then from that, you[br]subtract this evaluated at 0.

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But this is just all[br]0, so you get nothing.

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So this is what the yellow[br]stuff simplified to.

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And then this purple stuff,[br]or this magenta stuff,

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or mauve, or whatever color this[br]is, first you evaluate it at 2.

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You get 6 minus-- let's see,[br]2 squared over 2 is 2, minus 8

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over 12.

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And then from that,[br]you're going to subtract

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this evaluated at 1.

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So it's going to be 3 times 1--[br]that's 3-- minus 1/2 minus 1

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over 12.

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And now what we're[br]essentially left with

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is adding a bunch of fractions.

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So let's see if we can do that.

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It looks like 12 would[br]be the most obvious

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common denominator.

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So here you have 8/12[br]minus 1/12 plus 12/12.

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So this simplifies[br]to-- what's this?

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This is 19/12, the part[br]that we have in yellow.

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And then this business,[br]let me do it in this color.

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So 6 minus 2, this is[br]just going to be 4.

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So we can write this as[br]48/12-- that's 4-- minus 8/12.

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And then we're going to have to[br]subtract a 3, which is 36/12.

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Then we're going to add to[br]1/2, which is just plus 6/12,

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and then we're[br]going to add a 1/12.

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So this is all going to simplify[br]to-- let's see, 48 minus 8

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is 40, minus 36 is 4, plus[br]6 is 10, plus 1 is 11.

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So this becomes plus 11/12.

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Let me make sure[br]I did that right.

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48 minus 8 is 40,[br]minus 36 is 4, 10, 11.

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So that looks right.

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And then we're ready[br]to add these two.

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19 plus 11 is equal to 30/12.

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Or if we want to simplify[br]this a little bit,

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we can divide the numerator[br]and the denominator by 6.

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This is equal to[br]5/2, or 2 and 1/2.

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And we're done.

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We figured out the area[br]of this entire region.

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It is 2 and 1/2.

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