Let g(x) = ∫^x0 f(t) dt , where f is the function whose...

Let $ \displaystyle g(x) = \int^x_0 f(t) \,dt $, where $ f $ is the function whose graph is shown. (a) Evaluate $ g(0) $ and $ g(6) $ (b) Estimate $ g(x) $ for $ x $ = 1, 2, 3, 4, and 5. (c) On what interval is $ g $ increasing? (d) Where does $ g $ have a maximum value? (e) Sketch a rough graph of $ g $. (f) Use the graph in part (e) to sketch the graph of $ g'(x) $. Compare with the graph of $ f $.

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

(a) Evaluate $ g(0) $ and $ g(6) $
(b) Estimate $ g(x) $ for $ x $ = 1, 2, 3, 4, and 5.
(c) On what interval is $ g $ increasing?
(d) Where does $ g $ have a maximum value?
(e) Sketch a rough graph of $ g $.
(f) Use the graph in part (e) to sketch the graph of $ g'(x) $. Compare with the graph of $ f $.

Transcript

00:01 We are given a function f, or given its graph, and we're told that g is the function defined to be the integral from 0 to x of f of t d t.

00:13 In part a, we're asked to evaluate g of 0 and g of 6.

00:22 Well, by definition, g of 0 is the integral from 0 to 0 of f of t d t.

00:34 Since the limits of the integral are the same, it follows that this is just zero.

00:44 Likewise, g of 6 by definition, is the integral from 0 to 6 of f of t d t.

00:59 However, notice that the graph of f has rotational symmetry about the point three zero.

01:36 So the part from x equals 0 to 3 is equal an area but opposite in sign to the part from x equals 3 to 6 and therefore this is equal to g of 6 is in fact 0 as well.

01:59 Then in part b we're asked to estimate g of x for x from 1 to 5 so to do this you want to count the squares between the graph of f and the x -axis.

02:20 And so these are going to be pretty rough estimates.

02:24 Therefore, the squares above the x -axis be counted as positive.

02:28 The squares below the x -axis be count as negative.

02:35 And so, for example, you might calculate g of 1.

02:40 Well, this is the area under f from 0 to 1, which is about 2 .8.

02:50 Likewise, counting squares again, g of 2 is approximately g of 1, which was 2 .8, plus the area from x equals 1 to 2 under the graph of f, which is about 1 .9.

03:07 And so this is approximately 4 .7.

03:20 Likewise, g of 3 is approximately the area from x equals 0 to 2 under the curve 4 .7, plus the area from x equals 2 to 3 under the curve, which is about 0 .6.

03:36 And so this is about 5 .3.

03:47 And we can do these same processes for g of 4 and g of 5.

03:54 G of 4 is about 5 .3, but now we subtract about 0 .6, which is about 4 .7.

04:05 And likewise, g of 5 is about, gf4, which was 4 .7 minus about 1 .9, which is about 2 .8.

04:29 Then in part c, we're asked on what interval is the function g increasing? well, from x equals 0 to 3, the area between the curve and the x -axis is all positive, and then negative afterwards.

04:53 So as x goes from 0 to 2 ,000, 3, it follows that the cumulative area increases.

05:08 On the other hand, past x equals 3, total cumulative area decreases, and therefore it follows that the function g is increasing on the interval 03.

05:56 Then in part d, we're asked where the function g has a maximum value.

06:03 Well, as we pointed out in part c, g is increasing on the interval 0 to 3 and is then decreasing on the interval 36.

06:18 Therefore, it follows that g has a maximum value at x equals.

06:28 Three...

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