g

f

g g

f

Connecting AB to BC: Area Between Curves With a slight modification, we can change the concept of finding the area of a region under a curve (that lies above the x-axis) to finding the area of a region between two curves.

Consider the following graphs of and that are continuous on the interval .

Connecting AB to BC: Area Between Curves Quick Check Quick Check 1: Find the area bound by the curves

as shown in the graph above. Shade the region that is bounded by those given equations.

( )y f x= ( )y g x= [ ]2,4

[ ]-ò3

1

( ) ( )f x g x dx ò3

1

( )f x dx ò3

1

( )g x dx

2 2, 1,x y y= - = -1, and yy x e= =

Topic: 10.1 Defining Convergent and Divergent Infinite Series

Date: March 30, 2020 AP CALCULUS BC YouTube Live Virtual Lessons Mr. Bryan Passwater Mr. Anthony Record

Topic: Unit 8

Connecting AB to BC

Area and Volume

Date: April 23, 2020

Area of region between f and g

Area of region under f = Area of region

under g

_ =

Area of a Region Between Two Curves

If f and g are continuous on [a, b] and for all x in [a, b], then the area of the region bounded by the graphs of f and g and the vertical lines and is

Connecting AB to BC: Volumes Using Known Cross Sections Our focus will now shift from using the definite integral to find areas of two-dimensional shapes to finding volumes of three-dimensional shapes. Our first technique, however, will rely on using simple two-dimensional shapes and “stacking” several of them together to form a three-dimensional figure.

Quick Check 2: The base of a solid is the region formed by the graph of , the x-axis and the line x = 4.

For each of the following, set up an integral expression that will compute the volume of the solid described.

a.) The cross sections perpendicular to the x-axis are squares.

b.) The cross sections perpendicular to the y-axis are semi-circles.

c.) The cross sections perpendicular to the x-axis are equilateral triangles.

d.) The cross sections perpendicular to the y-axis are rectangles whose heights are 4y .

y x=

x⎡⎣

⎤⎦2dx

0

4

⌠⌡⎮

12π 1

24− y2( )⎡

⎣⎢

⎤

⎦⎥

2

dy

0

2

⌠

⌡⎮

12

x( ) 32

x⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥dx

0

4

⌠

⌡⎮⎮ = 3

4x( )2⎡

⎣⎢⎢

⎤

⎦⎥⎥dx

0

4

⌠

⌡⎮⎮

4− y2( ) 4y( )⎡⎣

⎤⎦dy

0

2

⌠⌡⎮

Volumes of Solids with Known Cross Sections

1. For cross sections of area A(x) taken perpendicular to the x-axis.

2. For cross sections of area A(y) taken perpendicular to the y-axis.

Kaufmann Center for the Performing Arts, Kansas City, MO. This building is located across the street from the Kansas City Convention Center where many AP Exams, including Calculus are graded each summer.

Connecting AB to BC: Volumes by Disks and Washers

Disk Method Horizontal Axis of Revolution

Vertical Axis of Revolution

Washer Method

Horizontal Axis of Revolution

Vertical Axis of Revolution

Quick Check 3: Set up, but do not evaluate the integrals that would compute the volume of the solid described in each of the following problems.

a.) The region enclosed by the graph of , the x-axis, the line x = 1, and the line x = 4 is rotated about the x – axis.

[ ]2Volume ( )b

a

V R x dxp= = ò [ ]2Volume ( )d

c

V R y dyp= = ò

[ ] [ ]( )2 2Volume ( ) ( )b

a

V R x r x dxp= = -ò [ ] [ ]( )2 2Volume ( ) ( )d

c

V R y r y dyp= = -ò

2y x=

π 2 x⎡⎣

⎤⎦2dx

1

4

⌠⌡⎮

= π 4x dx1

4

∫

b.) The region enclosed by the graph of , the y-axis, the line y = 1, and the line y = 4 is rotated about the y – axis.

Quick Check 4: Set up, but do not evaluate the integrals that would compute the volume of the solid described in each of the following problems.

a.) The region enclosed by the graphs of and from x = 0 to x = 1 is rotated about the x – axis.

b.) The region enclosed by the graphs of and the line y = x is rotated about the line .

2/3y x=

y = x2 3 ⇒ x = y3 2

π y3 2⎡⎣ ⎤⎦2dy

1

4

⌠⌡⎮

/2xy e= 2( 1)y x= -

π ex 2( )2 − x −1( )2( )2⎡

⎣⎢

⎤

⎦⎥dx

0

1

⌠

⌡⎮

23y x x= -1y = -

π 3x − x2( )+1( )2 − x +1( )2⎡⎣⎢

⎤⎦⎥dx

0

2

⌠⌡⎮

Quick Check 5:

Consider the graph of shown above where a and b are positive real numbers. Match the integral setup on the left with its corresponding axis of revolution on the right.

(a) (iv) (i) x - axis

(b) (i) (ii) y -axis

(c) (ii) (iii) y = b

(d) (iii) (iv) x = a

( )y f x=

[ ]20

( )b

V a f y dyp= -ò

[ ]20

( )a

V f x dxp= ò

[ ]( )22

0

( )b

V a f y dyp= -ò

[ ]( )22

0

( )a

V b b f x dxp= - -ò

( )y f x= ( )x f y=

Free Response Practice: Connecting AB to BC 2020 FRQ Practice Problem BC1

BC1 The functions f and g are twice differentiable for all real values of x where g(x) = x. A portion of the graphs of f and g are given in the figure above. It is known that the average value of over the interval is equal to c. (a) Find the area of the region R in terms of c.

(b) The region R is the base of a solid whose cross sections perpendicular to the x-axis form rectangles with height . Find the volume of this solid in terms of c.

f[ ]0,c

f x( )− x⎡⎣ ⎤⎦dx0

c

⌠⌡⎮

= f x( )⎡⎣ ⎤⎦dx0

c

⌠⌡⎮

− x dx0

c

∫ = c2 − 12c2 = 1

2c2

average value = 1c

f x( )⎡⎣ ⎤⎦dx0

c

⌠⌡⎮

= c⇒ f x( )⎡⎣ ⎤⎦dx0

c

⌠⌡⎮

= c2 x dx0

c

∫ = 12x2⎡

⎣⎢

⎤

⎦⎥

0

c

= 12c2

( )f x¢

volume = f x( )− x( ) ′f x( )( )⎡⎣

⎤⎦dx

0

c

∫ = f x( ) ′f x( )− x ′f x( )⎡⎣ ⎤⎦dx0

c

∫

= f x( ) ′f x( )⎡⎣ ⎤⎦dx0

c

∫ − x ′f x( )⎡⎣ ⎤⎦dx0

c

∫ = 12f x( )⎡⎣ ⎤⎦

2

0

c

− x f x( )⎡⎣ ⎤⎦ 0

c+ f x( )⎡⎣ ⎤⎦dx

0

c

∫

= 12f c( )⎡⎣ ⎤⎦

2− 1

2f 0( )⎡⎣ ⎤⎦

2− c f c( )⎡⎣ ⎤⎦ + c

2 = 12c2 − 1

2c2

⎛⎝⎜

⎞⎠⎟

2

− c2 + c2 = 38c2

u = f x( )⇒ du = ′f x( )dxdv = ′f x( )dx⇒ v = f x( )

u = x⇒ du = dxdv = ′f x( )dx⇒ v = f x( )

f x( ) ′f x( )⎡⎣ ⎤⎦dx∫ = f x( )⎡⎣ ⎤⎦2− f x( ) ′f x( )⎡⎣ ⎤⎦dx∫ x ′f x( )⎡⎣ ⎤⎦dx∫ =

2 f x( ) ′f x( )⎡⎣ ⎤⎦dx∫ = f x( )⎡⎣ ⎤⎦2

x f x( )⎡⎣ ⎤⎦ − f x( )⎡⎣ ⎤⎦dx∫

f x( ) ′f x( )⎡⎣ ⎤⎦dx∫ = 12f x( )⎡⎣ ⎤⎦

2

The problem has been restated. BC1 The functions f and g are twice differentiable for all real values of x where g(x) = x. A portion of the graphs of f and g are given in the figure above. It is known that the average value of over the interval is equal to c.

(c) The function H is defined by . Find and in terms of c. Does the

graph of H have a relative minimum, relative maximum, or neither at . Give a reason for your

answer.

(d) For , the function f is linear and intersects the x-axis at . Find

2 points on the function f are

(e) Write but do not evaluate an expression involving one or more integrals that gives the perimeter of R .

f[ ]0,c

H (x) = 2x2 − f (t)dt0

2x

∫ ′Hc2

⎛⎝⎜

⎞⎠⎟ 2

cH æ ö¢¢ç ÷è ø

x = c2

H (x) = 2x2 − f (t)dt0

2x

∫ ′H x( ) = 4x − 2 f 2x( ) ′′H x( ) = 4− 4 ′f 2x( )

′Hc2

⎛⎝⎜

⎞⎠⎟= 4

c2

⎛⎝⎜

⎞⎠⎟− 2 f c( ) = 2c − 2c = 0 ′′H

c2

⎛⎝⎜

⎞⎠⎟= 4− 4 ′f c( ) > 0 because ′f c( ) < 0

The graph of H has a relative minimum at x = c2

because ′Hc2

⎛⎝⎜

⎞⎠⎟= 0

and ′′Hc2

⎛⎝⎜

⎞⎠⎟> 0 or concave up.

x c³ 53cx = ( ).f c¢

c,c( ) and 5c3

,0⎛⎝⎜

⎞⎠⎟⇒ ′f c( ) = c − 0

c − 5c3

= 3c3c −5c

= − 32

Perimeter = c2+ c2 + c2 + 1+ ′f x( )⎡⎣ ⎤⎦

2dx

0

c

⌠⌡⎮

= c2+ c 2 + 1+ ′f x( )⎡⎣ ⎤⎦

2dx

0

c

⌠⌡⎮

The problem has been restated. BC1 The functions f and g are twice differentiable for all real values of x where g(x) = x. A portion of the graphs of f and g are given in the figure above. It is known that the average value of over the interval is equal to c. (f) For , . Find the second degree Maclaurin polynomial for H(x) and use it to

approximate .

(g) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line .

f[ ]0,c

0x £ ( ) 2cos( )f x x=

2cH æ ö

ç ÷è ø

f x( ) = 2cos x( ) ′f x( ) = −2sin x( )P2 x( ) = H 0( )+ ′H 0( )x + ′′H 0( )

2!x2 H (x) = 2x2 − f (t)dt

0

2x

∫ ⇒ H (0) = 0

P2 x( ) = 0+ −4( )x + 42!x2 ′H (x) = 4x − 2 f 2x( )⇒ ′H (0) = −2 f 0( ) = −4

P2 x( ) = −4x + 2x2 ′′H (x) = 4− 4 ′f 2x( )⇒ ′′H (0) = 4− 4 ′f 0( ) = 4

H c2

⎛⎝⎜

⎞⎠⎟≈ P2

c2

⎛⎝⎜

⎞⎠⎟= −4 c

2⎛⎝⎜

⎞⎠⎟+ 2 c

2⎛⎝⎜

⎞⎠⎟

2

= −2c + 2 c2

4⎛⎝⎜

⎞⎠⎟= c

2

2− 2c

2y = -

volume = π f x( )+ 2( )2− x + 2( )2⎡

⎣⎢⎤⎦⎥dx

0

c

⌠⌡⎮

2020 FRQ Practice Problem BC2

BC2 A cellular relay tower has a height of 96 feet. The horizontal cross sections of the tower at height h feet form circles with diameters D(h), measured in feet. The function D is differentiable and decreases as h increases. Selected values for D(h) are given in the table above.

(a) Explain why there must be a value for h on the interval such that .

(b) Is there a value for h on the interval such that ? Explain your reasoning.

(c) Approximate the rate of change of the radius of the tower with respect to h at h = 85.

(d) Write an integral expression in terms of D(h) that represents the volume of the cell tower.

0 96h£ £ 1( )30

D h¢ = -

On the interval 0 ≤ h ≤ 60 which is part of the interval 0 ≤ h ≤ 96,

D h( ) has an average rate of change of D 60( )− D 0( )

60− 0= 3.0−5.0

60= −2

60= − 1

30D h( ) is differentiable and is therefore continuous so you can apply the Mean Value Theorem.

There is a least one number h in 0 < h < 60 such that ′D h( ) = D 60( )− D 0( )60− 0

= − 130

0 96h£ £ ( ) 3.3D h =

D h( ) is continuous on the interval 40 ≤ h ≤ 60 so by the Intermediate Value Theorem,

D 60( ) < 3.3< D 40( ) then there is number h, 40 < h < 60 such that D h( ) =3.3

R x( ) = 12 D x( )⇒ ′R x( ) = 12 ′D x( )

′R 85( ) = 12 ′D 85( ) ≈ 12D 90( )− D 80( )90−80

⎛

⎝⎜

⎞

⎠⎟ =

2.5( )− 2.7( )20

= −0.220

= −0.01

V = π 12D h( )⎡

⎣⎢

⎤

⎦⎥

2

dx

0

96

⌠

⌡⎮

h (feet) 0 20 40 60 80 90 96

D(h) (feet) 5.0 4.0 3.4 3.0 2.7 2.5 2.4

h

The problem has been restated.

BC2 A cellular relay tower has a height of 96 feet. The horizontal cross sections of the tower at height h feet form circles with diameters D(h), measured in feet. The function D is differentiable and decreases as h increases. Selected values for D(h) are given in the table above.

(e) Approximate the volume of the tower between h = 0 and h = 96 using a midpoint Riemann sum with three subintervals as indicated in the table.

(f) The diameter, in feet, of the horizontal cross section at height h feet is modeled by the function given by

. The height h of a bug walking vertically up the side of the tower is changing at a

constant rate of feet per second . Find the rate that the diameter of the horizontal cross sections is changing with respect to time when the bug is 44 feet above ground.

V = π 12D h( )⎡

⎣⎢

⎤

⎦⎥

2

dx

0

96

⌠

⌡⎮ ≈ π 1

2D 20( )⎛

⎝⎜⎞⎠⎟

2

40( )+ 12D 60( )⎛

⎝⎜⎞⎠⎟

2

40( )+ 12D 90( )⎛

⎝⎜⎞⎠⎟

2

16( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= π 124( )⎛

⎝⎜⎞⎠⎟

2

40( )+ 123( )⎛

⎝⎜⎞⎠⎟

2

40( )+ 122.5( )⎛

⎝⎜⎞⎠⎟

2

16( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= π 160( )+ 90( )+ 25( )⎡⎣ ⎤⎦ = 275π = 863.9379…

f

32 2 20( )5hf h - +

=

110

dDdt

= dfdh

⋅ dhdt

⇒ dDdt

= ′f h( ) 110

⎛⎝⎜

⎞⎠⎟= 110

′f h( )

′f h( ) = 15 −2( ) 12 h+ 20

⎛⎝⎜

⎞⎠⎟= − 15 h+ 20

⇒ dDdt h=44

= − 15 44+ 20

110

⎛⎝⎜

⎞⎠⎟= − 1400

h (feet) 0 20 40 60 80 90 96

D(h) (feet) 5.0 4.0 3.4 3.0 2.7 2.5 2.4

h

2020 FRQ Practice Problem BC3

BC3 Let R be the shaded region given in the figure above. The region R is bounded by the graph of the differentiable function where . The graph of contains the points .

(a) Is there a value c on such that . Explain your reasoning.

(b) If R is rotated about the x-axis to form a solid, use a right Riemann sum with four equal subintervals as indicated in the graph to approximate the volume of the solid.

(c) Using a midpoint Riemann sum with three equal subintervals indicated in the graph, approximate the length of the curve of f (x) from x = 3 to x = 9.

( )y f x= (2) (10) 0f f= = ( )y f x=(4,2), (6,3), and (8,4)

2 10x< < 2( )3

f c¢ =

f x( ) is differentiable and is therefore continuous so you can apply the Mean Value Theorem.

There is a least one number c on 2 < x < 8 which is part of the interval 2 < x <10

such that ′f c( ) = f 8( )− f 2( )8− 2

= 4− 06

= 46= 2

3

V = π f x( )⎡⎣ ⎤⎦2dx

0

10

∫ ≈ π f 4( )( )2 2( )+ f 6( )( )2 2( )+ f 8( )( )2 2( )+ f 10( )( )2 2( )⎡⎣⎢

⎤⎦⎥

= π 2( )2 2( )+ 3( )2 2( )+ 4( )2 2( )+ 0( )2 2( )⎡⎣⎢

⎤⎦⎥ = π 8+18+ 32⎡⎣ ⎤⎦ = 58π

L = 1+ ′f x( )⎡⎣ ⎤⎦2dx

3

9

⌠⌡⎮

≈ 1+ ′f 4( )⎡⎣ ⎤⎦22( )+ 1+ ′f 6( )⎡⎣ ⎤⎦

22( )+ 1+ ′f 8( )⎡⎣ ⎤⎦

22( )⎡

⎣⎢⎤⎦⎥

= 1+ 38⎡

⎣⎢

⎤

⎦⎥

2

2( )+ 1+ 35⎡

⎣⎢

⎤

⎦⎥

2

2( )+ 1+ 0⎡⎣ ⎤⎦22( )

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

= 1+ 964⎡

⎣⎢

⎤

⎦⎥ 2( )+ 1+ 9

25⎡

⎣⎢

⎤

⎦⎥ 2( )+ 2( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 7364

2( )+ 3425

2( )+ 2( )⎡

⎣⎢⎢

⎤

⎦⎥⎥= 1473 + 2

534 + 2

x 3 ? 4 5 6 7 ? 8 0 9 ?

( )f x¢

381235

BC3 Let R be the shaded region given in the figure above. The region R is bounded by the graph of the differentiable function where . The graph of contains the points . (d) Use Euler’s method, starting at x = 6 with three steps of equal size to approximate .

( )y f x= (2) (10) 0f f= = ( )y f x=(4,2), (6,3), and (8,4)

(3)f

x f x( ) dy = ′f x( ) −1( )6 3 − ′f 6( ) = − 3

5= −0.6

5 2.4 − ′f 5( ) = − 12= −0.5

4 1.9 − ′f 4( ) = − 38= −0.375

3 1.525

x 3 ? 4 5 6 7 ? 8 0 9 ?

( )f x¢

381235