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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(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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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.