Graph of f 3. The continuous function f is defined on the...

Graph of f 3. The continuous function f is defined on the closed interval -6 ? x ? 5. The figure above shows a portion of the graph of f, consisting of two line segments and a quarter of a circle centered at the point (5, 3). It is known that the point (3, 3 - ?5) is on the graph of f. (a) If ?_{-6}^{5} f(x) dx = 7, find the value of ?_{-6}^{-2} f(x) dx. Show the work that leads to your answer. (b) Evaluate ?_{3}^{5} (2f'(x) + 4) dx. (c) The function g is given by g(x) = ?_{-2}^{x} f(t) dt. Find the absolute maximum value of g on the interval -2 ? x ? 5. Justify your answer. (d) Find lim_{x ? 1} rac{10^x - 3f'(x)}{f(x) - arctan x}

Transcript

00:01 So there's a lot to know about this problem.

00:04 And i could write all of these out, but i'll probably just use them when i need them.

00:09 So f of negative 2 equals 1.

00:12 We have, you know, i could even write 0 negative 1.

00:16 Like this is negative 1, 0.

00:19 But then we have this slope here.

00:21 It's a little bit steeper because it gets up to 2, 3.

00:24 And then you have this circle down here.

00:28 There's a lot of other information you need.

00:30 Like it says the 3 -3 minus, so this point right here, is 3 -3 -5, if i'm reading that correctly.

00:47 And, yeah, i think we can answer some of these questions.

00:51 Like when they ask us for the, or they tell us, i guess, the integral from negative 6 to 5 of f -fx -d -x is equal to 7, and they're now asking us to evaluate the integral from negative 6 to negative 2 of f of x d x.

01:09 Well, i guess my thought process is that what you can do is figure out that integral from negative 6 to 5.

01:19 Or how about this? let me change that to a plus sign and change that negative 6 to a negative 2 of f of x d x is equal to the integral.

01:36 From negative 6 to positive 5 f of x d x.

01:43 So what i can do to try to solve for negative 6 to negative 2 of f of x dx is subtract over.

01:53 So that integral from negative 6 to 5 of f of x d x minus minus that's a minus sign there, the integral from negative 2 to 5 of f of x d x.

02:08 I hope that makes sense what i just did, just algebra of manipulation.

02:12 So since negative 6 to 5 is equal to 7 and just plug that in.

02:18 But now i need to figure out the integral from negative 2 to 5.

02:23 And the first thing i would point out is the negative area, the area that's below, that's negative, will cancel out with this area above that's positive.

02:34 But i still have a negative area here, which is actually negative 1 4th, actually cancels out with this area right.

02:44 Here.

02:45 So maybe i just make a trapezoid from this point to this point.

02:52 So i got to figure out the area of a trapezoid, which is one half times the base times the heights.

02:58 Well, i guess i should say times the height times the two bases, one and three.

03:04 But then i have to add to it this area over here, which i think what i would do is find the area of that little rectangle, which is a three by 3 so that's 9 but i have to subtract off this semicircle well it's not even a semicircle it's a quarter of a circle um when i write pi there so it's one -fourth of pi r squared so nine minus what would that be 9 pi over 4 and so looking at that one third one plus 3 is 4 half of which is 2 if i add this 9 to it that will be what am i looking at 7 minus 2 plus 9 would be 11 minus 9 pi over 4 but then i have to distribute that minus in there so i'm actually going to write as 9 pi over 4 7 minus 11 give me negative 4 i would just leave my answer like that so then the next one is when they ask you for the integral from 2 to 5 of 2 times f prime of x plus 4 dx.

04:25 What do i have to think about here? well, the fundamental theorem of calculus would say that that's the same thing as 2 f of 5.

04:36 I'll put in parentheses minus f of 2.

04:39 So i just put a little note there that that is the fundamental theorem of calculus.

04:45 But you still have to do the integral of 4.

04:48 So that would be 4x and plug in your two bounds for that one from 2 to 5.

04:54 I think i can get this answer because i know f of 5 is 0 and f of 2 and the graph was 3.

05:02 And then over here i'm going to get 20 minus.

05:05 I'm just plugging in my bounds 4 times 5 and 2 times 4.

05:13 So i'm looking at negative 6 plus 12, which gives me 6 on that one.

05:19 So there's b.

05:20 A was up here.

05:22 This is b.

05:24 So what do we have next? so g is a new function defined from negative 2 to x, and they want the absolute max with a justification.

05:37 Okay, so i guess what i need to think about is the integral from negative 2 to x of f of t d t.

05:45 I should really use a different color.

05:48 It is to find the derivative.

05:50 It's just the fundamental theorem of calculus.

05:52 I need to see where that equals zero...

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