APR6 - Solutions
1998 A P Calculus AB Scoring Guidelines
1. Let R be the region bounded by the x-axis, the graph of y = &, and the line x = 4.
(a) Find the area of the region R.
(b) Find the value of h such that the vertical line x = h divides the region R into two regions
of equal area.
(c) Find the volume of the solid generated when R is revolved about the x-axis.
(d) The vertical line x = k divides the region R into two regions such that when these two
regions are revolved about the x-axis, they generate solids with equal volumes. Find the
value of k.
8
(b) l oh & d x = - 3
i,
:
lh&dx=A4&dx
1: limits and constant
1: integrand
1: answer
k
(d) a
J0 (&12
k2
.
dx = 4a
a
Jn
k
4
( 6 1 2 dx = a
2r-
:
k2
_
k2
(&12
dx
1
1: equation in k
I 2
-
I:
answer
APR6
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AP@CALCULUS AB
2003 SCORING GUIDELINES (Form B)
APR6 - Solutions
Question 1
Let f be the function given by f ( x ) = 4x2 - x3, and let
e
be the line y = 18 - 32, where I is tangent to the
graph of f. Let R be the region bounded by the graph of
f and the x-axis, and let S be the region bounded by the
graph of f, the line
I, and the 5axis, as shown above.
(a) Show that e is tangent t o the graph of
the point x = 3.
IJ
= f(x) at
(b) Find the area of S.
(c)
Find the volume of the solid generated when R is revolved about the x-axis.
1 : finds f1(3) and f ( 3 )
(a) f l ( x ) = 8 2 - 3 x 2 ; f 1 ( 3 )= 24 - 27 = -3
f(3) = 36 - 27 = 9
or
2 : .
Tangent line at x = 3 is
y = -3(x - 3 )
-finds equation of tangent line
1: -
+ 9 = -32 + 18,
which is the equation of line
e.
( b ) f ( x ) = 0 at x = 4
The line intersects the x-axis at x = 6.
1
4
Area = - ( 3 ) ( 9 )- J3 ( 4 r 2 - :I;'' ) dx:
2
= 7.916 or 7.917
shows (3,9) is on bot h the
graph o f f and line
e
2 : integral for non-triangular region
1 : limits
4 : .
1 : integrand
1 : area of triangular region
1 : answer
OR
Area = L 4 ( ( 1 8 3 x 3 - ( 4 x 2 - x 3 ) ) d x
( c ) Volume =
=
11h4( 4 x 2
-
x3
dx
1 5 6 . 0 3 8 ~or 490.208
3:
I
1 : limits and constant
1 : integrand
1 : answer
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APR6
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