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AP Calculus Testbank(Chapter 6)
(Mr. Surowski)
Part I. Multiple-Choice Questions
1. Suppose that f is an odd differentiable function. Then1∫−1xf ′(x) dx =
(A) f(1);
(B) f ′(1)
(C) f(1)− f(−1)
(D) 0
(E) 2.
2. The slope field indicated below most likely depicts the differen-tial equation
(A)dy
dx= x+ y
(B)dy
dx= x+ 5
(C)dy
dx= y + 5
(D)dy
dx= −x+5
(E)dy
dx= −y + 5
3. You are given differentiable functions f and g. Which of thefollowing is a consequence of the integration by parts formula?
(A)∫f ′′(x)g(x) dx+
∫f(x)g′′(x) dx = f ′(x)g′(x)
(B)∫f ′′(x)g(x) dx+
∫f ′(x)g′(x) dx = f(x)g(x)
(C)∫f ′′(x)g(x) dx+
∫f ′(x)g′(x) dx = f ′(x)g(x)
(D)∫f ′′(x)g(x) dx+
∫f ′′′(x)g(x) dx = f ′(x)g(x)
(E)∫f ′′(x)g(x) dx+
∫f(x)g′(x) dx = f(x)g(x).
4. Given the slope field below, what is the most plausible behavior
of the solution of the IVPdy
dx= f(x, y), y(−4) = 4?
(A) limx→∞
y = −∞
(B) limx→∞
y = 0
(C) limx→∞
y = +∞
(D) limx→∞
y = does notexist.
(E) limx→∞
y = −2
5. You are given the differential equationdy
dx= −1
5
√y, where y(0) =
25. For which value(s) of x is y = 0?
(A) ±5 (B) 10 (C) 50 (D) ±10 (E) 5
6.∫
tan6 x sec2 x dx =
(A)tan7 x
7+ C
(B)tan7 x
7+
sec3 x
3+ C
(C)tan7 x sec3 x
21+ C
(D) 7 tan7 x+ C
(E)2
7tan7 x secx+ C
7.∫ 1/2
0
2√1− x2
dx =
(A)π
6(B)
π
3(C) −π
3(D)
2π
3(E) −2π
3
8.∫
7xe3x2
dx =
(A)1
42e3x2
+ C
(B)6
7e3x2
+ C
(C)7
6e3x2
+ C
(D) 7 e3x2
+ C
(E) 42 e3x2
+ C
9.∫ex(e3x) dx =
(A)1
3e3x + C
(B)1
4e4x + C
(C)1
4e5x + C
(D) 4 e4x + C
(E) 5 e5x + C
10.∫x√
5− x dx =
(A) −10
3(5− x)
32 + C
(B)
√5x2
2− x3
3+ C
(C)10
3
√5x2
2− x3
3+ C
(D) 10(5− x)12 +
2
3(5− x)
32 + C
(E) −10
3(5− x)
32 +
2
5(5− x)
52 + C
11. Ifdy
dx=x3 + 1
yand y = 2 when x = 1, then when x = 2, y =
(A)
√27
2(B)
√27
8(C) ±
√27
8(D) ±3
2(E) ±
√27
2
12.∫
lnx
3xdx =
(A) 6 ln2 x+ C
(B)1
6ln(lnx) + C
(C)1
3ln2 x+ C
(D)1
6ln2 x+ C
(E)1
3lnx+ C
13.∫
sin5(2x) cos(2x) dx =
(A)sin6 2x
12+ C
(B)sin6 2x
6+ C
(C)sin6 2x
3+ C
(D)cos5 2x
3+ C
(E)cos5 2x
6+ C
14.∫ π/2
0sin(2x)esin2 x dx =
(A) e (B) e− 1 (C) 1− e (D) e+ 1 (E) 1
15.∫ π/2
0sin5 x cosx dx =
(A)1
6(B) −1
6(C) 0 (D) −6 (E) 6
16.∫ 1
0sin−1 x dx =
(A) 0 (B)π + 2
2(C)
π − 2
2(D)
π
2(E) −π
2
17. Ifdy
dx= 3y cosx, and y = 8 when x = 0, then y =
(A) 8e3 sinx (B) 8e3 cosx (C) 8e3 sinx + 3 (D)3y2
2cosx+ 8 (E)
3y2
2sinx+ 8
18.∫
dx√9− x2
=
(A) sin−1 3x+ C
(B) ln |x+√
9− x2|+ C
(C)1
3sin−1 3x+ C
(D) sin−1 x
3+ C
(E) 13 ln |x+
√9− x2|+ C
19.∫x sin(2x) dx =
(A) −x2
2cos(2x) + C
(B) −x2
4cos(2x) + C
(C) −x2
cos(2x) +1
4sin(2x) + C
(D) −x2
cos(2x) +1
2cos(2x) + C
(E) −1
2cos(2x) +
1
4sin(2x) + C
20. Given the differential equationdz
dt= z
(4− z
100
), where z(0) =
50, what is limt→∞
z(t)?
(A) 400 (B) 200 (C) 100 (D) 50 (E) 4
Part II. Free-Response Questions
1. Compute the following indefinite integrals:
(a)∫
(1− x)5 dx = −(1− x)6
6+ C
(b)∫x2√1− x dx = −2
7(1− x)7/2 + 4
5(1− x)5/2− 2
3(1− x)3/2 +C
(c)∫
x√x2 − 1
dx =√x2 − 1 + C
(d)∫ex sin(ex) dx = − cos(ex) + C
(e)∫ √
ln ex dx = 23x
3/2 + C
(f)∫ (
sin2 x− 1)d(sinx) = 1
3 sin3 x− sinx+ C
(g)∫
x dx
x4 + 1= 1
2 tan−1 x2 + C
(h)∫x sec2 (x2) dx = 1
2 tanx2 + C
(i)∫
sec 2x tan 2x dx = 12 sec 2x+ C
(j)∫
dx√x√
1−√x
= −4√
1−√x
(k)∫
dx√x√
1− x= 2 sin−1√x+ C
(l)∫x lnx dx =
x2
2lnx− x2
4+ C
(m)∫xex
2
dx = 12ex2
+ C
(n)∫xex dx = xex − ex + C
(o)∫ex sinx dx = 1
2ex(sinx− cosx) + C
(p)∫x3 3√
1− x2 dx
= −34x
2(1− x2)4/3 − 956(1− x
2)7/3 + C
2. Compute the following definite integrals:
(a)∫ e2
e
dx
x lnx= ln(lnx)
∣∣∣e2e
= ln 2
(b)∫ √π/4
0
x dx
1 + x4 = 12 tan−1 x2
∣∣∣√π/4
0= 1
2
(c)
π/2∫0
x2 sinx dx = (−x2 cosx+ 2x sinx+ 2 cosx)∣∣∣π/20
= π − 2
(d)∫ π
−π
cosx dx√4 + 3 sinx
= 23
√4 + 3 sinx
∣∣∣π−π
= 0
3. Suppose that f is an even differentiable function. Compute1∫−1xf ′(x) dx.
This can’t be computed explicitly, but we can simplify a bit.Using integration by parts with u = x, dv = f ′(x) dx, we getdu = dx, v = f(x), and so
1∫−1
xf ′(x) dx = xf(x)∣∣∣1−1−∫ 1
−1f(x) dx = 2f(1)− 2
∫ 1
0f(x) dx.
4. Below is sketched the slope field for the differential equationdy
dx= f(x, y). Sketch a possible solution of the above differential
equation satisfying the initial value y(−5) = −2.
-5 -4 -3 -2 -1 1 2 3 4 5
-3
-2
-1
1
2
3
x
y
Equation 1: y'=a(y+3)
5. Given the slope field indicated below, sketch solutions of theIVPs
(i)dy
dx= f(x, y), y(−1) = 4, x ≥ −1
(ii)dy
dx= f(x, y), y(−1) = −1, x ≥ −1
-1 1 2 3
-4
-2
2
4
x
y
Equation 1: y'=ax/(x+3)
6. Sketch the slope field describing the solutions of the ODEdy
dx= x− y.
-6 -5 -4 -3 -2 -1 1 2 3 4 5 6
-4
-2
2
4
x
y
Equation 1: y'=x−y
7. Solve the initial value problemdy
dx= 2 cos x+ sinx,
y(π) = 0. y = 2 sin x− cosx− 1.
8. Solve the initial value problemd2y
dx2 = 2 cos x+ sinx,
y(π) = 0, y′(π) = 0. y = −2 cosx− sinx− x+ π − 2
9. Solve the initial value problemdy
dx= −2y, y(0) = 100.
y = 100e−2x
10. Solve the initial value problemd2y
dx2 = −2y, y(0) = 1, y′(0) = 0.
y = cos√
2x
11. The (continuous) growth of money is modelled by the IVPdy
dt= ry, y(0) = y0, where y0 is an initial amount (of money).
(i) Solve the IVP, given that there is initially $10,000 in the bank,and that the interest rate is 6.2%.y = 10000e.062t
(ii) How much is in the bank after 2 12 years?
y(2.5) = 10000e(.062)·(2.5) ≈ $11, 676.58
(iii) How long will it take for the money to double in value?Solving 20000 = y(t) = 10000e.062t leads to
0.062t = ln 2⇒ t ≈ 11.18 years.
12. Suppose that we have the initial value problemdy
dt= −ky,
y(0) = y0 and that we know that y(150) = 12y0. Find k. We have
that y(t) = y0e−kt, and that 1
2y0 = y(150) = y0e−150k. This gives
150k = ln 2, and so k ≈ 4.6× 10−3.
13. (Carbon-14 Dating). The initial value problem of importance inCarbon-14 dating is
dy
dt= −ky, y(0) = y0.
It is typically assumed that the half-life of the radioactive nucleiof carbon 14 is roughly 5,700 years.
(i) Using the above, show that k = (ln 2)/5700. We are giventhat y(5700) = 1
2y0. This implies that 12y0 = y0e
−5700k ⇒ k =(ln 2)/5700.
(ii) After how many years of radioactive decay will the originalcarbon 14 have decayed by 20%? We are given that y0e
−kt =y(t) = 0.8y0 ⇒ t = (ln 0.8)/(−k) = (ln 0.8)/(−(ln 2)/5700) ≈1, 835 years.
14. Newton’s Law of Cooling says that the IVP governing the tem-perature T of a heated object immersed in an environment at atemperature of Ts is
dT
dt= −k(T − Ts), T (0) = T0,
(where k is a positive constant depending on the various media).
(i) Show by differentiation that the solution of the above IVP
is given by T (t) = Ts + (T0 − Ts)e−kt. We have that
dT
dt=
−k(T0−Ts)e−kt = −k(T−Ts).Also, T (0) = Ts+(T0−Ts)e0 =T0, so the initial condition is satisfied.
(ii) Compute limt→∞
T (t) = Ts.
(iii) Does your answer to part (ii) make sense? Of course thismakes sense, as we expect the temperature of a heated objecteventually to cool to the temperature of the environment.
15. Assume, as in Problem 14 above, that Newton’s Law of Cool-ing has solution given by T (t) = Ts + (T0 − Ts)e
−kt. We applythis model to a bowl of hot noodle soup, where the soup wasoriginally at 90◦C and cooled to 60◦C after 10 minutes in a roomwhose ambient temperature was 20◦C.
(i) Find the value of k in the above model. We’re given the solu-tion T (t) = Ts+(T0−Ts)e−kt = 20+70e−kt. Also, we’re giventhat 60 = T (10) = 20 + 70e−10k ⇒ k = (ln(7/4))/10 ≈ 0.056.
(ii) What is the temperature of the soup after half an hour?Using k ≈ 0.056 we infer that T (30) = 20 + 70e−30k ≈ 20 +70e−30·0.056 ≈ 33.05◦C.
(iii) After how many minutes will the soup cool to 30◦C?Here,we’re trying to solve the equation 30 = 20 + 70e−0.056t for t.This leads to t = (ln 7)/0.056 ≈ 34.75 minutes (or close to 35minutes). (Note that this is consistent with the result of part(b).)
16. Assume, as in Problem 14 above, that Newton’s Law of Cool-ing has solution given by T (t) = Ts + (T0 − Ts)e
−kt. We applythis model to a bowl of hot noodle soup, where the soup wasoriginally at 90◦C and cooled to 60◦C after 10 minutes in a room,and then cooled to 50◦C after another 10 minutes. Determine theambient temperature of the room. 1 We have the two equationsin the two unknowns Ts and k:
60 = Ts + (90− Ts)e−10k and 50 = Ts + (90− Ts)e−20k.
We havee−10k =
60− Ts90− Ts
, e−20k =50− Ts90− Ts
.
Since e−20k =)e−10k)2, we see that
50− Ts90− Ts
=(60− Ts)2
(90− Ts)2 .
This simplifies slightly to
(90− Ts)(50− Ts) = (60− Ts)2.
1This problem is a bit more difficult than Problem 15, above.
The above ultimately reduces to a linear equation in Ts, withsolution Ts = 45◦C.
17. (Mr. S on roller skates.) Believe it or not, Mr. S is pretty goodat roller skating! One of his favorite activities is skating as fastas he can and then stop, seeing how far he can “coast.” Once hestarts coasting, the differential equation that governs his motion
is mdv
dt= −kv, where v is his velocity in km/hr, where m is
Mr. S’s mass (in kg), and where k > 0 is a constant representingfriction. Assume that k = 60, 000, that m = 80 kg, and that at theinstant he starts to coast, Mr. S is moving at 30 km/hr.
(a) Determine Mr. S’s velocity, in km/hr, as a function of time.The solution of the initial-value problem is v(t) = v0e
−kt/m =30e−750t.
(b) Determine how far Mr. S will coast (in meters). Sincedx
dt=
v = 30e−750t, we have that the total distance traveled is
x =
∫ ∞0
(dx
dt
)dt =
∫ ∞0
30e−750t dt = − 1
25e−750t
∣∣∣∞0
=1
25km.
Therefore, Mr. S will coast for 40 meters.
18. The Logistic Differential Equation is given by
dP
dt=
k
MP (M − P ) ,
where k is a positive constant.
(i) Show by differentiation that the solution of the above differ-ential equation is given by
P =M
1 + Ae−kt.
This is routine but a bit tedious. The details have been omit-ted.
(ii) Compute limt→∞
P . = M
19. Suppose that the spread of measles in a given school is predictedby the logistic function
P (t) =200
1 + 199e−t,
where t is the number of days after a student comes into contactwith an infected student.
(i) Write down the corresponding logistic differential equation.
This isdP
dt= .005P (200− P ).
(ii) Compute P (0) and explain what this means. P (0) = 1,which means that initially one person is infected with measles.
(iii) After how many days will half of the students have becomeinfected? We solve the equation
100 =200
1 + 199e−t,
which reduces to e−t =1
199, so t = ln 199 ≈ 5.3days.
(iv) After how many days will 90% of the students have becomeinfected? In this case we are to solve
180 =200
1 + 199e−t,
which leads to
t = ln 1990 ≈ 7.6 days.
20. Sketch the slope field describing the solutions of the logistic dif-
ferential equationdP
dt= P (5− P ), P ≥ 0.
21. Consider the differential equationdy
dx=
3− xy
.
(a) Let y = f(x) be the particular solution to the given differ-ential equation for 1 < x < 5 such that the line y = −2 istangent to the graph of f . Find the x-coordinate of the pointof tangency, and determine whether f has a local maximum,local minimum, or neither at this point. Justify your answer.The line whose equation is y = −2 is horizontal; the only
value of x for whichdy
dx= 0 is x = 3. Next, we have, using
implicit differentiation and the quotient rule, that
y′′ =−y − (3− x)y′
y2 .
At the point in question, x = 3, y = −2, anddy
dx= 0. There-
fore, y′′(3,−2) =2
4> 0, and so a local minimum occurs at
this point.
(b) Let y = g(x) be the particular solution to the given differ-ential equation for −2 < x < 8, with the initial conditiong(6) = −4. Find y = g(x). Here we must actually solve thedifferential equation. We separate variables and integrateboth sides: ∫
y dy =
∫(3− x) dx
which leads toy2
2= 3x− x2
2+ C. Since y = −4 when x = 6,
we get 8 = 18− 18+C and so C = 8. Writing y as a functionof x gives
y = ±√
16 + 6x− x2;
however, as y is negative (at x = 6), we infer that, in fact,
y = g(x) = −√
16 + 6x− x2.
22. Consider the differential equationdy
dx= x2(y − 1).
(a) On the axes provided, sketch a slope field for the given dif-ferential equation at the twelve points indicated.
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2
3
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(b) While the slope field in part (a) is drawn at only twelvepoints, it is defined at every point in the xy-plane. Describeall points in the xy-plane for which the slopes are positive.
The expressiondy
dx= x2(y − 1) > 0 at all values (x, y) where
x 6= 0 and y > 1.
(c) Find the particular solution y = f(x) to the given differentialequation with the initial condition f(0) = 3. Separating thevariables and integrating leads to∫
dy
y − 1=
∫x2 dx⇒ ln |y − 1| = x3
3+ C.
Write this as y − 1 = Kex3/3; since y = 3 when x = 0, we get
2 = K. Therefore, the solution is y = 2ex3/3 + 1.
23. Consider the differential equationdy
dx=
3x2
e2y .
(a) Find a solution y = f(x) to the differential equation satisfy-
ing f(0) =1
2. Separating the variables and integrating:∫e2y dy =
∫3x2 dx⇒ e2y = 2x3 + C.
Since y = 12 when x = 0, get e = C. Therefore e2y = 2x3 + e
so thaty = f(x) =
1
2ln(2x3 + e).
(b) Find the domain and range of the function f found in part(a). The domain consists of all real numbers x satisfying2x3 + e > 0 ⇒ x > 3
√−e/2; the range consists of all real
numbers y.
The graph of the solution is shown below:
24. The function f is differentiable for all real numbers. The point(3,
1
4
)is on the graph of y = f(x), and the slope at each point
(x, y) on the graph is given bydy
dx= y2(6− 2x).
(a) Findd2y
dx2 and evaluate it at the point(
3,1
4
).
First, we haved2y
dx2 = 2ydy
dx(6−2x)−2y2.Next, since
dy
dx
(3,
1
4
)=
0, we see thatd2y
dx2 = −2
(1
4
)2
= −1
8.
(Note that this implies that the function f has a local maxi-mum at the point
(3, 1
4
).)
(b) Find y = f(x) by solving the differential equationdy
dx= y2(6− 2x) with the initial condition f(3) =
1
4.
Separating the variables and integrating yields∫dy
y2 dy =
∫(6− 2x) dx⇒ −1
y= 6x− x2 + C.
Since y =1
4when x = 3 we obtain −4 = 18 − 9 + C, which
implies that C = −13. Therefore, we obtain
y = f(x) =1
x2 − 6x+ 13.
25. Let f be the function satisfying f ′(x) = x√f(x) for all real num-
bers x, where f(3) = 25.
(a) Find f ′′(3). f ′′(x) =√f(x) +
xf ′(x)
2√f(x)
⇒ f ′′(3) =√
25 +
3 · 15
2 · 5= 5 +
9
2=
19
2.
(b) Write an expression for y = f(x) by solving the differential
equationdy
dx= x√y with the initial condition f(3) = 25.
Separating the variables and integrating:∫dy√y
=
∫x dy ⇒ 2
√y =
x2
2+ C;
since y = 25 when x = 3, we infer that 10 =9
2+ C, and so
C =11
2. This implies that 2
√y =
x2
2+
11
2and so
y = f(x) =1
16(x2 + 11)2.
26. Consider the differential equationdy
dx= −2x
y.
(a) On the axes provided, sketch a slope field for the given dif-ferential equation at the twelve points indicated.
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−2
x
y
(b) Let y = f(x) be the particular solution to the differentialequation with the initial condition f(1) = −1. Write anequation for the line tangent to the graph of f at (1,−1) and
use it to approximate f(1.1). We have thatdy
dx(1,−1) = 2,
and so the line tangent to the graph of f at (1,−1) is y + 1 =2(x− 1), which can be re-written as y = 2x− 3. Using this toapproximate f(1.1) we have
f(1.1) ≈ (2x− 3)∣∣∣x=1.1
= 2.2− 3 = −0.8.
(c) Find the particular solution y = f(x) to the given differentialequation with the initial condition f(1) = −1. Separatingthe variables and integrating:
∫y dy = −
∫2x dy ⇒ y2
2= −x2 + C.
Since y = −1 when x = 1 we infer quickly that C =3
2and
we have the solution
y = f(x) = −√
3− 2x2.
(Note the necessity of the minus sign when taking the squareroot!)
27. You are given the initial value problemdy
dx= x− y, y(0) = 0.
(i) Use Euler’s method to approximate y(0.2) (use dx = 0.1).
From y(0.1) ≈ y(0) +dy
dx(0, 0)(0.1) = 0, we then have
y(0.2) ≈ y(0.1) +dy
dx(0.1, 0)(0.1) = 0 + 0.01 = 0.01.
(ii) Show by differentiation that the solution of this IVP isy(x) = e−x+x− 1. From y = e−x+x− 1 get y′ = −e−x+1 =x− (e−x + x− 1) = x− y.
(iii) Compare the relative error in the estimation of y(0.2). (Usethe definition
Relative Error =
∣∣∣∣exact value− approximate valueexact value
∣∣∣∣ .)The exact value of y(0.2) = e−0.2 + 0.2 − 1 = 0.019 (accurateto three decimal places), whereas the approximate value wasshown above to be y(0.2) ≈ 0.01. Therefore,
Relative Error =
∣∣∣∣0.019− 0.010
0.019
∣∣∣∣ = 0.47,
i.e., we have incurred a 47% error in the approximation.
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