Be Prepared
for the
Calculus
Exam
Mark Howell
Gonzaga High School, Washington, D.C.
Martha Montgomery
Fremont City Schools, Fremont, Ohio
Practice exam contributors:
Benita Albert
Oak Ridge High School, Oak Ridge, Tennessee
Thomas Dick
Oregon State University
Joe Milliet
St. Mark's School of Texas, Dallas, Texas
Skylight Publishing
Andover, Massachusetts
Copyright © 2005-2011 by Skylight Publishing
Chapter 10. Annotated Solutions to Past Free-Response Questions
This material is provided to you as a supplement to the book
Be Prepared for the AP Calculus Exam.
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2011 AB
AP Calculus Free-Response
Solutions and Notes
Question AB-1
(a)
v ( 5.5 ) ­ ≈ −0.453 and a ( 5.5 ) ­ ≈ −1.359 . The velocity is decreasing because
v′ ( 5.5 ) = a ( 5.5 ) < 0 . Since v ( 5.5 ) and a ( 5.5 ) have the same sign (both negative),
the speed is increasing.
1 6
v ( t ) dt ­ ≈ 1.949 .
6 ∫0
(b)
Average velocity =
(c)
Total distance traveled = ∫ v ( t ) dt ­ ≈ 12.573 .
(d)
v ( t ) changes sign at ­ t ≈ 5.19552 . Let b = 5.19552 1.
6
0
x ( b ) = x ( 0 ) + ∫ v ( t ) dt = 2 + ∫ v ( t ) dt ­ ≈ 14.135 .
b
b
0
0
Notes:
1.
Store this number in your calculator, and use the stored value in the subsequent
calculation.
3
4
FREE-RESPONSE SOLUTIONS ~ 2011 AB
Question AB-2
(a)
(b)
(c)
H ′ ( 3.5 ) ≈
H ( 5 ) − H ( 2 ) 52 − 60
8
=
= − degrees Celsius/minute.
5−2
3
3
1 10
H ( t ) dt is the average temperature of the tea in degrees Celsius from t = 0 to
10 ∫ 0
t = 10 minutes.1 The trapezoidal approximation gives
1 ⎛ 66 + 60
60 + 52
52 + 44
44 + 43 ⎞ 2
+ 3⋅
+ 4⋅
+ 1⋅
⎜ 2⋅
⎟.
10 ⎝
2
2
2
2 ⎠
By the FTC,
∫
10
0
∫
10
0
H ′ ( t ) dt = H (10 ) − H ( 0 ) = 43 − 66 = −23 degrees Celsius.
H ′ ( t ) dt represents the net change in temperature of the tea, in degrees Celsius,
over the 10-minute time interval.3
(d)
The temperature of biscuits at t = 10 minutes
B (10 ) = B ( 0 ) + ∫ B′ ( t ) dt = 100 + ∫ B′ ( t ) dt ­ ≈ 34.183 . The biscuits are
10
10
0
0
43 – 34.183 = 8.817 degrees Celsius cooler than the tea.
Notes:
1.
2.
3.
It is important to mention the units both for the variable t (minutes) and the average
temperature (degrees Celsius).
1
No need to evaluate. For the curious, the value is ≈ ( 529.5 ) = 52.950 .
10
An alternative meaning might be “The tea cooled by 23°C during the 10 minutes,”
but that would describe the meaning of the value of the expression, while the
problem clearly asks for the meaning of the expression itself.
FREE-RESPONSE SOLUTIONS ~ 2011 AB
Question AB-3 1
(a)
(b)
(c)
1⎞
⎛1⎞
⎛1⎞
⎛
f ⎜ ⎟ = 1 . f ′ ( x ) = 24 x 2 so f ′ ⎜ ⎟ = 6 . The tangent line is y − 1 = 6 ⋅ ⎜ x − ⎟ .2
2⎠
⎝2⎠
⎝
⎝2⎠
Area =
∫
1
2
0
( sin (π x ) − 8 x ) dx = ⎛⎜⎝ − π1 cos (π x ) − 2 x4 ⎞⎟⎠
Volume = π ∫
3
1/ 2
0
1
2
0
1 ⎛ 1⎞ 1 1
= 0− −⎜− ⎟ = − .
8 ⎝ π⎠ π 8
((1 − f ( x )) − (1 − g ( x )) ) dx .
2
2
3
Notes:
1.
2.
3.
This is the third consecutive year that an area-volume problem has appeared in the
closed calculator part of the free response section. This trend is likely to continue.
1⎞
⎛
Or y = 1 + 6 ⋅ ⎜ x − ⎟ . Don’t waste time to simplify.
2⎠
⎝
Use the names of the functions given in the problem to save time and prevent
transcription errors.
5
6
FREE-RESPONSE SOLUTIONS ~ 2011 AB
Question AB-4
−3
g ( −3) = −6 + ∫
(b)
g ′ ( x ) changes sign from positive to negative only once on [−4, 3] , so g has an
0
f ( t ) dt = −6 −
9π 1
.
g ′ ( x ) = 2 + f ( x ) , so g ′ ( −3) = 2 + 0 = 2 .
4
(a)
absolute maximum when this happens. For −4 ≤ x ≤ 0 , g ′ ( x ) = 2 + f ( x) > 0 . For
0 ≤ x ≤ 3 , g ′ ( x ) = 0 when f ( x ) = −2 . f ( x ) = 3 − 2 x , so f ( x ) = −2 when x =
Since g ′ ( x ) > 0 for −4 < x <
maximum is at x =
(c)
5
.
2
5
5
and g ′ ( x ) < 0 for < x < 3 , the absolute
2
2
5 2
.
2
Only x = 0. g ′ ( x ) = 2 + f ( x ) changes from increasing to decreasing only at x = 0,
and g ′ ( x ) never changes from decreasing to increasing. So the graph of g changes
from concave up to concave down, at x = 0.
(d)
f ( 3 ) − f ( −4 ) − 3 + 1
2
=
= − . There is no
3 − ( −4 )
7
7
contradiction with the Mean Value Theorem because the condition that f is
differentiable on the interval is not satisfied: f is not differentiable at x = 0.
The average rate of change is
Notes:
1.
2.
The integral from 0 to −3 is negative.
Since g ′ ( x ) changes sign from positive to negative only once in the interval, it is
easier to use this argument than the candidate test.
7
FREE-RESPONSE SOLUTIONS ~ 2011 AB
Question AB-5
(a)
dW
dt
=
t=
1
4
1
1
⋅ (1400 − 300 ) = 44 . The tangent line is y = 1400 + 44t . At t = ,
25
4
W ≈ 1411 tons.
(b)
(c)
d 2W
1 ⎛ dW
= ⋅⎜
2
dt
25 ⎝ dt
1
1
⎞ 1 ⎛ 1
⎞
(W − 300 ) . For 0 < t < ,
⎟ = ⋅ ⎜ ⋅ (W − 300 ) ⎟ =
4
⎠ 25 ⎝ 25
⎠ 625
d 2W
W ≥ 1400 > 300 , so
> 0 . Therefore the tangent line lies under the graph of
dt 2
y = W (t ) . Therefore, 1411, the estimate based on the tangent line, underestimates
the actual value.
1
1
1
t + C . Substituting W = 1400 at
25
t = 0 we get ln (1100 ) = C . Since W (t ) is increasing, W − 300 > 0 . Solving for W,
∫ W − 300 dW = ∫ 25 dt
we get W − 300 = e
⇒ ln W − 300 =
t
+ ln (1100 )
25
Notes:
1.
No need to simplify further.
⇒ W = 300 + e
t
+ ln (1100 )
25
1
t
25
or W = 300 + 1100e .
8
FREE-RESPONSE SOLUTIONS ~ 2011 AB
Question AB-6
(a)
f ( x) is continuous when x ≠ 0 . lim+ f ( x ) = 1 and lim− f ( x ) = 1 , so
x →0
x →0
lim f ( x ) = 1 = f ( 0 ) . Therefore, f is continuous at x = 0, too.
x →0
(b)
(c)
⎧−2 cos x, for x < 0
1 ⎛3⎞
3
. −4e −4 x = −3 ⇒ e −4 x =
f ′( x) = ⎨
⇒ x = − ln ⎜ ⎟ .
−4 x
4 ⎝4⎠
4
⎩−4e , for x > 0
Average value =
1
1
1
f ( x ) dx =
∫
1 − ( −1) −1
2
1⎛
⎜ ( x + 2 cos x )
2⎜
⎝
(∫
(1 − 2sin x ) dx + ∫ 0 e−4 x dx
−1
0
1
)=
1
⎛ e −4 1 ⎞ ⎞ 1
⎛ e −4 x ⎞ ⎞ 1 ⎛
⎟
2
1
2
cos
1
−
−
+
−
+
+
=
( )) ⎜ − ⎟ ⎟ .
(
⎜
⎜
⎟
−1
⎝ −4 −4 ⎠ ⎠
⎝ −4 ⎠ 0 ⎟⎠ 2 ⎝
0
(
)
Notes:
1.
No need to simplify further. If you insist, =
13
1
− cos ( −1) − 4 .
8
8e
2011 BC
AP Calculus Free-Response
Solutions and Notes
Question BC-1
2
(a)
Speed =
⎛ dx ⎞ ⎛ dy ⎞
⎜ ⎟ +⎜ ⎟
⎝ dt ⎠ ⎝ dt ⎠
⎛d x d y⎞
G
a ( 3) = ⎜ 2 , 2 ⎟
⎝ dt dt ⎠
2
2
2
= 132 + sin 2 ( 9 ) ­ ≈ 13.007 .
t =3
= ( 4, 6 cos ( 9 ) ) ­ ≈ ( 4, − 5.467 ) . 1
t =3
dy dy
=
dx dt
sin 9
dx
. At t = 3, this is
­ ≈ 0.032 .
dt
13
(b)
The slope is
(c)
x ( 3) = x ( 0 ) + ∫
(d)
Total distance traveled is
3
dx
dt = ∫ ( 4t + 1) dt ­ = 21 ;
0 dt
0
3 dy
3
y ( 3) = y ( 0 ) + ∫
dt = −4 + ∫ sin ( t 2 ) dt ­ ≈ −3.226 .
0 dt
0
3
∫ 0 ( 4t + 1)
3
2
(
+ sin ( t 2 )
)
2
dt ­ ≈ 21.091
Notes:
1.
You could use numeric derivatives of
dx
dy
and
instead.
dt
dt
Question BC-2
See AB Question 2.
9
10
FREE-RESPONSE SOLUTIONS ~ 2011 BC
Question BC-3 1
k
1 + ( 2e 2 x ) dx
(a)
Perimeter = 1 + k + e 2 k + ∫
(b)
Volume = π ∫
(c)
dV π
dk
1 dV π
1 π e2
. At k = ,
= ⋅ 4e 2 ⋅ =
.
= ⋅ 4e 4 k
4
3
3
dt
dt
2 dt
4
k
0
(e )
2x 2
0
2
k
π
0
4
dx = π ∫ e 4 x dx =
e4 x
k
0
=
π
(e
4
4k
− 1) .
Notes:
1.
This is the third consecutive year that an area-volume problem has appeared in the
closed calculator part of the free response section. This trend is likely to continue.
Question BC-4
See AB Question 4.
Question BC-5
See AB Question 5.
FREE-RESPONSE SOLUTIONS ~ 2011 BC
11
Question BC-6
(a)
sin x = x −
(b)
cos x = 1 −
x2 x4 x6
+ − + ... .
2! 4! 6!
⎛
⎞ ⎛ x 2 x4 x6
⎞
x 6 x10 x14
f ( x) = ⎜ x 2 − +
−
+ ... ⎟ + ⎜1 − + − + ... ⎟ =
3! 5! 7!
⎝
⎠ ⎝ 2! 4! 6!
⎠
1+
(c)
(d)
x3 x5 x 7
x 6 x10 x14
+ − + ... . sin ( x 2 ) = x 2 − +
−
+ ...
3! 5! 7!
3! 5! 7!
f(
x2 x4 ⎛ 1 1 ⎞ 6
+ − ⎜ + ⎟ x + ...
2 4! ⎝ 3! 6! ⎠
6)
1
(0) = − ⎛ 1 + 1 ⎞
⎛ 6! ⎞
( 6)
⎜
⎟ ⇒ f ( 0 ) = − ⎜ + 1⎟ = −121 .
⎝ 3! 6! ⎠
⎝ 3! ⎠
6!
⎛ 1⎞
on the interval ⎜ 0, ⎟ is less
⎝ 4⎠
The Lagrange error bound is less than
From the graph, the maximum value of f (
than 40. 2
5)
( x)
5
40 ⎛ 1 ⎞
1
1
1
1
⎛1⎞
⎛1⎞
. So P4 ⎜ ⎟ − f ⎜ ⎟ <
.
⋅⎜ ⎟ =
=
<
5
5! ⎝ 4 ⎠ 3 ⋅ 4
3 ⋅1024 3000
⎝4⎠
⎝ 4 ⎠ 3000
Notes:
1.
Leave it at that.
2.
⎛ 1⎞
We chose 40 because undoubtedly f ( 5) ( x ) < 40 on ⎜ 0, ⎟ , 40 is convenient to use
⎝ 4⎠
in calculations, and 40 gives the desired error bound of less than
1
. 35 would
3000
work as well, but the arithmetic would be slightly more difficult. The choice of 30
⎛1⎞
doesn’t work because f ( 5) ⎜ ⎟ is slightly greater than 30.
⎝4⎠
2011 AB (Form B)
AP Calculus Free-Response
Solutions and Notes
Question AB-1 (Form B)
∫
60
S ′ ( t ) dt ­ ≈ 171.813 millimeters.
(a)
S (0) = 0 . S (60) = S (0) +
(b)
The average rate of change is
(c)
The volume at time t, V ( t ) = π r 2 S ( t ) = 100π S ( t ) . The rate of change of the
0
S ( 60 ) − S ( 0 )
171.813
≈­
≈ 2.864 millimeters/day.
60 − 0
60
volume is V ′ ( t ) = 100π S ′ ( t ) . V ′ ( 7 ) = 100π S ′ ( 7 ) ­ ≈ 602.218 mm3/day.
(d)
D ( 0 ) = M ′ ( 0 ) − S ′ ( 0 ) = 0.825 − 1.5 < 0 and
D ( 60 ) = M ′ ( 60 ) − S ′ ( 60 ) = 72.825 − 3.448 > 0 .
1
Since both M ′ ( t ) and S ′ ( t ) are
continuous functions, D(t) is continuous. By the Intermediate Value Theorem,
D (t ) = 0 for some t between 0 and 60. At that point in time, M ′ ( t ) = S ′ ( t ) .
Notes:
1.
Note that we are given expressions for S ′(t ) and M (t ) . It might be easier to enter
M (t ) into your calculator, then get M ′(0) and M ′(60) numerically.
13
14
FREE-RESPONSE SOLUTIONS ~ 2011 AB (FORM B)
Question AB-2 (Form B)
(a)
1000
3000
and lim− r ( t ) =
. Since lim+ r ( t ) ≠ lim− r ( t ) ,
t
→
5
t →5
t →5
e
8
lim r ( t ) does not exist, so r is not continuous at t = 5.
lim r ( t ) = 1000e −0.2⋅5 =
t → 5+
t →5
(b)
The average value of r ( t ) is
8
1 8
1 ⎛ 5 600t
1
⎞
=
+
r
t
dt
dt
1000e −2t dt ⎟ 1­ ≈ (1234.5073 + 829.9146 ) ≈
(
)
⎜ ∫0
∫
∫
0
5
8
8
8⎝ t +3
⎠
258.053 liters per hour.
(c)
r ′(3) ­ = 50 liters per hour per hour. At t = 3 hours, the drainage rate is increasing
at the rate of 50 liters per hour per hour.
(d)
12000 − ∫ r ( t ) dt = 9000 .
A
0
Notes:
1.
Calculating an integral of a piecewise-defined discontinuous function is rare, if not
unprecedented, in a free response problem. Nonetheless, integrating such functions
is obviously a skill that is tested.
FREE-RESPONSE SOLUTIONS ~ 2011 AB (FORM B)
15
Question AB-3 (Form B)
∫ f ( x ) dx + ∫
4
6
0
4
2
g ( x ) dx = x3/2
3
4
1
16
22 1
+ ⋅2⋅2 = + 2 =
.
2
3
3
0
(a)
Area =
(b)
Volume =
(c)
The graph of g has the slope −1 . So we must have
1
1
1
=1 ⇒ x = , y = .
f ′( x) = 1 ⇒
4
2
2 x
∫
2
0
2 y (6 − y − y 2 ) dy .
Notes:
1.
It is easier to use geometry — the area of the triangle — for
∫ g ( x ) dx .
6
4
Otherwise,
6
⎛
36 ⎞ ⎛
16 ⎞
x2 ⎞
⎛
(6
−
)
=
6
−
x
dx
x
⎜
⎟ = ⎜ 36 − ⎟ − ⎜ 24 − ⎟ = 2 .
∫4
2⎠4 ⎝
2 ⎠ ⎝
2⎠
⎝
6
Question AB-4 (Form B)
(a)
f has a critical point where f ′ ( x ) = ( 4 − x ) x −3 = 0 ⇒ x = 4 . This is a relative
maximum for f since f ′ ( x ) changes sign from positive to negative at x = 4.
(b)
f ′ ( x ) = 4 x −3 − x −2 ⇒ f ′′( x) = −12 x −4 + 2 x −3 = x −4 (−12 + 2 x) .
f ′′( x) < 0 if and
only if 0 < x < 6. Therefore, the graph of f is concave down on (0, 6) .1
(c)
f ( x) = f (1) + ∫ f ′(t ) dt = 2 + ∫
x
1
x
1
( 4t
−3
− t −2 ) dt =
x
t −2 t −1
2 1
2 1
= 2 − 2 + − (−2 + 1) = 3 − 2 + .
2 + 4⋅ −
−2 −1 1
x
x
x
x
Notes:
1.
Or (0, 6] . x = 0 must be excluded, because it is not in the domain of f.
16
FREE-RESPONSE SOLUTIONS ~ 2011 AB (FORM B)
Question AB-5 (Form B)
(a)
(b)
a ( 5 ) = v′ ( 5 ) ≈
∫
60
0
v ( t ) dt is the total distance in meters Ben traveled on his unicycle during the
60 seconds.
(c)
(d)
v (10 ) − v ( 0 ) 2.3 − 2
=
= 0.03 meters per second per second.
10 − 0
10
∫
60
0
v ( t ) dt ≈ 2.0 ⋅ (10 − 0) + 2.3 ⋅ (40 − 10) + 2.5 ⋅ (60 − 40) meters.1
Yes. The Mean Value Theorem applied to the time interval [40, 60] guarantees a
B ( 60 ) − B ( 40 ) 49 − 9
=
= 2.
time t when v ( t ) = B′ ( t ) =
60 − 40
20
B ( t ) B′ ( t )
2 L ( t ) L′ ( t ) = 2 B ( t ) B ′ ( t ) ⇒ L′ ( t ) =
. At t = 40,
L (t )
( L ( 40 ) )
2
L′ ( 40 ) =
= 144 + ( B ( 40 ) )
2
⇒ L ( 40 ) = 144 + 81 = 225 = 15 . So
9 ⋅ 2.5
meters per second. 2
15
Notes:
1.
2.
And leave it at that to save time and avoid arithmetic mistakes. (This is equal to 139
meters.)
= 1.5 m/sec.
Courtesy John Mahoney, ap-calc EDG moderator,
http://www.nthf.org/inductee/05mahoney.htm.
FREE-RESPONSE SOLUTIONS ~ 2011 AB (FORM B)
17
Question AB-6 (Form B)
(a)
(b)
∫
4π
− 2π
1
⎛ x⎞
⎛ x⎞
f ( x ) dx = ∫ g ( x ) dx − ∫ cos ⎜ ⎟ dx = ⋅ 6π ⋅ 2π − 2sin ⎜ ⎟
− 2π
− 2π
2
⎝2⎠
⎝2⎠
4π
4π
4π
= 6π 2 .
− 2π
f has a critical point at x = 0 since g ′ ( 0 ) doesn’t exist and so f ′ ( 0 ) doesn’t exist.
1
⎛x⎞
For x ≠ 0 , f ′ ( x ) = g ′ ( x ) + ⋅ sin ⎜ ⎟ . On the interval (−2π , 0) ,
2
⎝2⎠
1 ⎛x⎞
f ′ ( x ) = 1 + sin ⎜ ⎟ , so the equation f ′( x) = 0 has no solutions. On the interval
2 ⎝2⎠
1 1
⎛x⎞
⎛x⎞
(0, 4π ) , f ′ ( x ) = − + sin ⎜ ⎟ . f ′( x) = 0 at x = π , where 1 = sin ⎜ ⎟ . The
2 2 ⎝2⎠
⎝2⎠
answer is: f has two critical points on (−2π , 4π ) , at x = 0 and at x = π .
(c)
⎛ π⎞
h′ ( x ) = 3 g ( 3 x ) ⇒ h′ ⎜ − ⎟ = 3g ( −π ) = 3π .
⎝ 3⎠
2011 BC (Form B)
AP Calculus Free-Response
Solutions and Notes
Question BC-1 (Form B)
See AB Question 1 (Form B).
Question BC-2 (Form B)
2
1 π
r (θ ) ) dθ
(
∫
2 π /2
­ ≈ 47.513 .
(a)
Area =
(b)
Solving r (θ ) cos (θ ) = −3 , we get­ θ ≈ 2.017 . At this angle,
y = r (θ ) sin (θ ) ­ ≈ 6.272 .
(c)
y = r (θ ) sin (θ ) = ( 3θ + sin θ ) sin θ ⇒
dy d
=
dt dθ
( ( 3θ + sin θ ) sin θ ) ddtθ .
At
2π dy
,
­ ≈ −1.40954 ⋅ 2 ≈ −2.819 . This means that when θ = 2π , the
3 dt
3
y-coordinate of the particle is decreasing at the rate of 2.819 units of length per unit
of time.
θ=
Question BC-3 (Form B)
See AB Question 3 (Form B).
19
20
FREE-RESPONSE SOLUTIONS ~ 2011 BC (FORM B)
Question BC-4 (Form B)
1 5
1
f ( x ) dx = ( −10 ) = −2 .
∫
0
5
5
(a)
The average value of f on [0, 5] is
(b)
∫
(c)
Since g ′ ( x ) = f ( x ) and f ( x) < 0 on (0, 5), g is decreasing on [0, 5]. The graph of
10
0
f ( x ) dx = −10 + 27 = 17 ⇒
∫ ( 3 f ( x ) + 2 ) dx = 3 ⋅17 + 20 = 71 .
10
0
g is concave up where g ′ = f is increasing, that is when 3 < x < 8. g is both concave
up and decreasing on (3, 5).1
(d)
The arc length is
∫
20
0
1 + ( h′ ( x ) ) dx = ∫
20
2
0
2
10
2
⎛ ⎛ x ⎞⎞
1 + ⎜ f ′ ⎜ ⎟ ⎟ dx = 2 ∫ 1 + ( f ′ ( u ) ) du , where
0
⎝ ⎝ 2 ⎠⎠
x
1
u = , du = dx . Therefore, the arc length of the graph of h from 0 to 20 is twice
2
2
the arc length of the graph of f on [0, 10]. The latter is 11 + 18 = 29 , so the arc length
of the graph of h from 0 to 20 is equal to 58.
Notes:
1.
Or [3, 5].
Question BC-5 (Form B)
See AB Question 5 (Form B).
FREE-RESPONSE SOLUTIONS ~ 2011 BC (FORM B)
21
Question BC-6 (Form B)
(a)
(b)
(c)
ln(1 + x3 ) = x3 −
3n
x 6 x9 x12
n +1 x
+ −
+ ... + ( −1) ⋅
+ ... .
n
2 3
4
1 1 1
n +1 1
At x = 1, the series is 1 − + − + ... + ( −1) ⋅ + ... . It converges by the
2 3 4
n
alternating series convergence test. At x = −1 , the series is
1 1 1
1
−1 − − − − ... − − ... . It diverges, because it is the negative of the harmonic
2 3 4
n
series. The interval of convergence is ( −1, 1] .
The first four terms of the series for f ′(t ) are
f ′(t ) = 3t 2 −
6t 5 9t 8 12t11
+
−
+ ... = 3t 2 − 3t 5 + 3t 8 − 3t11 + ... ⇒
2
3
4
1
⎛ 3t 5 3t11 ⎞
3 3
′
f (t ) = 3t − 3t + 3t − 3t + ... . g (1) ≈ ∫ ( 3t − 3t ) dt = ⎜
−
⎟ = − .
0
11 ⎠ 0 5 11
⎝ 5
2
(d)
4
10
16
22
1
4
10
The alternating series error does not exceed the absolute value of first omitted term,
1
3 1
16
∫ 0 3t dt = 17 < 5 .