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

Graph of f 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} (10^x - 3f'(x)) ? (f(x) - arctan x).

Transcript

00:01 Okay, so let's get started with part a of our exercise.

00:06 Well, here we have that the integral from negative 6 to 5 of f of x in the x.

00:18 Well, this guy is the integral from negative 6 to negative 2 of f of x in the x plus the integral from negative 2.

00:29 To 5 of f of x in the x well let's compute this guy first since we are going to need it okay so the integral from negative 2 to 5 of f of x in the x can be computed geometrically well this one is an integral from negative 2 to 0 of f of x in the x which geometrically is just 1 okay perfect then we are gonna have an integral from 0 to 2 okay so an integral from 0 to 2 2 of f of x in the x okay plus an integral from 2 to 5 of f of x in the x okay now let's compute the integral from 0 to 2 well this one is just what well well well we know that f is a line between 0 and 2.

01:34 So let's just find the question of our line.

01:38 Okay, now what is the question of our line between 0 and 2? okay, so let's compute the slope of this line.

01:48 Well, the slope of this line is 4 over 2, so is 2.

01:55 So let me write like this, f of x is 2 multiplied by x minus 1 with x belonging to the interval 0 2.

02:07 Okay perfect.

02:09 So at this point this integral here is easy to compute.

02:15 Well here an antiderivative of 2x minus 1 is x squared minus x.

02:22 So this guy is just 2.

02:26 Okay perfect.

02:27 Finally for this one here well we just need to use the fact that f is a circle centered at the point five three for x belonging to the interval two five okay so in this case what are we gonna have well this guy is gonna be what well this one is gonna be the area of the square with side given by two five and 03 so 9 minus 1 fourth of the area of this circle with radius 3 which is okay so the radius is 3 divided by 4 so we are going to have 9 pi okay 9 pi over 4 perfect okay so what do we get well we get the that the integral from negative 2 to 5...

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