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Let $g(x)=\int_{0}^{x} f(t) d t,$ where $f$ is th…

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Problem 2 Medium Difficulty

Let $ \displaystyle g(x) = \int^x_0 f(t) \,dt $, where $ f $ is the function whose graph is shown.

(a) Evaluate $ g(x) $ for $ x $ = 0, 1, 2, 3, 4, 5, and 6.
(b) Estimate $ g(7) $.
(c) Where does $ g $ have a maximum value? Where does it have a minimum value?
(d) Sketch a rough graph of $ g $.

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Video Transcript

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

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This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

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James StewartCalculus: Early Transcendentals

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