WEBVTT
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we are given the function G. Fx which is
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the integral from zero to X. Of the function
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F. Of T. D. T. Where
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F. Is the function whose graph is given in
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exercise too of this chapter, in part A were
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asked to evaluate the function G for X equals 012345
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and six. In order to do this first,
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let's look at the figure for F. So going
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to reproduce a copy of that graph here. We
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have our X and Y axes. X axis will
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range from zero up to six, and the Y
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axis from zero up to three. Yeah. And
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the graph of F begins at the.01, Goes
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down to the.10 All the way to the point
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to-1. So I guess why actually ranges down
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the-1 as well. Then goes back up 230
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and continues all the way up to 63 We can
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divide this up into triangles and rectangles like this.
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So we have a triangle here. I'll call that
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triangle A. We have two more triangles here and
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here. I'll call these triangles, B and C
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. We have another triangle here. Call this triangle
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D. Then we have a rectangle which is actually
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a square here. I'll call this E another triangle
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above that, which I'll call F. Another rectangle
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, which I'll call G. And finally another triangle
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. H. Now as G fx is defined,
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it is the area between the graph of F and
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the X axis where we have areas below the axis
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being negative. Therefore for example, we have that
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G of zero. Well this is by definition The
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integral from zero to itself, zero of F F
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T. D. T. Which is just zero
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. Because the limits of the integral are the same
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. Now G. F. one, This is
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by definition the integral from 0 to 1 of Fft
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. And we see that this is the same as
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the area of A. Which because it's a triangle
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, this is one half times one times one Which
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is just 1/2 G F two. By definition is
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the integral From 0 to 2 of F F T
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. D. T. Which is looking at our
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graph, the areas of triangles A and B added
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together. Remember that these are signed areas. Well
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, we saw that this is in fact one half
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plus one half times one times negative one. Their
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height is below the X axis. And so this
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is zero. Likewise, GF three Is the integral
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from 0 to 3 of Fft. This is the
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area of triangles A. B and C, which
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is just the area of triangle C, Which again
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is negative 1/2. And so this is negative 1/2
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. Likewise, We find that g. f.
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four is areas of A. B. And C
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plus the area of D. As well. And
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this turns out to be zero G. F.
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Five. This is the areas A. B,
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C. And D. Plus. Now the area
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of E. And the area of F. Now
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we see that he is a rectangle in. So
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it's area is simply the base times the height.
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And so this is simply area of E. Which
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is one plus the area of F, which is
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one half, Which is 3/2. And finally G
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. Of six is the area's A plus B plus
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C. Plus D. Plus E. Plus F
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. Plus the last two areas G.& H
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. And so this is going to be 3/2 plus
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the area of G. Which again is a rectangle
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. So this is going to be base times height
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, which is to plus another half and this is
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four. So those are the values of G for
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X between zero and 6. In part B,
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we are asked to estimate G of seven. Well
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, by definition, GF seven is the integral from
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0 to 7 of Fft. Which Using integral properties
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, this is the same as the integral from 0
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to 6 of Fft Plus the integral from 6 to
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7 of fft. Now we saw in part a
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that G of six was four. Now looking at
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the area from X equals 6 to 7, there
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appears to be about 2.2 Square units. So this
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is approximately four plus 2.2 or 6.2 in part C
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. Were asked to determine where the function G has
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a maximum value and where it has a minimum value
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. Well noticed that if we begin at T equals
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zero from our results from parts A and B,
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the maximum amount of positive area is added When T
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reaches seven. Therefore, it follows that G has
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maximum value at seven. This maximum value is in
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fact, G F seven Which we saw from part
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two we estimated was about 6.2. Now, Beginning
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A. T Equals zero. The maximum amount of
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negative area is added when T equals three. So
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it follows that G reaches a minimum at T equals
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three. This minimum is G of three, which
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we saw in part a was negative one half.
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Finally, in Part D were asked to sketch a
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rough graph of the function G. To do this
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, we'll use parts A and B. So we
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essentially have a table of values we can generate from
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those parts and from those parts, we see that
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well, if we set up our X and Y
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axes, X is going to range from zero up
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to seven and Y from zero up to about well
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six. And as we saw from Part A,
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we started out at zero, move up to about
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a half. Then we start moving down Until I
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reach zero again At about two. Like this,
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we reach a minimum value at X equals three,
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About negative 1/2 from part C. We know this
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. And then from that point we start increasing like
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this Until we get to x equal seven where She
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is about 6.2. We start leveling off there.