Expert: Paul Klarreich - 4/9/2006

Question
Hi Paul,
Please go to:
http://apcentral.collegeboard.com/repository/ap03_sg_calculus_ab_b_26473.pdf
This is the collegeboard website
I just took this AP exam and have two questions which I'm hoping you can answer.
By question#1, part b, I don't understand any of their methods and am hoping you can shed light on what they did over there. I did the integral of (18-3x)-(4x^2-x^3) from 4 to 6 which equals 63.333,which is not the correct answer.
My second question is on #3 parts c and d. By c, how did they know that the integral of area is volume?And by c, what did they do?
Thank you for your time.

Answer
Hi, Jeff,

You wrote:
Subject:  AP Exam 2003 form B
Question:  Hi Paul,
Please go to:
http://apcentral.collegeboard.com/repository/ap03_sg_calculus_ab_b_26473.pdf
This is the collegeboard website
I just took this AP exam and have two questions which I'm hoping you can answer.
By question#1, part b, I don't understand any of their methods and am hoping you can shed light on what they did over there. I did the integral of (18-3x)-(4x^2-x^3) from 4 to 6 which equals 63.333,which is not the correct answer.
My second question is on #3 parts c and d. By c, how did they know that the integral of area is volume?And by c, what did they do?

>> [you mean d?]

Thank you for your time.
---------------------------

For question 1b, you are finding the area between two 'curves'.  While the upper curve is the line L, y = 18 - 3x, the lower curve is not always the same.

It is the graph of  f = 4x^2 - x^3 for the region to the left of its x-intercept.  It is the x-axis (or  y = 0) from that point to the x-intercept of the line, which they find to be  x = 6.  

The first step will be to find the x-intercept of the graph of f, because that is the dividing point.  They found  x1 = 4 for this.

So you are integrating

}x=6
|    (top - bottom) dx
{x=3

but, while 'top' is always  18-3x, 'bottom' varies and you must write:

}x=4
|    (18 - 3x - (4x^2 - x^3)) dx
{x=3

PLUS

}x=6
|    (18 - 3x - zero) dx
{x=4


The first integral is:

}x=4
|    (18 - 3x - 4x^2 + x^3) dx
{x=3

= 18x - 3x^2/2 - 4x^3/3 + x^3/4  from 3 to 4

and the second is:

18x - 3x^2  from 4 to 6.

Hopefully you can work those out and get their answer.  HOWEVER, you were asking me if I can decipher their sneaky ways of doing it.

Yes, I can.  In their diagram, put these labels on points:

Let A be the point at the top where L and f(x) are tangent.  Drop a VERTICAL LINE from A down to the x-axis and call that point B.  Call the x-intercept of f(x) = C, and the x-intercept of the line D.  So B,C,D, are all on the x-axis.

B is ((3,0), C is (4,0), and D is (6,0).

Now there is a portion of area R which is between that VERTICAL line and the function.  It is triangle-ish and bounded by AB, BC, and the curve AC.  You could determine its area by integrating f(x) from 3 to

4.  If you subtracted that area from the triangle ABD, which is easily found, you get the answer for S.

OR, you could do this:  From C, draw a vertical line UP to meet the line at E.  This DIVIDES area R into a part to the left, that you integrate, (it is the first of our integrals, above.) and a triangle CED, whose area you can easily find.

So the (1/2)(3)(9) and the (1/2)(2)(18-12) things you see are just computations for the area of a triangle, 1/2 base * height.
-------------------------
About 3c,d:

3c:  You are doing a volume of revolution by the method of disks.  The way to do that is to find the volume of one disk element, which is a (thin) cylinder :

V = pi r^2 h

dV = pi r^2 dx, and in this case, the radius is diameter/2,

so, since they make B(x) the diameter,

dV = pi[B/2]^2 dx

and if that is integrated from x = a to x = b, it is the volume of revolution between a and b.

3d: If a function B(x) behaves itself (continuous, differentiable,.. you know the drill) on an interval, then if you want to prove that B'(x) = 0 somewhere, find two points where B has the same values.  In this case, you could do it twice.  

The table gives B = 30 at two places, x = 60 and 180.  So B' must be zero at some c1 between them.  Also you find B = 26 at x = 240 and 360.  So B' must be zero at some c2 between them, too.  And the c1 and c2 are not the same, because these intervals don't overlap.

But now we have B' having the same value (zero) at two different points, c1 and c2.  So B'' must be zero somewhere between c1 and c2.

You use the MVT (actually it's Rolle's Theorem, but who's counting?) twice in this problem.