This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Use the technique of antiderivatives by parts with u = x 2 and dv = sin x dx . It will take 2 iterations with a different choice of u and dv for the second iteration. ∫x 2 sin x dx = − x 2 cos x + ∫ 2 x cos x dx ( = − x 2 cos x + 2 x sin x − ∫ 2sin x dx ) = − x 2 cos x + 2 x sin x + 2 cos x + C 85. D f (3) − f (1) 5
= . True
3 −1
2
II. Not enough information to determine the average value of f. False
I. Average rate of change of f is III. Average value of f ′ is the average rate of change of f. True
86. A 1
A
B
; 1 = A( x + 3) + B( x − 1)
=
+
( x − 1)( x + 3) x − 1 x + 3
1
1
Choose x = 1 ⇒ A =
and choose x = −3 ⇒ B = − .
4
4 Use partial fractions. ∫
87. B 88. C 1
1⎡ 1
1
⎤ 1 x −1
dx = ⎢ ∫
dx − ∫
dx ⎥ = ln
+C
( x − 1)( x + 3)
4 ⎣ x −1
x+3 ⎦ 4 x+3 Squares with sides of length x. Volume =
f ( x) = ∫ x2
0 22
x dy
0 ∫ 2 = ∫ (2 − y ) dy
0 sin t dt ; f ′( x) = 2 x sin( x 2 ); For the average rate of change of f we need to determine f (0) and f ( π ) . f (0) = 0 and f ( π ) = ∫ 0π sin t dt = 2 . The average rate of 2
. See how many points of intersection there are for the
π
2
on the interval ⎡ 0, π ⎤ . There are two.
graphs of y = 2 x sin( x 2 ) and y =
⎣
⎦
π change of f on the interval is AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 226 1997 Calculus BC Solutions: Part B
89. D x t2 1 1+ t f ( x) = ∫ dt ; f (4) = ∫
5 Or, f (4) = f (1) + ∫ 4
1 x2
1 + x5 4
1 t2
1+ t5 dt = 0.376 dx = 0.376 Both statements follow from the Fundamental Theorem of Calculus. 90. B 6
65
5
5
F ( x) = kx; 10 = 4k ⇒ k = ; Work = ∫ F ( x)dx = ∫
x dx = x 2
0
02
2
4 AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 6
0 = 45 inchlbs 227 1998 Calculus AB Solutions: Part A
1. D y′ = x 2 + 10 x ; y′′ = 2 x + 10; y′′ changes sign at x = −5 2. B ∫ −1 f ( x)dx = ∫ −1 f ( x)dx + ∫ 2 4 2 4 f ( x)dx = Area of trapezoid(1) – Area of trapezoid(2) = 4 − 1.5 = 2.5
2 1 2 dx = ∫ x −2 dx = − x −1 2 1
2 3. C ∫1 4. B This would be false if f was a linear function with nonzero slope. 5. E ∫ 0 sin t dt = − cos t 0 = − cos x − (− cos 0) = − cos x + 1 = 1 − cos x 6. A Substitute x = 2 into the equation to find y = 3. Taking the derivative implicitly gives
d2
x + xy = 2 x + xy′ + y = 0 . Substitute for x and y and solve for y′ .
dx
7
4 + 2 y′ + 3 = 0; y′ = −
2 x 2 1 x x ( e 1 = ) e e
x2 − 1
1
3
⎛1
⎞
⎛1
⎞ ⎛1
⎞1
dx = ∫ x − dx = ⎜ x 2 − ln x ⎟ = ⎜ e 2 − 1⎟ − ⎜ − 0 ⎟ = e 2 −
1
2
x
x
⎝2
⎠1 ⎝2
⎠ ⎝2
⎠2 7. E ∫1 8. E h( x) = f ( x) g ( x) so, h′( x) = f ′( x) g ( x) + f ( x) g ′( x) . It is given that h′( x) = f ( x) g ′( x) .
Thus, f ′( x) g ( x) = 0 . Since g ( x) > 0 for all x, f ′( x) = 0 . This means that f is constant. It
is given that f (0) = 1 , therefore f ( x) = 1 . 9. D Let r (t ) be the rate of oil flow as given by the graph, where t is measured in hours. The total
number of barrels is given by 24 ∫0 r (t )dt . This can be approximated by counting the squares below the curve and above the horizontal axis. There are approximately five squares with
area 600 barrels. Thus the total is about 3, 000 barrels.
10. D
11. A f ′( x) = ( x − 1)(2 x) − ( x 2 − 2)(1)
( x − 1) 2 ; f ′(2) = (2 − 1)(4) − (4 − 2)(1)
(2 − 1) 2 =2 Since f is linear, its second derivative is zero. The integral gives the area of a rectangle with
zero height and width (b − a) . This area is zero. AP Calculus MultipleChoice Question Collection
Copyright © 2005 by College Board. All rights reserved. Available at apcentral.collegeboard.com. 228 1998 Calculus AB Solutions: Part A
1...
View
Full
Document
This note was uploaded on 12/29/2010 for the course MATH 214 taught by Professor Smith during the Fall '10 term at Oregon Tech.
 Fall '10
 smith
 Calculus

Click to edit the document details