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let r be the region of the first quadrant enclosed by the graphs of f of x is equal to 8 x to the third,
and g of x is equal to sine of pi x as shown in the figure above, and they drew the figure right over
here. Part A: Write the equation for the line tangent to the graph of f if x is equal to 1/2. So let
me just redraw it here, just so that, I like drawing on a black background, I guess that's the main reason im
why I'm redrawing it, so the function f of x is 8, is equal to 8 x to the third, looks like this, it looks
like that, this is our f of x axis this is our x axis, and we want an equation for the line tangent at
x is equal to 1/2, so this is x is equal to 1/2, go up here, if you evaluate f of 1/2, you get you f-you
get 8 times 1/2 to the third, which is 8 times one-eighths, which is 1, they actually gave us
that already on this point, which is 1/2, comma 1, and we need to find the equation for the tangent line.
The tangent line will look something like that, and to figure out its equation, we just really figure
out its slope, and then we know a point that its on, and we can either use point-slope, or you can just
use your kind of standard slope intercept form to give an equation for that line. So the first part,
lets figure out its slope, and the slope of the tangent line is going to be the same slope as the slope
of our function at that point, or another way to think about it, it is going to be f prime, f prime of
1/2, or the derivative evaluate at 1/2. The derivate gives us the slope of that line at any point, so
what is f prime of x? F prime of x is just the derivative of this, so 3 times 8 is 24, times x squared,
24 x squared, f prime of 1/2 is equal to 24 times 1/2 squared, which is equal 24 times 1/4, which is
equal to 6. So the slope of this line is equal to 6, I'll use m for slope which will be, we mentioned
that we would be using algebra, we first learned it in algebra, so the slope is going to be 6, so the
general equation of this line is y is equal to mx plus b, this is the slope, this is the y-intercept,
we already know that the slope is 6, and then we can use the fact that the line goes through the point
1/2, 1 to figure out what b is, so when y is 1, 1 is equal to our slope times x, x is 1/2, or another way
to say it, when x is 1/2, y is 1 plus some, plus some y-intercept, or if I take x is 1/2, I multiply
it by the slope, plus the y-intercept, I should get 1. So I get 1 is equal to 3 plus b, I can subtract
3 from both sides, and I get negative 2 is equal to b. So the equation of the line is going to be y is
equal to 6 x minus 2. That is the equation of the tangent line. Now part B: Find the area of r. So r
is this region right over here, is this region right over here, it's bounded on-above by g of x, which
is the sine of pi x, it's bounded below, it's bounded below by f of x, or 8 x to the third. So the area,
the area is going to be, so I'll just scroll down a litle bit, I still want to be able to see this graph
over here. Part b, part b. The area of r is going to be equal to the integral, from 0, from 0, that's
this point of intersection right over here, to 1/2, so I'll make it here, this is 0 to 1/2, 0 to 1/2,
and then the function on the top, so we could just take the area of that, but we'll have to subtract
from that the area underneath the function below, or one way to think about it, the integral from 0 to
1/2 of the top function is g of x, which is sine of pi x, so the sine of pi x, the sine of pi x, but
if we just evaluated this integral, let me just put a d x over here, if we just evaluted this we would
get the entire area of this entire region, but what we need to do is subtract out the area underneath
the second, underneath f of x, let me just subtract the area under that, so we just subtract from that
f of x, and f of x we already saw is 8 x to the third, is 8 x to the third power. And now we can just
evaluate this, so let me draw a little line here, it's getting a little messy, it's getting a little bit messy,
I'll just do it down here, so we need to take the anti-derivative of sine of pi x. Well, the derivative
of, the derivative of cosine of x is negative sine x, the derivative of cosine of pi x is negative pi
cosine of pi x, so the derivate- the anti-derivative of sine of pi x is going to be negative 1 over pi
cosine, cosine of pi x. And you can verify it for yourself, and you might say, "How do you know it was
a negative?" Well, I put the negative there, so that when I take the derivative of the cosine of pi x,
I would get a negative sign, but that negative will cancel out the negative to get a postive here, and
you say, "Why did you put a one over pi here?" Well, when you take the derivative of this this, using
the chain rule, you take the derivative of the pi x, you'll get pi, and you would multiply everything
by, and then you would get negative sine of pi x, and that pi doesn't show up here, so I need something
for it to cancel out with, and that's what this one over-one over pi is going to cancel out with. And
you can do u substitution, and all the rest, if, if, if you found that, if you found something like that
useful, but it's in general a good habit, or I guess it's a good-it's good to be able to do this almost,
almost by sight, so...
and you can verify that this derivative is equal to sine of pi x
so the anti-derivative of sine of pi x is this,
the anti-derivative of negative 8x to the 3rd power is negative 8
I'm going to divide it by 4, so negative 2x to the 4th power
and I just, all I did is I incremented the 3 to a 4
and then I divided the 8 by the 4
and you can take the derivative of this
to verify that it is the same thing as negative 8x to the 3rd power
Now we're going to have to evaluate that from 0 to 1/2.
When you evaluate at 1/2...so I'm going to get...I'm going to get
negative 1 over pi, cosine of pi over 2
cosine of pi over 2 minus 2 times 1/2 to the 4th power is 1/16
so that's it evaluated at 1/2
and then from that I'm going to subtract negative 1 over pi cosine of 0
let me just write it out...
so minus negative 1 over pi, negative 1 over pi cosine of pi times 0
cosine of pi, let me just write, cosine of 0 pi I could write or pi times 0
minus 2 times 0 to the 4th
so that's just going to be minus 0
so let's evaluate this...
so to simplify it, we have a cosine of pi over 2
cosine of pi over 2, is just going to be 0. So this whole thing just becomes 0
and then you have a negative 2 divided by 16, that's negative 1/8
negative 1/8...
and then from that I'm going to subtract this business here
cosine of 0, this is 1. So this is just a negative 1 over pi, a negative 1 over pi
and then I have a 0 there, so I can ignore that
so this is equal to negative 1 over 8, plus, plus 1 over pi
and we are done.
This part of it, you're not allowed to use a calculator
so this is about as far as I would expect them to get
to, to, for, eh... about as far as I would expect them, to expect you, uh, to get
and so I'll leave you there
in the next video we will do part... we will do part C