Math 101 Fall 2006 Exam 2 Solutions
Instructor: S. Cautis/M. Simpson/R. Stong
Tuesday, November 14, 2006
Instructions: This is a closed book, closed notes exam. Use of calculators is
not permitted. You have one hour and fifteen minutes. Do all 8 problems.
Please do all your work on the paper provided. You must show your work to
receive full credit on a problem. An answer with no supporting work will receive
no credit.
Please print you name clearly here.
Print name:
Upon finishing please sign the pledge below:
On my honor I have neither given nor received any aid on this exam.
Grader’s use only:
1.
/15
2.
/10
3.
/20
4.
/10
5.
/5
6.
/10
7.
/10
8.
/20
1. [15 points] Evaluate the following limits, if they exist.
(a) limu→1
tan(πu)
u2 −1
The limit is indeterminate of type 0/0 so L’Hôpital gives
lim
u→1
tan(πu)
π sec2 (πu)
π · (−1)2
π
=
lim
=
= .
2
u→1
u −1
2u
2
2
(b) limx→0+ x ln(sin x)
The limit is indeterminate of type 0·∞, so rewriting and applying L’Hôpital
gives
cos x
ln(sin x)
= lim+ sin x 2
(1/x)
x→0
x→0 (−1/x )
2
−2x cos x + x2 sin x
−x cos x
= lim+
= lim+
sin x
cos x
x→0
x→0
0−0
= 0.
=
1
lim+ x ln(sin x) = lim+
x→0
1
(c) limx→0 (x + 1) 2 sin x
The limit is indeterminate of type 1∞ , so rewriting and applying L’Hôpital
gives
1
ln(x + 1)
2
sin
x
= lim exp
lim (x + 1)
x→0
x→0
2 sin x
ln(x + 1)
= exp lim
x→0 2 sin x
!
= exp
lim
1
x+1
x→0 2 cos x
1
= e1/2 .
= exp
2
2. [10 points] Find the first three derivatives of the following function
f (x) = cos(x2 + 1).
f ′ (x) = −2x sin(x2 + 1)
f ′′ (x) = −2 sin(x2 + 1) − 4x2 cos(x2 + 1)
f ′′′ (x) = −4x cos(x2 + 1) − 8x cos(x2 + 1) + 8x3 sin(x2 + 1)
= 8x3 sin(x2 + 1) − 12x cos(x2 + 1).
3. [20 points] Evaluate the following integrals:
R
(a) (1 + x3 )2 dx
Z
R
(b)
3 2
(1 + x ) dx =
Z
(x6 + 2x3 + 1)dx =
1 7 1 4
x + x + x + C.
7
2
x sec2 (3x2 + 5)dx
Substitute u = 3x2 + 5, so du = 6xdx (or xdx = 16 du) to get
Z
Z
1
1
1
2
2
x sec (3x + 5)dx =
sec2 udu = tan u + C = tan(3x2 + 5) + C.
6
6
6
R1
2x+ex
0 x2 +ex dx
(c)
Substitute u = x2 + ex , so du = (2x + ex )dx and x = 0 ⇒ u = 1,
x = 1 ⇒ u = 1 + e to get
Z
1
0
(d)
2x + ex
dx =
x2 + ex
R2
−3
Z
1
1+e
1
1+e
du = ln u|1 = ln(1 + e) − ln 1 = ln(1 + e).
u
|x − 1|dx
For x < 1, we have |x − 1| = 1 − x and for x > 1 we have |x − 1| = x − 1.
Hence
Z 2
Z 1
Z 2
|x − 1|dx =
(1 − x)dx +
(x − 1)dx
−3
−3
1
1
2
1 2
1 2 x − x = x− x +
2
2
−3
1
1
9
1
= 1−
− −3 −
+ (2 − 2) −
−1
2
2
2
17
.
=
2
Alternately, one can use areas. The region under the curve is a triangle of
height 4 and base 4 and a second triangle of height 1 and base 1. Hence
the total area is 21 · 4 · 4 + 21 · 1 · 1 = 17
2 .
R3
4. [10 points] Evaluate 1 (x − 1)2 + 1 dx from the definition of the integral. That is, compute
n
X
f (xi )∆x
lim
n→∞
i=1
for a standard partition of the interval. The following formulas may be
useful:
n
X
n
n
n(n + 1) X 2
n(n + 1)(2n + 1) X 3 n2 (n + 1)2
i =
i =
;
;
2
6
4
i=1
i=1
i=
i=1
Since a = 1, b = 3, and f (x) = (x − 1)2 + 1, we have ∆x = b−a
n =
.
Hence
xi = a + i∆x = 1 + 2i
n
!
2
n
n
n
X
X
2i 2 X
2i
2
f 1+
f (xi )∆x =
+1
=
n
n
n
n
i=1
i=1
i=1
!
!
n
n
n 2
X
2 X
8 X 2
8i
2
1
i +
= 3
+
=
n3
n
n
n i=1
i=1
i=1
8 1 3 1 2 1
2
= 3
n + n + n + ·n
n
3
2
6
n
8
14 4
4
4
4
= + + 2 +2=
+ + 2.
3 n 3n
3
n 3n
Hence
Z
1
3
n
X
f (xi )∆x
(x − 1)2 + 1 dx = lim
n→∞
= lim
n→∞
i=1
14
4
4
+ + 2
3
n 3n
=
14
.
3
2
n
and
5. [5 points] Let F (x) =
R cos x
−1
ln(t + 2)dt. What is F ′ (x)?
F ′ (x) = ln(cos x + 2)
d cos x
= − sin x ln(2 + cos x).
dx
6. [10 points] A particle moves along the x-axis with acceleration function
a(t) = 1 − e−t , the initial position is x(0) = 2 and the initial velocity is
v(0) = 0. Find the particle’s position function x(t) as a function of time.
The velocity is an antiderivative of accelaration, hence
v(t) = t + e−t + C.
Plugging in t = 0, we see 0 = v(0) = 0 + 1 + C so C = −1 and
v(t) = t + e−t − 1.
The position is an antiderivative of velocity, hence
x(t) =
1 2
t − e−t − t + C.
2
Plugging in t = 0, we see 2 = x(0) = 0 − 1 − 0 + C, so C = 3 and
x(t) =
1 2
t − e−t − t + 3.
2
7. [10 points] Find the area of the region bounded by the curves y =
and y = x2 .
√
8x
√
The curve y = 8x is the upper half of a parabola pointed in the positive
x-direction and the curve y = x2 is an ordinary parabola. The
√ two curves
intersect when their y coordinates are equal, i.e., when 8x = x2 , or
8x = x4 or x(x3 − 8) = 0. Hence the intersections are at x = 0 and x = 2
and these are the
x vlaues in the region. Plugging in
√ lowest and highest
2
x = 1, we get
√ 8 = 2.828 · · · > 1 = 1, thus we see2 the higher curve is
y = f (x) = 8x and the lower curve is y = g(x) = x . Hence
b
Area =
Z
=
Z
=
a
2
0
(f (x) − g(x))dx =
Z
2
0
√
(2 2x1/2 − x2 )dx =
16 8
−
3
3
8
−0= .
3
√
( 8x − x2 )dx
!2
√
4 2 3/2 1 3 x − x 3
3
0
√x−18 , the first
x2 +36
−36(x−3)(x+6)
. YOU
(x2 +36)5/2
8. [20 points] For the function f (x) =
′
18(x+2)
(x2 +36)3/2
′′
f (x) =
and f (x) =
IFY THESE FORMULAS.
two derivatives are
NEED NOT VER-
(a) Find (and justify) all horizontal and vertical asymptotes of the graph
y = f (x). At any vertical asymptotes compute both the left and right
hand limits of f (x).
The denominator of f is never zero, so there are no vertical asymptotes.
Since
1 − 18
1
x − 18
x
= lim q
= = 1,
lim √
2
x→∞
1
x + 36 x→∞ 1 + 362
x
there is a horizontal asymptote at y = 1, approached as x → ∞. Since
1 − 18
1
x − 18
x
= lim − q
= − = −1,
lim √
2
x→−∞
1
x + 36 x→−∞
1 + 362
x
there is a horizontal asymptote at y = −1, approached as x → −∞.
(b) Find the intervals on which f (x) is increasing and those on which it
is decreasing.
The denominator of f ′ (x) is positive so f ′ (x) > 0 for x > −2 and f ′ (x) <
0 for x < −2. Hence f is increasing on (−2, ∞), or [−2, ∞) and f is
decreasing on (−∞, −2), or (−∞, −2].
CONTINUED ON THE NEXT PAGE
(c) Find the critical points of f (x) and classify them as local maxima,
local minima or neither.
The denominator of f ′ (x) is positive (hence never zero), so the only critcal
points are where f ′ (x) = 0, i.e., at x = −2. Since f ′ (x) changes from
negative to positive at x = −2, the first derivative test shows x = −2 is a
local minimum. (It also shows that x = −2 is a global minimum, but this
was not asked.) Alternately, since
f ′′ (−2) =
720
> 0,
(40)5/2
the second derivative test shows x = −2 is a local minimum.
(d) Find the intervals on which f (x) is concave upward and those on which
it is concave downward.
. Since the denominator is always posiRecall that f ′′ (x) = −36(x−3)(x+6)
(x2 +36)5/2
tive, f ′′ (x) can only change sign at x = −6 and x = 3. For x < −6, both
x − 3 and x + 6 (and −36) are negative so f ′′ (x) < 0. For −6 < x < 3,
x − 3 and −36 are negative, but x + 6 is positive, so f ′′ (x) > 0. For x > 3,
both x − 3 and x + 6 are positive, so f ′′ (x) < 0. (Alternately, one can plug
in one value in each interval such as
360
<0
(85)5/2
648
1
f ′′ (0) = 5 =
>0
6
12
360
<0
f ′′ (4) = −
(52)5/2
f ′′ (−7) = −
to get these conclusions.)
Hence f is concave down of (−∞, −6), f is concave up on (−6, 3) and f
is concave down on (3, ∞).
√x−18
x2 +36
showing the results
√
of (a)-(d). √
(The following values may
√ be helpful f (−6) = −2 2 ≈ −2.83,
f (−2) = − 10 ≈ −3.16, f (3) = − 5 ≈ −2.24.)
(e) On the next page sketch the graph of y =