MATH 105 203
Midterm 2 Solutions
MATH 105 203 Midterm 2 Solutions
1. (a) Find the derivative of the function:
x2
Z
cos(t2 ) dt
f (x) =
x
at the point x = 0.
Solution: We have that:
x2
Z
cos(t2 ) dt
f (x) =
x
0
Z
Z
2
=
x2
cos(t2 ) dt
cos(t ) dt +
x
0
Z
x
=−
x2
Z
2
cos(t2 ) dt
cos(t ) dt +
0
0
Apply the Fundamental Theorem of Calculus part I, we get:
f 0 (x) = − cos(x2 ) + 2x cos(x4 )
⇒ f 0 (0) = − cos(0) + 0 = −1
(b) Use Simpson’s rule to approximate:
Z
1
2
dx
x
with n = 4 subintervals. Find a bound on the error. No need to simplify your
answers!
1
b−a
1
Solution: We have a = 1, b = 2, n = 4, and f (x) = . So, ∆x =
= .
x
n
4
There are 5 grid-points using the formula xk = a + k∆x:
x0 = 1,
x1 = 1.25,
x2 = 1.5,
x3 = 1.75,
x4 = 2
Using Simpson’s Rule, we get:
∆x
(f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + f (x4 ))
3
1 1
4
2
4
1
=
+
+
+
+
12 1 1.25 1.5 1.75 2
S4 =
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MATH 105 203
Midterm 2 Solutions
To find the error bound, we want to compute f (4) (x):
f 0 (x) = −x−2 ,
f 00 (x) = 2x−3 ,
f (3) (x) = −6x−4 ,
f (4) (x) = 24x−5
We want to find the maximum value of |f (4) (x) on [1, 2], so we want to find
its critical points. Since f (5) (x) = −120x−6 6= 0 for any value of x, there are
no critical points and the maximum occurs at one of the endpoints. We have
that |f (4) (1)| = 24 and |f (4) (2)| = 43 . Thus, for K = 24, |f (4) (x)| ≤ K for all x
between 1 and 2. Therefore, the error bound for S4 is:
E4 ≤
K(b − a)(∆x)4
24
=
180
180(4)4
(c) Find the definite integral
Z
π
sec2 x dx.
0
π
Solution: Since sec2 x is undefined at x = , the above definite integral is an
2
improper integral. Thus,
Z
π
Z
2
sec x dx =
0
π
2
Z
2
π
sec2 x dx,
sec x dx +
π
2
0
if both improper integrals on the right hand side converge. We have that:
Z
0
π
2
Z
2
sec x dx = limπ
b→ 2
b
secx dx
0
= lim
tan x |b0
π−
b→ 2
= lim
tan b = ∞
π−
b→ 2
Z
π
2
Since
Z
2
sec x dx diverges, we get that the definite integral
0
π
sec2 x dx also
0
diverges.
(d) A student randomly guesses at each answer in a true/false quiz consisting of 3 questions. Let X be the random variable representing the number of correct answers.
Find the probability density function of X.
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MATH 105 203
Midterm 2 Solutions
Solution: All possible outcomes are: X = 0, X = 1, X = 2 and X = 3.
1
Moreover, each question has a probability of of being guessed correctly. We
2
write C for a correct answer and I for an incorrect answer in the same order
as that of the questions. Then, there are 8 different possible combinations of
answers: CCC, CCI, CIC, ICC, CII, ICI, IIC, and III.
So, the probability density function f of X is:
1
f (X = 0) = ,
8
3
f (X = 1) = ,
8
3
f (X = 2) = ,
8
1
f (X = 3) = .
8
(e) Find the indefinite integral
Z
sin3 (x) cos10 (x) dx.
Solution: Since the exponent of sin(x) is odd, we want to use substitution
with u = cos x, and du = − sin(x) dx, we get:
Z
Z
3
10
sin (x) cos (x) dx = sin2 (x) cos10 (x) sin(x) dx
Z
= (1 − cos2 (x)) cos10 (x) sin(x) dx
Z
= −(1 − u2 )u10 du
Z
= (−u10 + u12 ) du
1
1 11
u + u13 + C
11
13
Z
1
1
⇒ sin3 (x) cos10 (x) dx = − cos11 (x) +
cos13 (x) + C
11
13
=−
(f) Solve the initial value problem
t
e−t y 0 = ,
y
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y(0) = −5.
MATH 105 203
Midterm 2 Solutions
Solution: Rewrite the differential equation into its separable form:
1
dy
=
tet
dx
y
Since
1
6= 0 for any values of y, there are no constant solutions. Then,
y
Z
y dy = tet dt
Z
y dy = tet dt
1 2
y + C = tet − et
2
where the integral on the right hand side is solved using integration by parts
with u = t, du = dt, and dv = et dt, v = et . Since y(0) = −5, we can solve for
C:
1
25 + C = −1
2
27
C=−
2
So, we get:
1 2 27
y −
= tet − et
2
2
√
⇒ y = ± 2tet − 2et + 27
Thus, since y(0) = −5, we get that the solution to the initial value problem is:
√
y = − 2tet − 2et + 27
2. Evaluate the definite integral:
ln
Z
√
3
2
e2t
3
0
(1 + e4t ) 2
dt.
Solution: First, using a direct substitution with u = e2t , and du = 2e2t dt, when
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MATH 105 203
Midterm 2 Solutions
√
√
ln 3
t = 0, u = 1, and when t =
, u = 3, we get:
2
Z
ln
√
3
√
e2t
2
Z
(1 + e4t )
0
3
2
dt =
3
1
3
1
2(1 + u2 ) 2
du
Using inverse trigonometric substitution with u = tan s, that is, s = arctan u, and
√
π
π
du = sec2 s ds, when u = 1, s = , and when u = 3, s = , we get:
4
3
√
Z π
Z 3
3
1
1
2
du
=
3
3 sec s ds
2
2
π
2
2
2(1 + u )
2(1 + tan s)
1
4
Z π
3
sec2 s
=
3 ds
2 s) 2
π
2(sec
4
Z π
3
1
=
ds
π
2
sec
s
4
Z π
3 1
=
cos s ds
π
2
4
√
√
π
1
1
π
1
π
3
2
= sin s | π3 = sin − sin =
−
4
2
2
3 2
4
4
4
Therefore,
Z
0
ln
√
2
3
√
√
3− 2
.
3 dt =
4
(1 + e4t ) 2
e2t
3. The time to failure of a transistor (in years) is a continuous random variable whose
cumulative distribution function is given by:
(
1 − e−mx if x ≥ 0,
F (x) =
0 otherwise,
where m is an unknown constant.
(a) Find the probability density function of X.
Solution: Let f be the probability density function of X. Then, f (x) =
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MATH 105 203
Midterm 2 Solutions
d
F (x). Thus, we get:
dx
(
me−mx if x ≥ 0,
f (x) =
0 otherwise,
where m is a constant.
(b) If the expected time to failure of a transistor is 10, find m.
Solution: We have that:
Z
∞
10 =
xf (x) dx
Z−∞
∞
mxe−mx dx
0
Z b
mxe−mx dx
= lim
=
b→∞
0
Using integration by parts with u = x, du = dx, and dv = me−mx dx, v =
−e−mx , we get:
Z b
10 = lim
mxe−mx dx
b→∞
0
1 −mx b
e
|0
b→∞
m
1
1
= lim −be−mb − e−mb +
b→∞
m
m
1
=
m
= lim −xe−mx −
Therefore, m =
1
.
10
(c) What is the probability that a transistor will last for at least 15 years?
Solution: Recall that F (x) represents the probability of X ≤ x, so the probability that a transistor will last for at least 15 years (ie. X ≥ 15 ) is:
1
1 − F (15) = 1 − (1 − e−( 10 )15 ) = e− 2 .
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3
MATH 105 203
Midterm 2 Solutions
4. (Extra credit) The monthly average price of silver has been growing at a rate proportional to the square root of the price since November 2011. The average price in
November 2011 of one gram of silver was $16 and the average price of the same in
February 2012 was $25. Write down the initial value problem of the monthly value of
silver as a function of time. Do not solve this problem!
Solution: Let y(t) represent the monthly value of silver as a function of time since
November 2011. Since the monthly average price of silver has been growing at a
rate proportional to the square root of the price since November 2011, we get:
dy
√
= k y,
dt
for some constant k. The information that the average price in November 2011 of
one gram of silver was $16 and the average price of the same in February 2012 was
$25, can be translated into:
y(0) = 16,
y(3) = 25
Thus, the initial value problem of the monthly value of silver as a function of time
is:
dy
√
= k y,
dt
y(0) = 16,
where k is some constant.
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y(3) = 25,