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Calculus Acrostic – AB Integrals Place your answer (CAPS ONLY) in the appropriate cell on the included spreadsheet. The grid containing a famous quote and author will automatically register. Every attempt to answer the question will register as a guess. The goal is to solve the puzzle in as few guesses as you can. When you think you have the puzzle solved, you can enter the words (IN CAPS) in the box on the right. Only when you get every word correct will you be told that the puzzle has been solved. If your teacher gives you the correct code, you can enter it to get instant feedback for every letter and word. No graphing calculators.

QUIZ 18 – Area and Volume 50. The shaded region in the figure to the right is the area enclosed by the graph of

and the line tangent to f at x = 2. Find this area.

D) 32 G)

S) T) 16

51. In the figure to the right, the line is tangent to the curve at x = 1. What is the area

of the shaded region?

A) E)

I) O)

52. The graph of is rotated about the x-axis on the interval . Find the volume of the solid generated.

D) S) T) Y)

53. Which of the following gives the shaded area as shown on the figure to the right?

i. ii.

H) i and ii I) i only O) ii only U) neither

f x( ) = x3 − 2x2 + 4x +1643

163

y = ex

e2− 1e

2e− 2e

e− 1e

e− 2e

y = sin x( )+ cos x( ) 0,3π 4⎡⎣ ⎤⎦

π 2 π 2 −1( ) 2π π 1+ 2( )

1

e

∫ ln xdx +e

2e

∫ 2−xe

⎛⎝⎜

⎞⎠⎟dx

0

1

∫ 2e− ex − ex( )dx

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54. Find a continuous function f such that the area under the graph of f from 0 to t is for all values of t.

A) E)

I) O) 55. The region bounded by the curves and the y-axis is shaded in the figure

to the right. If this region is rotated about the line , which of the following represents the volume of the solid?

L) M)

S) T)

56. Region R is bounded by the functions and the y-axis, as

shown by the figure to the right. Region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a semi-circle. Which of the following represents the volume of the solid?

C) D)

M) V)

t sin t2

f x( ) = −cos x2

2f x( ) = 1− cos x

2

2f x( ) = 2x2 cos x2 + sin x2 f x( ) = −2cos x2

y = tan x, y = 3y = 3

V = π0

π /3

∫ 3− tan x( )2 − 3− 3( )2⎡⎣⎢⎤⎦⎥dx V = π

0

π /6

∫ 3 − tan x( )2⎡⎣⎢⎤⎦⎥dx

V = π0

π /3

∫ 3− 3( )2 − tan2 x⎡⎣⎢⎤⎦⎥dx V = π

0

π /3

∫ 3 − tan x( )2⎡⎣⎢⎤⎦⎥dx

f x( ) = sin−1 x, g x( ) = cos−1 x,

π8 0

π /4

∫ cos−1 x − sin−1 x( )2 dx π4 0π /4

∫ cos−1 x( )2 − sin−1 x( )2⎡⎣⎢⎤⎦⎥dx

π8 0

2/2

∫ cos−1 x − sin−1 x( )2 dx π4 02/2

∫ cos−1 x( )2 − sin−1 x( )2⎡⎣⎢⎤⎦⎥dx