3. Let R be the region in the first quadrant enclosed by...

3. Let R be the region in the first quadrant enclosed by the graphs of f(x) = 8x^3 and g(x) = sin(?x), as shown in the figure above. (a) Write an equation for the line tangent to the graph of f at x = 1/2. [2] (b) Find the area of R. [4] (c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is rotated about the horizontal line y = 1. [3]

Transcript

00:01 So, to answer letter a, all you need to do is utilize f of x.

00:09 So, they tell you in the problem f of x is equal to 8x cubed.

00:15 And if you look at the graph, they even tell you that when x is 1 half, you have the y coordinate of 1, which makes perfect sense because 1 half cubed is 1 eighth times 8 is 1.

00:27 So, now what you need is the derivative of this, which is going to be your power rule, 8 times 3 is 24, and then you subtract 1 from the exponent.

00:38 And now what you need is the derivative when x is 1 half.

00:42 So, you're plugging in 1 half in for x, and 1 half squared is 1 fourth, and 1 fourth of 24 is 6.

00:53 So, your tangent line, i'm a big fan of point slope form, is y minus the y coordinate, the y coordinate is 1, equals the slope we just found, and then x minus the x coordinate of 1 half.

01:06 Now, to me, this is a good enough answer, but if you have a picky, you know, grader, you might have to distribute that 6, and then to get that minus 1 to the other side, add 1 over.

01:20 So, as long as my arithmetic is correct, i got y equals 6x minus 2, but either of these answers are correct.

01:27 So, both are correct.

01:32 So, if i go on to letter b, what i'm seeing there is the area.

01:38 Well, we're going to do for the area, the integral from 0 to 1, because i think you can tell they intersect, sorry, 0 to 1 half, and i'm just a big fan of doing the upper function, which should be sine of pi x, minus the lower function, which is the 8x cubed function, dx.

02:06 So, in order to evaluate this integral, you need to know that the integral of cosine, of sine is negative cosine, and think the chain rule for a second that this would be pi x, but with the pi x, with the chain rule, we typically multiply by the derivative of the inside.

02:24 Well, now we're going to multiply by the reciprocal of the derivative of the inside.

02:29 And then this thing, now what you're doing is you're adding 1 to the exponent, and then 8 divided by 4 would give you 2, and then you have to do both of these from 0 to 1 half.

02:40 I usually put in parentheses in here.

02:41 So now what you need to do is plug in 1 half for both those bounds, and hopefully you know the unit circle pretty well, that cosine of pi over 2 is 0...

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