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Unformatted text preview: AP Calculus AB
Review: Week 3: April 21" 2003 —April 25‘” 2003. Sample Review Questions: Non Calculator:
Integrals & Antiderivatives: Concepts and Techniques
#1, 3, 6,18, 24, 25. 1. r Multiple Choice: Approximately 2 minutes per question 2. Multiple Choice: A calculator may be used: Approximately 2.6 minutes per question. Integrals & Antiderivatives:
Concepts & techniques
# 78, 88, 89, 90 3. Free Response: 3 Questions: Approximately 15 minutes each.
Integral as an accumulator: [Calculator]
Function deﬁned as an Integral [Calculator]
Approximation of an Integral [Calculator]  Average Value of a function
Fundamental Theorem of Calculus
Deﬁnite Integral Approximations 4. Explorations: E
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l A,w<,.__I._._W_.._._____M...,ﬂ__mm.w.~u...44_...W._.M.n. 4.4.4.». REVIEW: WEEK 3: MULTIPLE CHOICE: NON CALCULATOR 1. [12(4x3 —6x)dx= OA. 2 QB. 4 Oc. 6 OD. 36 QB. 42 3. If]: f(x)dx=a+2b, then I: (f[x)+5)dx= . , Op; a+2b+5
OB. Sb—Sa
QC. 7b—4a
OD. 712—54 OE. Tb—t’m O
tn
(b
+
Q 24. The expression 1 + 4% +1’% +...+'1 is a Riemann sum approximation for QB". J:ng OB. LIJJ—cdx ‘ Oc. on LJdVEdX
‘ so 0 OE ‘mfozx
. EU X REVIEW: WEEK 3: MULTIPLE CHOICE WITH A CALCULATOR V
I I I I
I I l I
3 —‘_T—T‘T_—T‘"'
I I l I
I I I
I I I 2 "f‘_“‘T““‘T‘' 0& 0.3 on. 1.3 Qc. 3.3 OD. 4.3 OE. 5.3 88. Let Jﬁx) = k(:)dt, where I: has ﬁne graph shown above. Which of the following
could be the graph of f? ' ‘I DA. [I], 2] are used, which of the followhg is the trapezoidal approximation of I: f[x)dx? OA. 8 013‘ 12 0c. 16 OD. 24 OE. V 90. Which of the following are anﬁdenvatives of f(x) = sin xcos x?  2
1. F[x) “12" 2 II. F[x) C0: x cos[2x) III. F(x) = 4 O A. I onlyr
O B. II only
Q C. III only 0 D. I and III only
0 E. II and III only 32 a I. The temperature outside a house during a 24hour period is given by I F(t)=80—10cos 052524, where F0) is measured in degrees Fahrenheit and t is measured in hours. (a) Sketch the graph of F on the grid to the right. .K 100
‘32 (b) Find the average temperature, to the nearest
’1  degree Fahrenheit, between r= 6 and r= l4. :5. 90
(+1, (c) An air conditioner cooled the house Whenever the outside temperature was at or above 78 degrees i: so
13 Fahrenheit. For What values of r was the air Q . . .
E conditioner cooling the house? 5’ 7o
 (d) The cost of cooling the house accumulates at the
3 rate of $0.05 per hour for each degree the outside 60
‘3 temperature exceeds 78 degrees Fahrenheit. What 0 was the total cost, to the nearest cent, to cool the house for this 24hour period? I
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I Time in ﬂours Cinema/Iron: Z. The graph of a differentiable function f on the closed interval [1, 7] is shown above. Let
W): fwd: for 15x57. (3) Find ha). Co) Find h'(4). (c) On what interval or intervals isthe graph of k concave upward? Justify your answer. (01) Find the value of x at which h has its minimum on the closed interval [1, 7]. Justify your
answer. 5 I 52025 3035404550
“ma(seconds) The graph of the velocity v(:), in ftfsec, of a car traveling on a straight road, for D S: S 50, is shown
above. Atable of values for v(r), at 5 second intervals of time t, is shown to the right of the graph. (a) During what intervals of time is the acceleration of the car positive? Give a reason for your answer. (b) Find the average acceleration of the car, in ﬁlsecz, over the interval 0 St S 50. (c) Find one approximation for the acceleration of the car, in ﬁfsecz, at t = 40. Show the computations
you used to arrive at your answer. so .
(d) Approximate I0 v(:)altwith a Riemann sum, using the midpoints of ﬁve subintervals of equal length. Using correct units,'explain the meaning of this integral. 3
x
z
7
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i *‘kEVIEW: WEEK 3. SOLUTIONS 5 Evaluating this deﬁnite integral requires the use of
antiderivatives and the Fundamental Theorem of
Calculus.
2 4 2 2
,3 4X 6X
L[4X “6X)Q?X=IVT— 1
2
= x4 432::2 ' '
[ J 11
= (16 —12)— [I —3)
= 6
The correct answer (C) was chosen by 86% of test
takers.
:3. This question can be answered using the
property of deﬁnite integrals that
I; f d b b
L ( (20+ goo) x= L mm [a goodx
Therefore,
6 b V b
I (f(x)+ 5) dx=J f(x)dx+j 5 dz
0 (2 d
‘ 2b 5.
—(a+ )+[5x] '0
= [a+ 230+ [532 5a)
=7b—4a
Choice Aisincorrect because it treats the 5 as if .
it were outsmle the original integral expression. L Using substitution ofvariables let u ._. 3..
The correct answer (C) was chosen by 38% of V  1 2
test takers. K Then, air = Edt and a?! = 2du.
Thus, g9"; 03:: [ 20hr) .
=16” dz; =e"+C’
i
=92+C Choice D results from a Chain Rule error,
while Choice E neglects the ﬁnal step in the substitution process. The correct answer (C) was chosen by 69% of
test takers. I9. 24 ~ A Riemarm sum for a function f that has been partitioned into n equal subintervals on the l COS’X The limits of integration must be converted to appropriate values of the new
variable. The lower limit, x: 0, becomes u = tan!) = 0 and the upper limit, dx. Using substitution of variables, let u = tan 2:. Then, at; = sec"V mix = x=£, becomes u=tan(£)=l.
4 4
Therefore,
E tan 5
4 x 4 1
Q 2 dx: etanx. 2 dz.
0 cos x 0 cos x
‘ 1
=I Haiti
0
=6” 1
o
=el—eo
=e—1 Choice D omits the 2° term, and Choice E results from a sign error.
The correct answer (C) was chosen by 29% of test takers. ,2 .
interval [(1, b] has the form 2 f[c,.)£tx, where AX: b—gz and f[c,.) represents any value of
"—1 z...
the function f on the ith subinterval. Another way to write this formula onId be
f'(c:'1 )AJH f[c‘2 )Ax+ + f(cn or
. (f[C1)+ fl¢2)+ ~+ f(cn))Ax
Here, 3—10 is a factor of every term, suggesting that Ax: —1— 50' Because the terms inside the parentheses are values of the function J; evaluated at x= it appears that this is a rightendpoint Riemann sum, using 50 equal subintervals, for the function J; on the interval [0, 1]. Therefore, the Riemann sum
approximates the integral The correct answer (B) was chosen by 7% of test takers. z ‘L 3' 3 \"
QR?
:0:
3x 6 w}? w :. 7Q. 96’: Because F is an antiderivative of f: the
Fundamental Theorem of Calculus tells us that F(3)—F[D)=I: f(x)dx Because f is a continuous function on [0, 3], F(3)— F(U)=I:f(x)dx+ff[x)dx
=(2)(1)+ 2.3
=4.3 The correct answer (D) was chosen by 31% of
test takers. _By deﬁnition, Ia k(r)dt = 0. Therefore, f(a)= 0. This eliminates the graphs in Choices B, C, and D. The graph of h shows that f'(x)=h(x) is
deﬁned for all x on the interval a <1 <6. This eliminates Choice A, since the derivative of this
function is not deﬁned at x= 2). Choice E is the only graph for which
I: h(:)dz= o and f'[x)=h(x) is deﬁned for
all x onthe interval not <c.' The correct answer (E),was chosen by 49% of
test takers. ‘3‘]. (it). The formula for a trapezoidal approximation using four equal
subintervals is i b 1 . ‘
L f(x)dX“2"Ax(flxo)+2f(xl)+2f(x2)+2fix3)+f(x4)) Where Ax= 3);“.
For this question, Ax=—2:—D=—;, SO
2 .
Io f(x)¢m%.%.[3+2.3+2.5+28+13)
l
_Z.(4g)
=12 Each incorrect choice results from an error in the formula for the
i trapezoidal approximation. The correct answer (B) was chosen by 47% of test takers. If F(x) is an antiderivative of for), then F'(x) = f(x). Statement I is TRUE. Statement 11 is false. Statement III is TRUE.
' 2 2 — cos 2;:
menFiszmz’x, WhenF(x)= co: x, WhenF(x)=—4[——l,
. 25inx id . , 2cosx d , sin[2x) a?
F[x)= 2 E(mnx) F (x): 2 .Eﬁosx) F(x)= 4 .303»)
=sinxcosx =cosx(—smx) _ sin[2x)‘2
= — sin xcos x _ 4
= ésin[2x)
= 23in xcosx
2
= sinxcosx Note that the trigonometric identity sin(2x) = 2 sin 1003 x is needed to verify that statement III
is correct. The correct answer (D) was chosen by 21% of test takers. .___.~.. . ._ v/‘Ifw . I: It weirTErV . 1 point was awarded in part (a) for abellshaped ‘ l __ graph which included each of the following:
% oarninimurnof7ﬂatt=ﬂandr=24 only,
3 and ‘ ’
go  amaximumof90 atr=12 only. Time in Hours 3 points were awarded in part (b), as folloWs: l 14 If:
(h) Avg.=14 _ 6]; [80 10 005(ﬁﬂd! = %(697.2957795) =87.162 or 87.161
a: 87°F 1 point was awarded for the limits of
integration and for the coefﬁcient ( 1 0:1)
146 8’ 1 point was awarded for the correct integrand,
and 1 point was awarded for the answer (note that
a solution was not eligible for this point if the' bid}: mm). integral was not of the form 2 points were awarded in part (c), as follows: (c) [so 10 cos(—l’%)]7san 1 point was awarded for a correct inequality
2 — 10 cos (3)20 (or equation), and 1 point was awarded for the solutions as 5'33” 5: 5 18359 endpoints ofaninterval. 5.231 18.770
M
(d) r 185170 3 hoints were awarded in part (d), as follows: 18.769 ' . . .
c: 0.05 [[80 _ m cos(rr_1‘)] _ 731d: 1 point was awarded for the limits of
5.231 . 12 integration and the coefﬁcient 0.05,
CI ‘ .
5.230 1 point was awarded for the correct integrand,
=00500192741):S.096z$5.10 . and 1 point was awarded for the answer (note that
a solution was not eligible for this point if the
integral was not of the form 6
kL (mg—73m). (a) h(1)=[11f(:)ar = o M . Cb) h'(4)=f(4)=2 (c) l<x<3 and 6<x<7 I: is concave up when 0 ﬁz' is increasing
OR 4 f is increasing
OR o ’h”(x)>0 (d) Minimum at x =1 because: It increases on [1, 5] and decreases on [5, 7], so
the minimum is at an endpoint Ih(7)=areaR1—areaR2 >0
and h(l)=0 1 point was awarded in part (a) for the correct
answer of U. 2 points were awarded in part (b), as follows: 1 point was awarded for observing that h'(x) =f(x), and 1 point was awarded for the correct answer
of 2. 3 points were awarded in part (c), as follows: 1 point was awarded for the interval
1 < 7: <3, 1 point was awarded for the interval
6 < x <7, and 1 point was awarded for a correct
justification when both intervals were correctly identified (note that only one of
the three bulleted reasons was needed to receive the justification point). 3.points were awarded in part (d) as follows: 1 point was awarded for the correct answer
of x: 1, 1 point was awarded for observing that the
minimum is at an endpoint of the interval [1, 7 and 1 point was awarded for observing that the
minimum is not at x = 7. 3. (a) Acceleration is positive on (U, 3 5) and (4 5, 50)
because the velocity v0) is increasing on
[U,35]and [45,50]. 3 points were awarded in part (a) as follows: 1 point was awarded for identifying the interval
(0, 3 5), ~ 1 point was awarded for identifying the interval _
(4 5, 50), and e 1 point was awarded for a correct reason. (Note that the awarding of points ignored the inclusion of endpoints on the intervals.) v(5U)—v(0) 72—0 72
.Ac.=————=___=_
09mg C 50—0 so 50 or1.44 fti'set:2 (c) Difference quotient; for example: W45) — W40) = on  75 5 5 =3 fw’sec2 or
v(40)—t(35)=7581__6 a, 2
———5 . 5 — g sec or
W45)v(35)_60—31 . 21 2 OR Slope of tangent line, for example, through
(35, 90) and (4 0,75): 90—75
35—40 = 3 ltfsec2 SJ
@Lma _
as 10 [v(5) + v(15)+ v(25) + v(3 5) + v(45)]
=IU(12+ 3U +7U+ 81+ 60)
= 2530 feet This integral is the total distance traveled, in
feet, over the time 0 to 50 seconds. 1 point was awarded in part (b) for the correct answer of B
50' 2 points were awarded in part (c), as follows: 1 point was awarded for identifying a valid
method, and 1 point was awarded for the correct answer
using that method. (Note that no points were awarded in part (c) if a
valid method was not indicated.) 3 points were awarded in part (d), as follows: 1 point was aWarded for the correct midpoint
Riemann sum," 1 point was awarded for the answer, and 1 point was awarded for a correct explanation
of the meaning of the integral. ...
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This note was uploaded on 02/04/2011 for the course MATH 116 taught by Professor John during the Spring '10 term at Saint Michael's College  Colchester, Vermont.
 Spring '10
 John
 Calculus

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