Review Session for Final Exam
April 17, 2015
All Problems in Final Exam in 2014 Spring
9
1. (a) Let Q : −x + 5y − 3z = 2 and R : 3x − y − 4z = 0 be two planes. Determined if Q and R are parallel,
5
orthogonal, or identical.
By the definitions ofQ and R, we know that one normal vector of Q is n = h−1, 5, −3i, and
9
0
one normal vector of R is n = 3, − , −4 . Notice that
5
9
0
n · n = h−1, 5, −3i · 3, − , −4
5
9
+ (−3) · (−4)
= −1 · 3 + 5 · −
5
= −3 − 9 + 12
Answer.
= 0.
So n is orthogonal to n0 , which implies that Q and R are orthogonal .
Remark 1. Plane: The plane passing through the point P0 = (x0 , y0 , z0 ) with a nonzero normal vector
n = ha, b, ci is described by the equation:
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0,
or
ax + by + cz = d,
where d = ax0 + by0 + cz0 .
Remark 2. For any two distinct planes P and Q with normal vectors n and m respectively, the angle
between these two planes P and Q is defined as the acute angle between n and m (i.e., between 0 and
π/2). In particular,
a. Two distinct planes are parallel if their respective normal vectors are parallel (that is, the normal
vectors are scalar multiples of each other, or the angle is either 0 or π).
b. Two distinct planes are orthogonal if their respective normal vectors are orthogonal (that is, the
dot product of the normal vectors is zero, or the angle is π2 ).
(b) The volume of a right circular cone of radius x and height y is V (x, y) =
V (x, y) = π.
Answer.
πx2 y
. Graph the level curve
3
The equation of the level curve V (x, y) = π is:
πx2 y
= π.
3
Then πx2 y = 3π, that is,
y=
3
,
x2
for all x > 0 .
Remark 3. Given a surface/function z = f (x, y), for any z0 , the curve f (x, y) = z0 in the xy-plane is
called a level curve of the surface z = f (x, y) with level z0 .
(c) Let f (x, y) = sin(xy). Find
Answer.
∂2f
.
∂x∂y
In fact, we have
fx = cos(xy) · y
By the chain rule
= y cos(xy)
fxy = cos(xy) + y · [− sin(xy) · x]
By the product rule and the chain rule
= cos(xy) − xy sin(xy).
So we have
∂2f
= fxy = cos(xy) − xy sin(xy) .
∂x∂y
Remark 4. To compute fx (x, y) is just to compute the derivative of f (x, y) with respect to x when
treating y is a constant; to compute fy (x, y) is just to compute the derivative of f (x, y) with respect to
y when treating x is a constant.
Remark 5. The Clairuat’s theorem, fxy = fyx , just says that we can switch the orders when computing partial derivatives.
(d) Use sigma notation to write the midpoint Riemann sum for f (x) = x8 on [5, 15] with n = 50. Do not
evaluate the midpoint Riemann sum.
Answer.
Since n = 50, then
∆x =
15 − 5
10
=
= 0.2,
50
50
and
xk = 5 + 0.2k,
for all k = 0, 1, 2, · · · , 50.
So the midpoints are:
mk =
xk−1 + xk
5 + 0.2(k − 1) + 5 + 0.2k
=
= 5 + 0.2(k − 0.5),
2
2
for all k = 1, 2, · · · , 50.
Hence the midpoint Riemann sum is:
n
X
f (mk )∆x =
k=1
50
X
[5 + 0.2(k − 0.5)]8 · 0.2 .
k=1
Remark 6. For a regular partition x0 = a < x1 < · · · < xn = b of [a, b], then
∆x =
b−a
,
n
and
xk = a + k∆x,
for all k = 0, 1, 2, · · · , n.
The Riemann sum of f (x) on [a, b] with n subintervals can be written:
f (x∗1 )∆x
+
f (x∗2 )∆x
+ ··· +
f (x∗n )∆x
=
n
X
f (x∗k )∆x.
k=1
For the choices x∗k ’s, we have
a. For the left Riemann sum, we have x∗k = xk−1 = a + (k − 1)∆x.
b. For the right Riemann sum, we have x∗k = xk = a + k∆x.
xk−1 + xk
a + (k − 1)∆x + a + k∆x
c. For the midpoint Riemann sum, we have x∗k = mk =
=
=
2
2
1
a+ k−
∆x.
2
5
Z
(e) Evaluate
f (x) dx, where f (x) =
1
Answer.
3, if x ≤ 3,
x, if x ≥ 3.
In fact, we have
Z
5
Z
3
Z
5
f (x) dx =
f (x) dx +
f (x) dx
1
3
Z 5
Z 3
x dx
3 dx +
=
1
3
1
=
3x|31 +
5
x2 2 3
= 3(3 − 1) +
52 − 32
2
= 14.
So we get
Z
5
f (x) dx = 14 .
1
Remark 7. The following are some simple properties of the definite integrals:
Z a
•
f (x) dx = 0 for any a.
Za b
Z a
•
f (x) dx = −
f (x) dx.
a
b
Z b
Z c
Z b
•
f (x) dx =
f (x) dx +
f (x) dx for any c.
a
a
c
(f) If
f 0 (1)
= 2 and
f 0 (2)
Z
2
= 3, find
f 0 (x)f 00 (x) dx.
1
Answer.
Use the substitution, we have
Z
Z
f 0 (x)f 00 (x) dx =
f 0 (x) df 0 (x)
Z
=
u du Let u = f 0 (x)
Z
=
u du
=
=
1 2
u +C
2
1 0
[f (x)]2 + C
2
Since u = f 0 (x).
So we get
Z
1
2
2
1 0
32 − 22
5
[f 0 (2)]2 − [f 0 (1)]2
2
f (x)f (x) dx = [f (x)] =
=
= .
2
2
2
2
1
0
00
Remark 8. Substitution: Let u = g(x), then du = g 0 (x) dx and
Z
0
f (g(x))g (x) dx =
Z
Z
f (u) du,
b
0
Z
g(b)
f (g(x))g (x) dx =
and
a
f (u) du.
g(a)
Z
(g) Evaluate
Answer.
cos−1 (y) dy.
Use the integration by parts, by the ‘ILATE’ rule, let
1
u = cos−1 (y) =⇒ du = − p
dy
1 − y2
Z
dv = dy =⇒ v =
dy = y.
Then we have
Z
cos
−1
Z
(y) dy =
u dv
Z
= uv − v du
Z
y
p
dy
1 − y2
Z
1
dw
−1
= y cos (y) −
Let w = 1 − y 2 , then dw = −2y dy
1
2
w2
1
1
1
w− 2 +1 + C
= y cos−1 (y) − ·
1
2 1− 2
= y cos
−1
(y) +
1
= y cos−1 (y) − w 2 + C
p
= y cos−1 (y) − 1 − y 2 + C
Since w = 1 − y 2 .
So we get
Z
cos−1 (y) dy = y cos−1 (y) −
p
1 − y2 + C .
Remark 9. Integration by parts: Let u and v be differentiable, then
Z
Z
u dv = uv −
Z
v du,
and
b
0
u(x)v (x) d =
a
u(x)v(x)|ba
Z
−
b
v(x)u0 (x) dx.
a
That is,
Z
0
uv dx = uv −
Z
vu0 dx.
where dv is the most complicated portion of the integrand that can be “ easily” integrated, and u is that
portion of the integrand whose derivative du is a “ simpler ” function than u itself. For the choice of u,
please follow the ‘ILATE’ order:
I = Inverse trigonometric function
L = Logarithmic function
A = Algebraic function
T = Trigonometric function
E = Exponential function
Z
cos3 x sin4 x dx.
(h) Evaluate
Answer.
In fact, we have
Z
Z
1
3
4
cos x sin x dx =
cos3 x sin4 ·
d sin x
cos x
Z
=
cos2 x sin4 d sin x
Z
=
(1 − sin2 x) sin4 d sin x Since sin2 x + cos2 x = 1
Z
=
[sin4 x − sin6 x] d sin x
Z
=
(u4 − u6 ) du Let u = sin x
=
=
u5 u7
−
+C
5
7
sin5 x sin7 x
−
+C
5
7
Since u = sin x.
So we get
Z
Z
Remark 10. I. Evaluate
cos3 x sin4 x dx =
sinm (x) cosn (x) dx:
Cases
m odd and positive, n any real number
n odd and positive, m any real number
m and n both even
Z
II. Evaluate
sin5 x sin7 x
−
+C .
5
7
Strategy
Split off sin(x), rewrite the resulting even powers of sin(x)
in terms of cos(x), and then use u = cos(x)
Split off cos(x), rewrite the resulting even powers of cos(x)
in terms of sin(x),and then use u = sin(x)
Use half-angle formulas to transform the integrand into a
polynomial in cos(2x), and apply the preceding strategies
once again to powers of cos(2x) greater than 1.
tanm (x) secn (x) dx:
Cases
n even
m odd
m even and n odd
Strategy
Split off sec2 (x), rewrite the remaining even power of sec(x) in terms of tan(x),
and use u = tan(x)
Split off sec(x) tan(x), rewrite the remaining even power of tan(x) in terms of
sec(x), and use u = sec(x)
Rewrite the even power of tan(x) in terms of sec(x) to produce a polynomial in
sec(x), apply reduction formula to each term
III. One does not need to remember the above two tables, just try all possible substitutions d cos(x),
d sin(x), d tan(x), d sec(x), otherwise, try half angle identities.
Z
(i) Evaluate
Answer.
√
dx
.
3 − 2x − x2
In fact, by the completing square, we have
3 − 2x − x2 = −[x2 + 2x − 3]
= −[(x + 1)2 − 1 − 3]
= −[(x + 1)2 − 4]
= 4 − (x + 1)2 .
So we get
Z
dx
√
3 − 2x − x2
Z
dx
p
4 − (x + 1)2
Z
du
√
=
Let u = x + 1
4 − u2
u
= sin−1
+C
2
x+1
+ C Since u = x + 1.
= sin−1
2
=
So we have
Z
√
dx
= sin−1
2
3 − 2x − x
x+1
2
+C .
Remark 11. The completing square of a quadratic polynomial:
b
c
ax2 + bx + c = a x2 + x +
a
a
"
=a
b
x+
2a
2
−
b
2a
2
#
c
+
.
a
Z
(j) Evaluate
Answer.
x − 13
dx.
x2 − x − 6
Since x2 − x − 6 = (x − 3)(x + 2), then the partial fraction decomposition of
the form:
x − 13
has
−x−6
x2
A
B
x − 13
=
+
.
x2 − x − 6
x−3 x+2
Both sides are multiplied by the denominator x2 − x − 6 = (x − 3)(x + 2), then
x − 13 = A(x + 2) + B(x − 3)
= Ax + 2A + Bx − 3B
= (A + B)x + 2A − 3B.
So we get
A+B = 1
(1)
2A − 3B = −13
(2)
By (1) × 2 − (2), we have 5B = 15, that is, B = 3. Plug B = 3 into (1), then A = −2. So we get
2
3
x − 13
=−
+
.
x2 − x − 6
x−3 x+2
Then
Z
Z
x − 13
2
3
dx = −
dx +
dx
x2 − x − 6
x−3
x+2
= −2 ln |x − 3| + 3 ln |x + 2| + C.
Z
So we get
Z
x − 13
dx = −2 ln |x − 3| + 3 ln |x + 2| + C .
−x−6
x2
p(x)
be a proper rational function in reduced form.
q(x)
Assume the denominator q(x has been factored completed over the real numbers and m is a positive
integer, and a factor (x − r)m is in the denominator , then the partial fraction decomposition requires the
partial fractions
A1
A2
Am
+
+ ··· +
.
2
x − r (x − r)
(x − r)m
Remark 12. After using long division, let f (x) =
For example, we have
1
(x − 1)(x + 2)
1
(x + 1)(x + 3)(x − 2)
x
(x + 1)2 (x − 1)
=
=
=
A
B
+
x−1 x+2
A
B
C
+
+
x+1 x+3 x−2
A
B
C
+
+
.
x + 1 (x + 1)2 x − 1
(k) The random variable X has probability density function
(
f (x) =
0,
if x < 1,
3 −5
x 2 , if x ≥ 1.
2
Find the expected value E(X) of the random variable X.
Answer.
In fact, we have
Z
∞
xf (x) dx
E(X) =
−∞
Z 1
Z
∞
=
xf (x) dx +
xf (x) dx
−∞
1
Z ∞
3 5
x · x− 2 dx
= 0+
2
Z ∞1
3
3
x− 2 dx
=
2 1
Z
3 b −3
x 2 dx
= lim
b→∞ 2 1
b
3
3
1
= lim
· 3
· x− 2 +1 b→∞ 2 − + 1
2
1
=
1
lim −3[b− 2 − 1]
b→∞
= 3.
So we get
E(X) = 3 .
Remark 13. Let X be a continuous random variable, then
I. Cumulative distribution function(CDF) F (x) := Pr(X ≤ x), we have
a. 0 ≤ F (x) ≤ 1 for all x.
b. F (x) is a non-decreasing continuous function.
c. lim F (x) = 0.
x→−∞
d. lim F (x) = 1.
x→∞
II. Probability density function (PDF) f (x) =
dF (x)
, we have
dx
a. f (x) ≥ 0 for all x.
Z ∞
b.
f (x) dx = 1.
−∞
Z x
c. F (x) =
f (t) dt.
−∞
Z b
d. Pr(a ≤ X ≤ b) =
f (t) dt.
a
III. Expected value: Let f (x) be the PDF for a continuous random variable X, then the expected
value, or expectation, or mean, of X is defined by:
Z ∞
E(X) =
xf (x) dx.
−∞
Remark 14. Let a and b be two real numbers, then
I. Improper integrals for infinite intervals:
a. If f is continuous on [a, ∞), then
∞
Z
b
Z
f (x) dx.
f (x) dx := lim
b→∞
a
a
b. If f is continuous on (−∞, b], then
Z
b
Z
f (x) dx := lim
f (x) dx.
a→−∞
−∞
b
a
c. If f is continuous on (−∞, ∞) and for any real number c, then
∞
Z
c
Z
f (x) dx := lim
f (x) dx.
f (x) dx + lim
a→−∞
−∞
b
Z
b→∞
a
c
For the integrals in the above three cases, if the integral is finite, we say the integral converges;
otherwise, the integral diverges.
II. Improper integrals for unbounded integrands:
a. If f is continuous on (a, b] with lim f (x) = ∞ or −∞, then
x→a+
b
Z
Z
b
f (x) dx := lim
f (x) dx.
c→a+
a
c
b. If f is continuous on [a, b) with lim f (x) = ∞ or −∞, then
x→b−
Z
b
Z
f (x) dx := lim
c→b−
a
c
f (x) dx.
a
c. Let a < p < b, if f is continuous on (a, b] with lim f (x) = ∞ or −∞, then
x→p
Z
b
Z
f (x) dx := lim
a
c→p−
c
Z
f (x) dx + lim
a
d→p+
b
f (x) dx.
d
For the integrals in the above three cases, if the integral is finite, we say the integral converges;
otherwise, the integral diverges.
Z
III. For the definition of Improper integrals for infinite intervals, you can first compute f (x) dx,
then plug the upper limit and lower limit.
IV. For CaseZa and Case b in Improper integrals for unbounded integrands, you also can first
compute
f (x) dx, then plug the upper limit and lower limit. But for Case c in Improper
integrals for unbounded integrands, you must split the integral into two integrals, then do the
same steps as before.
"
n−1 #
∞ n
X
1
2
(l) Evaluate
+ −
.
3
5
n=1
Answer.
In fact, we have
"
∞ n
X
1
n=1
3
n−1 #
∞ n
∞ X
X
2
1
2 n−1
+ −
=
+
−
5
3
5
=
=
=
n=1
1
3
n=1
1
3
1−
1 5
+
2 7
17
.
14
+
1
1 − − 25
So we get
"
n−1 #
∞ n
X
1
2
17
.
+ −
=
3
5
14
n=1
Remark 15. Geometric series: For the geometric sequence {ark }∞
k=0 , then the sequence of partial
∞
sums {Sn }n=1 can be given by:

 na,
if r = 1
1 − rn
Sn =
 a·
, if r 6= 1.
1−r
Then
∞
X
k=0
(
k
ar =
a
, if |r| < 1,
1−r
diverges,
if |r| ≥ 1.
To remember the sum of the geometric series
∞
X
ark with |r| < 1, you just remember that the sum is
k=m
equal to the quotient of the initial term and 1 − r. The partial sum Sn is equal to the quotient of the
difference of the first term and the last term∗r and 1 − r. That is,
the sum of the geometric series =
the partial sum of the geometric series =
the first term
1 − the ratio
the first term − the last term × the ratio
.
1 − the ratio
(m) Let k be a constant. Find the value of k such that f (x) = 1 + k|x| is a probability density function on
−1 ≤ x ≤ 1.
Answer.
In fact, since f (x) is a probability density function on −1 ≤ x ≤ 1, then
Z
1
f (x) dx
1 =
−1
Z 1
=
(1 + k|x|) dx
−1
Z 0
Z
1
(1 − kx) dx +
=
−1
(1 + kx) dx
0
0
1
kx2 kx2 =
x−
+ x+
2 −1
2 0
k
k
= 1+ +1+
2
2
= 2 + k.
So k + 2 = 1, that is, k = −1 .
Remark 16. Let X be a continuous random variable, then
I. Cumulative distribution function(CDF) F (x) := Pr(X ≤ x), we have
a. 0 ≤ F (x) ≤ 1 for all x.
b. F (x) is a non-decreasing continuous function.
c. lim F (x) = 0.
x→−∞
d. lim F (x) = 1.
x→∞
II. Probability density function (PDF) f (x) =
dF (x)
, we have
dx
a. f (x) ≥ 0 for all x.
Z ∞
b.
f (x) dx = 1.
−∞
Z x
c. F (x) =
f (t) dt.
−∞
Z b
d. Pr(a ≤ X ≤ b) =
f (t) dt.
a
III. Expected value: Let f (x) be the PDF for a continuous random variable X, then the expected
value, or expectation, or mean, of X is defined by:
Z ∞
E(X) =
xf (x) dx.
−∞
Z
(n) Let h(s) be a continuous function with h(10) = 2. If f (x, y) =
xy
h(s) ds, find fx (2, 5).
1
Answer.
By the chain rule, we have
fx (x, y) = h(xy) · y − h(1) · 0 = yh(xy).
Then
fx (2, 5) = 5h(10) = 5 · 2 = 10 .
Z
x
f (t) dt is an an-
Remark 17. Fundamental theorem of calculus: If f (x) is continuous, then
a
tiderivative of f (x), and
d
dx
Z
b(x)
!
f (t) dt
= f (b(x)) · b0 (x) − f (a(x)) · a0 (x).
a(x)
Remark 18. To compute fx (x, y) is just to compute the derivative of f (x, y) with respect to x when
treating y is a constant; to compute fy (x, y) is just to compute the derivative of f (x, y) with respect to
y when treating x is a constant.
2. (a) Determine whether the series
∞
X
k=1
√
3
k4 + 1
√
converges.
k5 + 9
√
√
√
4
3
3
3
k3
k4 + 1
k4 + 1
k4
1
Answer. Since √
> 0 for all k ≥ 1 and √
behaves like √ = 5 = 7 as k → ∞. Let’s
5
5
5
k +9
k +9
k
k2
k6
use the limit comparison test, in fact, we have
√
3 4
√ k +1
k5 +9
1
=
7
k6
√
7
k 6 · 3 k4 + 1
√
k5 + 9
1
7
=
k 6 · [(k 4 + 1)2 ] 6
1
[(k 5 + 9)3 ] 6
=
k 7 · (k 4 + 1)2
(k 5 + 9)3
16
→ 1,
as k → ∞.
∞
X
1
∞
X
7
1
Since
> 1, then
7 is a p-series with p =
7 converges. By the limit comparison test, we know
6
6
k6
k=1 k
k=1
∞ √
3
X
k4 + 1
√
that the series
also converges .
k5 + 9
k=1
Remark 19. Tests of Infinite series:
∞
X
I. Divergence test: If a series
ak converges, then lim ak = 0. That means, if lim ak 6= 0, then
the series
∞
X
k→∞
k=1
k→∞
ak diverges.
k=1
Remark 20. But if lim ak = 0, the divergence test does NOT work.
k→∞
II. Integral test: If f (x) is a continuous, positive, decreasing function for x ≥ 1, let ak = f (k) for
k = 1, 2, · · · , then
Z ∞
∞
X
ak
and
f (x) dx
1
k=1
either both converge or both diverge. In the case of convergence, the value of the integral is NOT
equal to the value of the series.
∞
X
|ak+1 |
ak be an infinite series and r = lim
, then
III. Ratio test: Let
k→∞
|ak |
k=1
a. If 0 ≤ r < 1, the series converges.
b. If r > 1, the series diverges.
c. If r = 1, the ration test is inconclusive.
Remark 21. If the general kth term of series involves k!, k k or ak , where a is a constant, the Ratio
Test is advisable.
∞
∞
X
X
IV. Comparison test: Let
ak and
bk be series with positive terms, then
k=1
a. If 0 < ak ≤ bk for all k ≥ 1, and
k=1
∞
X
bk converges, then
k=1
b. If 0 < bk ≤ ak for all k ≥ 1 and
∞
X
k=1
bk diverges, then
∞
X
ak also converges.
k=1
∞
X
k=1
ak diverges as well.
V. Limit comparison test: Let
∞
X
∞
X
ak and
k=1
then
a. If 0 < L < ∞, then
∞
X
ak and
k=1
b. If L = 0 and
∞
X
c. If L = ∞ and
k→∞
k=1
∞
X
bk
,
ak
bk either both converge or both diverge.
k=1
bk converges, then
k=1
∞
X
bk be series with positive terms and L = lim
∞
X
ak converges.
k=1
∞
X
bk diverges, then
k=1
ak diverges.
k=1
Remark 22. If the general kth term of the series is a rational function of k, use the Comparison
Test or the Limit Comparison Test.
∞
∞
X
X
VI. Absolutely convergence: If the series
|ak | converges, then the series
ak converges.
k=1
Remark 23. If
∞
X
ak converges but
∞
X
k=1
For example,
∞
X
(−1)k
k=1
k
k=1
|ak | diverges, we say that
k=1
∞
X
ak conditionally converges.
k=1
conditionally converges.
VII. p-series: Let p be a real number, then
∞
X
1
converges, if p > 1,
=
p
diverges,
if p ≤ 1.
k
k=1
∞
VIII. Geometric series: For the geometric sequence {ark }∞
k=0 , then the sequence of partial sums {Sn }n=1
can be given by:

 na,
if r = 1
1 − rn
Sn =
 a·
, if r 6= 1.
1−r
Then
∞
X
k=0
(
k
ar =
a
, if |r| < 1,
1−r
diverges,
if |r| ≥ 1.
Remark 24. To remember the sum of the geometric series
∞
X
ark with |r| < 1, you just remember
k=m
that the sum is equal to the quotient of the initial term and 1 − r. The partial sum Sn is equal to
the quotient of the difference of the first term and the last term∗r and 1 − r. That is,
the sum of the geometric series =
the partial sum of the geometric series =
the first term
1 − the ratio
the first term − the last term × the ratio
.
1 − the ratio
(b) Find the radius of convergence of the power series
∞
X
k=0
Answer.
Let ck =
xk
10k+1 (k
+ 1)!
.
1
for all k ≥ 0, then
10k+1 (k + 1)!
∞
X
k=0
∞
X
xk
=
ck xk .
10k+1 (k + 1)!
k=0
Notice that
|ck+1 |
|ck |
=
=
=
=
Then the radius of convergence is: R =
1
10k+2 (k+2)!
1
10k+1 (k+1)!
10k+1 (k +
1)!
+ 2)!
k+1
(k + 1)!
10
·
k+2
(k + 2)!
10
1
1
·
→ 0,
10 k + 2
10k+2 (k
as k → ∞.
1
= ∞.
0
Remark 25. The radius of convergence of power series: For the power series
∞
X
∞
X
ck (x − a)k , let
k=0
|ck+1 |
1
, then the radius of convergence of the power series
ck (x − a)k is R = . In particular,
|ck |
r
k=0
∞
∞
X
X
the power series
ck (x − a)k absolutely converges for all |x − a| < R, and the power series
ck (x − a)k
r = lim
k→∞
k=0
diverges for all |x − a| > R.
k=0
(c) Let
∞
X
bn xn be the Maclaruin series for f (x) =
n=0
∞
X
3
1
3
1
−
, i.e.,
bn xn =
−
. Find
x + 1 2x − 1
x + 1 2x − 1
n=0
bn .
Answer.
In fact, we have
3
x+1
1
2x − 1
∞
= 3·
∞
X
X
1
=3
(−x)n =
3(−1)n xn
1 − (−x)
n=0
= −
1
=−
1 − 2x
∞
X
n=0
(2x)n =
n=0
∞
X
(−2n )xn .
n=0
So we get
3
1
−
x + 1 2x − 1
∞
∞
X
X
n n
=
3(−1) x −
(−2n )xn
f (x) = =
=
n=0
∞
X
n=0
[3(−1)n + 2n ]xn .
n=0
So we get
bn = 3(−1)n + 2n for all n ≥ 0 .
Remark 26. Two important Maclaurin series:
1
1−x
=
ex =
∞
X
k=0
∞
X
k=0
xk = 1 + x + x2 + x3 + x4 + · · ·
x2 x3 x4
xk
=1+x+
+
+
+ ···
k!
2!
3!
4!
for all −1 < x < 1
for all −∞ < x < ∞.
Remark 27. Combining power series: Suppose the power series
∞
X
k=0
k
ck x and
∞
X
dk xk converge to
k=0
f (x) and g(x), respectively, on an interval I, then
∞
X
a. The power series
(ck ± dk )xk converges to f (x) ± g(x) on I.
k=0
X
X
b. Let m be an integer such that k + m ≥ 0 for all therms of the power series xm
ck xk =
ck xk+m .
Then this series converges to xm f (x) for all x 6= 0 in I. When x = 0, the series converges to
lim xm f (x).
x→0
c. If h(x) = bxm , where m is a positive integer and b is a nonzero real number, then the power series
∞
X
ck [h(x)]k converges to the composite function f (h(x)), for all x such that h(x) ∈ I.
k=0
3. (a) Use the method of Lagrange multipliers to find the maximum and minimum values of (x + 1)2 + (y − 2)2
on the circle x2 + y 2 = 125. A solution that does not use the method of Lagrange multipliers will receive
no credit, even if the answer is correct.
Answer. Let f (x, y) = (x+1)2 +(y −2)2 and g(x, y) = x2 +y 2 −125, then we should study the following
optimization problem:
max / min
f (x, y) −→ objective function
−→ constraint.
g(x, y) = 0
Notice that
fx = 2(x + 1) = 2x + 2
fy = 2(y − 2) = 2y − 4
gx = 2x
gy = 2y.
Use the method of Lagrange multipliers, we should solve ∇f = λ∇g and g(x, y) = 0, that is,
2
2x + 2 = 2λx
(3)
2y − 4 = 2λy
(4)
2
x + y − 125 = 0
(5)
If λ = 0, by (3) and (4), then x = −1 and y = 2, which contradict with (5), since (−1)2 + 22 − 125 =
−120 6= 0.
If λ 6= 0, by (3) and (4), then (2x + 2) · 2λy = (2y − 4) · 2λx. Since λ 6= 0, then (x + 1)y = (y − 2)x, that
is, xy + y = xy − 2x. So we get y = −2x. Plug y = −2x into (5), we have x2 + (−2x)2 − 125 = 0, that
is, 5x2 = 125. So x2 = 25, that is, x = ±5. Since y = −2x, then y = ∓10. So we have two solutions:
(x, y) = (5, −10),
and
(−5, 10).
Notice that
f (5, −10) = (5 + 1)2 + (−10 − 2)2 = 180,
and f (−5, 10) = (−5 + 1)2 + (10 − 2)2 = 80.
So the maximum value of (x + 1)2 + (y − 2)2 on the circle x2 + y 2 = 125 is 180 , and the minimum value
is 80 .
Remark 28. Consider the optimization problem:
max / min
f (x, y) −→ objective function
g(x, y) = 0
−→ constraint.
Method of Lagrange Multipliers:
If ∇g(x, y) 6= 0 for all (x, y) such that g(x, y) = 0, to find the maximum and minimum values of
f (x, y) subject to the constraint g(x, y) = 0, we have two steps:
Step 1. Find the values of x, y and λ such that
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
Step 2. Among the points (x, y) found in Step 1, select the largest and smallest corresponding
function values. These values are the maximum and minimum values of f (x, y) subject to
the constraint g(x, y) = 0.
Remark 29. For solving ∇f (x, y) = λ∇g(x, y) and g(x, y) = 0, fist, you should check whether λ = 0 or
not. Usually, λ 6= 0
(b) Find the point on the circle x2 + y 2 = 125 that has minimum distance from the point (−1, 2).
Answer.
The distance from the point (−1, 2) to the point (x, y) on the circle x2 + y 2 = 125 is:
p
d(x, y) = (x + 1)2 + (y − 2)2 ≥ 0.
Then we have f (x, y) = |d(x, y)|2 . By the result
p of part (a), we
√ know that
√ the minimum value of d(x, y)
2
2
on the circle x + y = 125 is d(−5, 10) = f (−5, 10) = 80 = 4 5, which implies that the point
(−5, 10) on the circle x2 + y 2 = 125 has the minimum distance from the point (−1, 2).
Remark 30. Distance between two points: Let P = (x1 , y1 , z1 ) and Q = (x2 , y2 , z2 ), then
p
−−→
distance between P and Q = |P Q| = (x1 − x2 )2 + (y1 − y2 )2 + (z1 − z2 )2 .
1
4. Let T (x, y) = y 3 + x2 − 2xy + 6x − 6y.
9
(a) Find all critical points of T (x, y) and classify each as a local maximum, local minimum or saddle point.
Answer.
Since
Tx = 2x − 2y + 6
1 2
Ty =
y − 2x − 6.
3
Let’s solve
2x − 2y + 6 = 0
(6)
1 2
y − 2x − 6 = 0
3
(7)
By (6), then x − y + 3 = 0, that is, x = y − 3. Plug x = y − 3 into (7), then 13 y 2 − 2(y − 3) − 6 = 0, that
is, y 2 − 6y = 0. So we get y = 0 or y = 6. Since x = y − 3, then x = −3 or x = 3. Hence we have two
critical points:
(−3, 0) and (3, 6) .
Use the second derivative test to classify critical points. Notice that
Txx = 2,
Txy = Tyx = −2,
2
and Tyy = y.
3
So we have
2 −2
, then the discriminant D(−3, 0) = 2 · 0 − (−2)2 =
• At (−3, 0), the Hessian matrix is
−2 0
−4 < 0, which implies that (−3, 0) is a saddle point .
2 −2
, then the discriminant D(−3, 0) = 2 · 4 − (−2)2 = 4 > 0.
• At (3, 6), the Hessian matrix is
−2 4
Since Txx (3, 9) = 2 > 0, then (3, 6) is a local minimum point .
Remark 31. I. Critical points: If f has a local maximum or local minimum value at point (a, b),
then (a, b) is a critical point of f , that is, ∇f (a, b) = 0 and fx (a, b) = fy (a, b) = 0.
fxx fxy
II. The Hessian matrix of f is defined as:
.
fyx fyy
III. The Discriminant of f , D(x, y), is defined as:
D(x, y) = fxx (x, y)fyy (x, y) − (fxy (x, y))2 .
IV. Second Derivative Test: Let the point (a, b) be a critical point of f , that is, fx (a, b) = fy (a, b) = 0,
then
a. If D(a, b) > 0 and fxx (a, b) < 0, then f has a local maximum value at (a, b).
b. If D(a, b) > 0 and fxx (a, b) > 0, then f has a local minimum value at (a, b).
c. If D(a, b) < 0, then f has a saddle point at (a, b).
d. If D(a, b) = 0, then the test is inconclusive.
Where D(a, b) := fxx (a, b)fyy (a, b) − (fxy (a, b))2 is the discriminant of f .
(b) Find the maximum and minimum values of T (x, y) on the region R =
1 2
(x, y) : y − 3 ≤ x ≤ 0 .
3
Answer.
By part (a), we know that T (x, y) has two critical points (−3, 0) and (3, 6). For any point
1
(x, y) ∈ R, we should have x ≤ 0, then (3, 6) is not a point in R. For (−3, 0), since · 02 − 3 = −3 ≤ 0,
3
then (−3, 0) is a boundary point of R. Hence the maximum and minimum values of T (, y) on R must be
attained on the boundary of R.
Notice that the boundary of R consists two parts: one part is the vertical line segment x = 0, one part
1
is the arc x = y 2 − 3.
3
1
• For the point (x, y) on vertical line x = 0, then x = 0. Since y 2 − 3 ≤ x = 0, then y 2 ≤ 9, that is,
3
−3 ≤ y ≤ 3. On this vertical line, we have
1
g(y) := f (x, y) = f (0, y) = y 3 − 6y,
9
−3 ≤ y ≤ 3.
√
√
1
Since g 0 (y) = y 2 − 6, then g 0 (y) has zero y = ± 18 = ±3 2, which implies that g 0 (y) does not have
3
zero in [−3, 3]. Hence the maximum/minimum values must be either g(3) or g(−3). Notice that
g(−3) =
1
· (−3)3 − 6 · (−3) = 15,
9
and g(3) =
1 3
· 3 − 6 · 3 = −15.
9
1
• For the point (x, y) on the arc x = y 2 − 3, then −3 ≤ y ≤ 3. On this arc, we have
3
h(y) : = f (x, y)
1 2
= f
y − 3, y
3
2
1 2
1 2
1 2
1 3
y +
y −3 −2·
y −3 y+6·
y − 3 − 6y
=
9
3
3
3
1 3 1 4
2
=
y + y − 2y 2 + 9 − y 3 + 6y + 2y 2 − 18 − 6y
9
9
3
1 4 5 3
=
y − y − 9.
9
9
4 3 5 2
y2
15
y − y = [4y − 15], then h0 (y) has zeros y = 0 and y =
, which implies that
9
3
9
4
0
h (y) only has zero y = 0 in [−3, 3]. Hence the maximum/minimum values must be either h(0), or
h(3) or h(−3). Notice that
Since h0 (y) =
h(0) = −9,
h(3) =
1 4 5 3
· 3 − · 3 − 9 = −15,
9
9
and h(−3) =
1
5
· (−3)4 − · (−3)3 − 9 = 15.
9
9
In summary, the maximum value of T (x, y) on R is 15, and the minimum value of T (x, y) on R is = 15.
Remark 32. Let R be a bounded closed subset in R2 , to find the absolute maximum and minimum
values of f on R:
Step 1: Find the maximum and minimum values of f on the boundary of R.
Step 2: Determine the values of f at all critical points in R.
Step 3: The greatest function value found in Step 1 and Step 2 is the absolute maximum value of f on R,
and the least function value found in Step 1 and Step 2 is the absolute minimum value of f on R
5. An endowment is an investment account in which the balance ideally remains constant and withdrawals are
made on the interest earned by the account. Such an account may be modeled by the initial value problem
B 0 (t) = aB − m for t > 0, with B(0) = B0 . The constant a reflects the annual interest rate, m is the annual
rate of withdrawal, and B0 is the initial balance in the account.
(a) Solve the initial value problem with a = 0.02 and B(0) = B0 = $30, 000. Note that your answer depends
on the constant m.
Answer.
problem:
Since a = 0.02, and B(0) = B0 = 30, 000, then we need to solve the following initial value
B 0 (t) = 0.02B − m
B(0) = 30, 000
Since B 0 (t) = 0.02B − m, then
m
for all t ≥ 0. Since B(0) = 30, 000, then m = 0.02·30, 000 = 600.
0.02
Hence when m = 600, then B(t) ≡ 30, 000.
B 0 (t)
• If 0.02B − m 6= 0, then
= 1, that is,
0.02B − m
dB
= dt.
0.02B − m
Then we get
1
ln |0.02B − m| = t + C.
0.02
Then we get
0.02B − m = ±e0.02t+0.02C .
• If 0.02B−m = 0, that is, B(t) ≡
So
B = 50m ± 50 · e0.02t+0.02C = 50 ± 50e0.02C · e0.02t .
Since C is arbitrary constant, then we can take k = ±50e0.02C to be arbitrary constant, that is,
B(t) = 50m + ke0.02t
Since B(0) = 30, 000, then 50m + k = 30, 000, then
k = 30, 000 − 50m.
hence we get
B(t) = 50m + (30, 000 − 50m)e0.02t .
In summary, we have
B(t) = 50m + (30, 000 − 50m)e0.02t .
dy
= f (x)g(y), then
dx
1) If g(a) = 0 for some constant a, then y(x) ≡ a is a solution.
2) If g(y) 6= 0, then
1
dy
·
= f (x).
g(y) dx
That is,
1
dy = f (x) dx.
g(y)
Hence we have
Z
Z
1
dy = f (x) dx.
g(y)
Remark 33. Differentiable equations: Let’s consider y 0 =
(b) If a = 0.02 and B(0) = B0 = $30, 000, what is the annual withdraw rate m that ensures a constant
balance in the account?
Answer. Approach I: If we want to have a constant balance in the account, that is, B 0 (t) = 0 for all
t > 0. Since B 0 = aB − m = 0.02B − m, then 0.02B(t) − m = 0 for all t ≥ 0, that is,
B(t) =
m
,
0.02
for all t ≥ 0.
Since B(0) = 30, 000, then m = 0.02 · 30, 000 = 600.
Approach II: If we want to have a constant balance in the account, that is, B 0 (t) = 0 for all t > 0. By
part (a), we have B(t) = 50m + (30, 000 − 50m)e0.02t , then if 30, 000 − 50m = 0, that is, m = 600, we
have B(t) = 50m = 30, 000 for all t > 0. Hence, when the annual withdraw rate m = $600, we have a
constant balance in the account.
6. (a) Evaluate the sum of the convergent series
∞
X
1
.
k
π k!
k=1
Answer.
Recall that
x
e =
∞
X
xk
k=0
k!
,
−∞ < x < ∞.
Then
∞
X
1
π k k!
=
k=1
=
∞
X
k=1
∞
X
k=0
1 k
π
k!
1 k
π
k!
−1
1
= e π − 1.
So we get
∞
X
1
1
= eπ − 1 .
π k k!
k=1
Remark 34. Two important Maclaurin series:
1
1−x
=
ex =
∞
X
k=0
∞
X
k=0
xk = 1 + x + x2 + x3 + x4 + · · ·
x2 x3 x4
xk
=1+x+
+
+
+ ···
k!
2!
3!
4!
for all −1 < x < 1
for all −∞ < x < ∞.
(b) Assume that the series
∞
X
nan − 2n + 1
n=1
n+1
converges, where an > 0 for all n ≥ 1. Is the following series
∞
X
− ln a1 +
ln
n=1
an
an+1
convergent or divergent. If your answer is YES, evaluate the sum of the series − ln a1 +
∞
X
ln
n=1
Answer.
m
X
For any m ≥ 1, let Sm = − ln a1 +
ln
n=1
Sm = − ln a1 +
m
X
ln
n=1
an
an+1
an
.
an+1
an
, then
an+1
= − ln a1 +
m
X
[ln an − ln an+1 ]
n=1
− ln a1
=
+ ln a1 − ln a2
+ ln a2 − ln a3
+ ln a3 − ln a4
..
.
+ ln am−1 − ln am
+ ln am − ln am+1
= − ln am+1
So we should compute lim ln am+1 . Since that the series
m→∞
test, we must have lim
n→∞
nan − 2n + 1
= 0. Hence we get
n+1
lim
n→∞
Since lim
n→∞
∞
X
nan − 2n + 1
n=1
n+1
converges, by the divergence
n(an − 2)
= 0.
n+1
n+1
= 1, then we have
n
lim
n→∞
n + 1 n(an − 2)
·
= 0.
n
n+1
That is, we have
lim an = 2.
n→∞
Hence we get
lim Sm = − lim ln am+1 = − ln 2.
m→∞
m→∞
In summary, we know that the series − ln a1 +
n=1
− ln 2.
Remark 35. Divergence test: If a series
0, then the series
∞
X
∞
X
k=1
∞
X
k=1
ak diverges.
ln
an
an+1
is convergent and − ln a1 +
∞
X
n=1
ln
an
an+1
=
ak converges, then lim ak = 0. That means, if lim ak 6=
k→∞
k→∞
Selected Problems in Sample Exam for Midterm Exam 1
7. The function f (x, y) obeys
f (x, y) + sin(f (x, y)) = 2x + 4xy,
and f (0, 0) = 0.
Find fx (0, 0) and fxy (0, 0).
Answer.
Since f (x, y) + sin(f (x, y)) = 3x + 4xy, take the partial derivative with respect to x on the both
sides, by using the chain rule, we have
fx (x, y) + cos(f (x, y)) · fx (x, y) = 2 + 4y.
Plug (x, y) = (0, 0) into the above equation, since f (0, 0) = 0, then
fx (0, 0) + cos(0) · fx (0, 0) = 2 + 4 · 0.
That is, 2fx (0, 0) = 2. So we get
fx (0, 0) = 1.
Since fx (x, y) + cos(f (x, y)) · fx (x, y) = 2 + 4y, take the partial derivative with respect to y on the both sides,
by using the chain rule and the product rule, we have
fxy (x, y) − sin(f (x, y)) · fy (x, y) · fx (x, y) + cos(f (x, y)) · fxy (x, y) = 4.
Plug (x, y) = (0, 0) into the above equation, since f (0, 0) = 0, then
fxy (0, 0) − sin(0) · fy (0, 0) · fx (0, 0) + cos(0) · fxy (0, 0) = 4.
That is, 2fxy (0, 0) = 4. So we get
fxy (0, 0) = 2.
Remark 36. During the computation of fxy (0, 0), we find that fx (0, 0) and fy (0, 0), in general, we can use
the equation that f (x, y) satisfies to compute these two terms. In the above problem, we are lucky to have
sin(0) · fy (0, 0) · fx (0, 0) = 0, so we do not need to compute fy (0, 0), but in general, we should compute it just
as the computation of fx (0, 0).
Remark 37. Implicit differentiation: Take the partial derivative on the both sides of the equation and
apply the chain rule.
Selected Problems in Midterm Exam 1
8. Let Q and R be two planes given by the equations:
Q : 2x − y = 0,
R : y + z = 0.
Let P be the plane which passes through the point (1, 0, 0) and is orthogonal to both Q and R. Find the
equation of the plane P .
Answer. Let n = ha, b, ci be a normal vector of the plane P . By the equations of Q and R, then a normal
vector of Q is h2, −1, 0i, and a normal vector of R is h0, 1, 1i. Since the plane P is orthogonal to both Q and
R, then
ha, b, ci · h2, −1, 0i = 2a − b = 0,
and ha, b, ci · h0, 1, 1i = b + c = 0.
Since b = 2a, then c = −b = −2a. So we get
n = ha, b, ci = ha, 2a, −2ai = ah1, 2, −2i.
So h1, 2, −2i is a normal vector of P . Since P passes through the point (1, 0, 0), then the equation of P is:
(x − 1) + 2y − 2z = 0 .
Remark 38. Plane: The plane passing through the point P0 = (x0 , y0 , z0 ) with a nonzero normal vector
n = ha, b, ci is described by the equation:
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0,
or ax + by + cz = d,
where d = ax0 + by0 + cz0 .
Remark 39. For any two distinct planes P and Q with normal vectors n and m respectively, the angle
between these two planes P and Q is defined as the acute angle between n and m (i.e., between 0 and π/2).
In particular,
a. Two distinct planes are parallel if their respective normal vectors are parallel (that is, the normal vectors
are scalar multiples of each other, or the angle is either 0 or π).
b. Two distinct planes are orthogonal if their respective normal vectors are orthogonal (that is, the dot
product of the normal vectors is zero, or the angle is π2 ).
Selected Problems in Sample Exam for Midterm Exam 2
9. Evaluate lim
n→∞
Answer.
n
X
6(k − 1)2
n3
k=1
r
1+2
(k − 1)3
.
n3
1
in the sum, then
n
First, we should split off a ∆x =
n
X
6(k − 1)2
k=1
n
r
1 X 6(k − 1)2
(k − 1)3
=
1+2
n3
n
n2
n3
r
k=1
1+2
(k − 1)3
.
n3
k−1
k
Then write down the resulting sum as a function of
(left Riemann sum) or (right Riemann sum), so
n
n
we get
s
r
n
n
X
(k − 1)3
1X
k−1 2
k−1 3
6(k − 1)2
1+2
=
6·
1+2·
.
n3
n3
n
n
n
k=1
k=1
k−1
k
or
by x, we get our integrand f (x). Then when n → ∞,
nZ
n
Now substitute ∆x by dx, and substitute
1
the sum will converge to the definite integral
f (x) dx. In our case, we have
0
lim
n→∞
Z
For
6x2
n
X
6(k − 1)2
r
(k − 1)3
1+2
n3
n3
k=1
Z
=
1
6x2
p
1 + 2x3 dx.
0
p
1 + 2x3 dx, we have
Z
Z
p
√
3
6x 1 + 2x dx =
u du
2
Let u = 1 + 2x3 , then du = 6x2 dx
1
1
+1
2
+C
1u
1+ 2
2 3
u2 + C
3
3
2
(1 + 2x3 ) 2 + C
3
=
=
=
Since u = 1 + 2x3 .
Then
lim
n→∞
n
X
6(k − 1)2
k=1
r
1
√
(k − 1)3
2
2 2 3
2
3 23 2 =
1+2
=
(1
+
2x
)
=
−
·
3
−
2
3.
n3
3
3 3
3
0
n3
Z
Remark 40. For the definite integral
b
f (x) dx, we have
a
Z
b
f (x) dx := lim
a
n→∞
n
X
f (x∗k )∆x.
(8)
k=1
Z b
For the formula (8), it has two meanings: one is to compute
f (x) dx by using the Riemann sum; the other
a
Z b
n
X
1
k
one is to compute some sum like
f
by using the integral
f (x) dx.
n
n
a
k=1
Selected Problems in Midterm Exam 2
1
Z
x2
dx.
( 4 − x2 )3
√
10. Evaluate
0
Z
Answer.
Let’s compute
x2
√
dx first, use trigonometric substitution, let x = 2 sin(θ), then
( 4 − x2 )3
4 − x2 = 4 − 4 sin2 (θ) = 4 cos2 (θ),
and dx = 2 cos(θ) dθ.
Then we get
Z
4 sin2 (θ)
· 2 cos(θ) dθ
[2 cos(θ)]3
Z
Z
8 sin2 (θ) cos(θ)
sin2 (θ)
=
dθ
=
dθ
8 cos3 (θ)
cos2 (θ)
Z
=
tan2 (θ) dθ
Z
=
[1 − sec2 (θ)] dθ Since tan2 (θ) + 1 = sec2 (θ)
Z
Z
=
dθ − sec2 (θ) dθ
x2
√
dx =
( 4 − x2 )3
Z
= θ − tan(θ) + C.
Since x = 2 sin(θ), then
θ = sin−1
x
2
,
and
tan(θ) = √
x
.
4 − x2
So we get
Z
x
x2
x
−√
dx = sin−1
+ C.
2
( 4 − x2 )3
4 − x2
√
Hence we get
Z
0
1
√
1
x
1
π
x2
3
−1 x
−1 1
√
dx = sin
−√
= sin
−√ = −
.
2
2
6
3
3
( 4 − x2 )3
4 − x2 0
Remark 41. Trigonometric substitution:
√
a. Integrals involving a2 − x2 : Let x = a sin(θ), then dx = a cos(θ) dθ and
q
p
a2 − x2 = a2 − a2 sin2 (θ) = a cos(θ).
b. Integrals involving
c. Integrals involving
√
√
a2 + x2 : Let x = a tan(θ), then dx = a sec2 (θ) dθ and
q
p
a2 + x2 = a2 + a2 tan2 (θ) = a sec(θ).
x2 − a2 : Let x = a sec(θ), then d sec(θ) = sec(θ) tan(θ) dθ and
p
p
x2 − a2 = a2 sec2 (θ) − a2 = a tan(θ).
Z
11. Evaluate
Answer.
sin4 x sec2 x dx.
In fact, we have
Z
Z
sin4 x
4
2
sin x sec x dx =
dx
cos2 x
Z
[1 − cos2 x]2
dx Since sin2 x + cos2 x = 1
=
cos2 x
Z
1 − 2 cos2 x + cos4 x
=
dx
cos2 x
Z 1
=
− 2 + cos2 x dx
cos2 x
Z
Z
Z
=
sec2 x dx − 2 1 dx + cos2 x dx
Z
1 + cos(2x)
dx Since cos(2x) = 2 cos2 x − 1
= tan x − 2x +
2
x sin(2x)
= tan x − 2x + +
+C
2
4
3
1
= tan x − x + sin(2x) + C.
2
4
So we get
Z
1
3
sin4 x sec2 x dx = tan x − x + sin(2x) + C .
2
4
Remark 42. Important trigonometric identities:
sin2 (θ) + cos2 (θ) = 1,
sin(2θ) = 2 sin(θ) cos(θ),
1 + tan2 (θ) = sec2 (θ)
cos(2θ) = 2 cos2 (θ) − 1 = 1 − 2 sin2 (θ) = cos2 (θ) − sin2 (θ)
Selected Problems in Sample Exam for Final Exam
12. Let


1
 1
f (x) = lim  +
n→∞
3 2+ 1+
x−1 3
n
 x − 1
,
3  
n
(n−1)(x−1)
1
+ ··· +
2+ 1+
n
where x ≥ 1. Find the equation of the tangent line to the graph y = f 0 (x) at x = 2.
Answer.
By the definition of f (x), then
f (x) =
lim
n→∞
n−1
X
1
k=0
2+ 1+
k(x−1)
n
3 ·
x−1
n
n−1
=
lim (x − 1) ·
n→∞
1
1X
n
2 + 1 + (x − 1) ·
k=0
k
n
3
1
1
= (x − 1)
dt
3
0 2 + [1 + (x − 1)t]
Z x−1
1
=
du Let u = (x − 1)t, then du = (x − 1) dt.
2 + (1 + u)3
0
Z
So we have
1
1
1
·1−
·0=
.
3
3
2 + (1 + x − 1)
2 + (1 + 0)
2 + x3
y = f 0 (x) =
Then we get
y 0 (x) = f 00 (x) = −
3x2
.
(2 + x3 )2
So we get
y(2) =
1
1
= ,
3
2+2
10
and y 0 (2) = −
3 · 22
3
=− .
3
2
(2 + 2 )
25
Hence the equation of the tangent line of the graph y = f 0 (x) at x = 2 is:
y−
Z
Remark 43. For the definite integral
1
3
= − (x − 2) .
10
25
b
f (x) dx, we have
a
Z
b
f (x) dx := lim
a
n→∞
n
X
f (x∗k )∆x.
(9)
k=1
Z
b
For the formula (9), it has two meanings: one is to compute
f (x) dx by using the Riemann sum; the other
a
Z b
n
X
k
1
one is to compute some sum like
f
by using the integral
f (x) dx.
n
n
a
k=1
Z
13. Compute the following indefinite integral
sin(ln x) dx.
Answer. In fact, we have
Z
Z
sin(ln x) dx =
sin(u) · eu du Let u = ln x, then x = eu and dx = eu du
Z
=
eu sin(u) du
Z
=
sin(u) deu Use the integration by parts, by ‘ILATE’ rule
Z
u
= e sin(u) − eu cos(u) du
Z
u
= e sin(u) − cos(u) deu Use the integration by parts, by ‘ILATE’ rule
Z
u
u
= e sin(u) − e cos(u) − eu sin(u) du.
Then we get
Z
u
u
Z
u
e sin(u) du = e sin(u) − e cos(u) −
Then we have
Z
eu sin(u) du.
1
eu sin(u) du = eu [sin(u) − cos(u)] + C.
2
So we get
Z
x
1
sin(ln x) dx = eln x [sin(ln x) − cos(ln x)] + C = [sin(ln x) − cos(ln x)] + C .
2
2
Remark 44. Substitution: Let u = g(x), then du = g 0 (x) dx and
Z
Z
Z b
Z
f (g(x))g 0 (x) dx = f (u) du, and
f (g(x))g 0 (x) dx =
a
g(b)
f (u) du.
g(a)
Remark 45. Integration by parts: Let u and v be differentiable, then
Z
Z
Z b
Z b
b
0
u dv = uv − v du, and
u(x)v (x) d = u(x)v(x)|a −
v(x)u0 (x) dx.
a
a
That is,
Z
0
uv dx = uv −
Z
vu0 dx.
where dv is the most complicated portion of the integrand that can be “ easily” integrated, and u is that
portion of the integrand whose derivative du is a “ simpler ” function than u itself. For the choice of u, please
follow the ‘ILATE’ order:
I = Inverse trigonometric function
L = Logarithmic function
A = Algebraic function
T = Trigonometric function
E = Exponential function
Selected Problems in Final Exam in 2013 Spring
14. Fill in the blanks with right, left, or midpoint; an interval; and a value of n.
3
X
f (1.5 + k) · 1 is a
Riemann sum for f on the interval [
,
] with n =
.
k=0
Answer.
In the sum
3
X
f (1.5 + k) · 1, we know that ∆x = 1. Since the sum
k=0
3
X
f (1.5 + k) · 1 has 4 terms,
k=0
then n = 4 . Notice that x∗k = 1.5 + (k − 1) = 1.5 + (k − 1) · ∆x for k = 1, 2, 3, 4, then
3
X
f (1.5 + k) · 1 is the
k=0
left Riemann sum with a = 1.5 and b = 1.5 + 3 + 1 = 5.5, that is, the interval is [1.5, 5.5] .
Remark 46. For a regular partition x0 = a < x1 < · · · < xn = b of [a, b], then
∆x =
b−a
,
n
and
xk = a + k∆x,
for all k = 0, 1, 2, · · · , n.
The Riemann sum of f (x) on [a, b] with n subintervals can be written:
f (x∗1 )∆x
+
f (x∗2 )∆x
+ ··· +
f (x∗n )∆x
=
n
X
f (x∗k )∆x.
k=1
For the choices x∗k ’s, we have
a. For the left Riemann sum, we have x∗k = xk−1 = a + (k − 1)∆x.
b. For the right Riemann sum, we have x∗k = xk = a + k∆x.
c. For the midpoint Riemann sum, we have x∗k =
1
∆x.
k−
2
xk−1 + xk
a + (k − 1)∆x + a + k∆x
=
= a+
2
2
15. Find the interval of convergence of the following series:
∞
X
(x + 1)2k
(a)
.
k 2 9k
k=1
Answer.
Notice that
∞
X
(x + 1)2k
k=1
k 2 9k
=
=
∞
X
[(x + 1)2 ]k
k=1
∞
X
k=1
Let ck =
1
k 2 9k
k 2 9k
1
k 2 9k
yk ,
Let y = (x + 1)2 .
, then
|ck+1 |
|ck |
=
=
1
(k+1)2 9k+1
1
k2 9k
k 2 9k
(k + 1)2 9k+1
2
k
9k
1
=
· k+1 → ,
k+1
9
9
Then the radius of convergence of
∞
X
as k → ∞.
ck y k is R = 9, which implies that |x + 1|2 < 9, that is, |x + 1| < 3.
k=1
So we get −4 < x < 2. Hence the interval of convergence of
∞
X
(x + 1)2k
k=1
k 2 9k
is (−4, 2) .
Remark 47. The radius of convergence of power series: For the power series
∞
∞
X
ck (x − a)k , let
k=0
X
|ck+1 |
1
r = lim
, then the radius of convergence of the power series
ck (x − a)k is R = . In particular,
k→∞ |ck |
r
k=0
∞
∞
X
X
the power series
ck (x − a)k absolutely converges for all |x − a| < R, and the power series
ck (x − a)k
k=0
diverges for all |x − a| > R.
k=0
(b)
∞
X
∞ X
a1
ak+1
ak
= .
−
ak+1 ak+2
a2
ak (x − 1)k , where ak > 0 for k = 1, 2, · · · , and
k=1
Answer.
k=1
Notice that
a1
a2
=
=
=
=
∞ X
ak
ak+1
−
ak+1 ak+2
k=1
n X
ak
ak+1
lim
−
n→∞
ak+1 ak+2
k=1
a1 an+1
lim
−
n→∞
a2 an+2
an+1
a1
− lim
.
a2 n→∞ an+2
Then
− lim
n→∞
So we get
lim
n→∞
So the radius of convergence of
∞
X
k=1
converges at x = 1 .
k
an+1
= 0.
an+2
an+2
= ∞.
an+1
ak (x − 1) is 0, which implies that the series
∞
X
k=1
ak (x − 1)k only
16. Let f (0) = 1, f (2) = 3 and
f 0 (2)
4
Z
= 4. Calculate
√
f 00 ( x) dx.
0
Answer.
In fact, we have
Z
Z
√
00 √
f ( x) dx =
f 00 (u) · 2u du Let u = x, then x = u2 and dx = 2u du
Z
= 2 uf 00 (u) du
Z
= 2 u df 0 (u)
Z
0
= 2uf (u) − 2 f 0 (u) du Use the integration by parts
= 2uf 0 (u) − 2f (u) + C
√
√
√
= 2 xf 0 ( x) − 2f ( x) + C
Since u =
√
x.
So we get
Z
0
Remark 48.
4
√
√
√
√
√ 4
f 00 ( x) dx = 2 xf 0 ( x) − 2f ( x) 0 = 2 · 4 · f 0 (2) − 2f (2) + 2f (0) = 12 .
I. Substitution: Let u = g(x), then du = g 0 (x) dx and
Z
Z
0
f (g(x))g (x) dx =
b
Z
f (u) du,
and
0
Z
g(b)
f (g(x))g (x) dx =
f (u) du.
a
g(a)
II. Integration by parts: Let u and v be differentiable, then
Z
Z
u dv = uv −
Z
v du,
and
a
b
u(x)v 0 (x) d = u(x)v(x)|ba −
That is,
Z
0
uv dx = uv −
Z
vu0 dx.
Z
a
b
v(x)u0 (x) dx.
Selected Problems in Final Exam in 2012 Spring
17. Consider the function

if x < 0,
 a,
k tan−1 (x), if 0 ≤ x ≤ 1,
F (x) =

b,
if x > 1.
Find the values of a, k and b for which F is a valid cumulative distribution function of a continuous random
variable X, and the probability density function of X.
Answer.
Since F is a cumulative distribution function of a continuous random variable, then lim F (x) = 1
x→∞
and F (x) is continuous, which implies that
1 = b = k tan−1 (1) =
kπ
,
4
and a = k tan−1 (0) = 0.
Then we get
a = 0, b = 1, and k =
4
.
π
Then we have

if x < 0,

 0,
4
−1
F (x) =
tan (x), if 0 ≤ x ≤ 1,

 π
1,
if x > 1.
Since f (x) = F 0 (x), then

0,
if x ≤ 0,

 4
1
f (x) =
·
, if 0 < x < 1, .
2

 π 1+x
0,
if x ≥ 1.
Remark 49. Let X be a continuous random variable, then
I. Cumulative distribution function(CDF) F (x) := Pr(X ≤ x), we have
a. 0 ≤ F (x) ≤ 1 for all x.
b. F (x) is a non-decreasing continuous function.
c. lim F (x) = 0.
x→−∞
d. lim F (x) = 1.
x→∞
II. Probability density function (PDF) f (x) =
dF (x)
, we have
dx
a. fZ (x) ≥ 0 for all x.
∞
b.
f (x) dx = 1.
−∞
Z x
c. F (x) =
f (t) dt.
−∞
Z b
d. Pr(a ≤ X ≤ b) =
f (t) dt.
a
III. Expected value: Let f (x) be the PDF for a continuous random variable X, then the expected value,
or expectation, or mean, of X is defined by:
Z ∞
E(X) =
xf (x) dx.
−∞
18. Find the limit, if it exists, of the sequence {ak }, where
ak =
Answer.
k! sin3 k
.
(k + 1)!
In fact, we have
|ak | =
=
=
≤
Then
−
k! sin3 k (k + 1)! k!
· | sin3 k|
(k + 1)!
1
· | sin3 k|
k+1
1
.
k+1
1
1
≤ ak ≤
,
k+1
k+1
for all k ≥ 1.
By the squeeze theorem of sequence, then
lim ak = 0 .
k→∞
∞
∞
Remark 50. Squeeze theorem: Let {an }∞
n=1 , {bn }n=1 and {cn }n=1 be sequences such that an ≤ bn ≤ cn
for all integers n greater than some positive integer N . If lim an = lim cn = L, then lim bn = L.
n→∞
n→∞
n→∞
Selected Problems in Quiz 3
Z
19. Use the (right) Riemann sum to compute
1
(2x + 1) dx.
0
Answer.
Since f (x) = 2x + 1 is continuous on [0, 1], so f is integrable on [0, 1]. Hence we can use the
Z 1
(2x + 1) dx. Now for any n, the regular partition using n tells us that
right Riemann sum to approximate
∆x =
0
1−0
1
= and
n
n
k
,
n
xk = 0 + k∆x =
So the right Riemann sum is: x∗k = xk =
n
X
for all k = 0, 1, 2, · · · , n.
k
and
n
f (x∗k )∆x
=
k=1
=
n
X
1
k
·
f
n
n
k=1
n 1 X 2k
·
+1
n
n
k=1
!
n
n
X
2X
1
k+
=
n
k=1
k=1
1
2 n(n + 1)
=
·
·
+n
n
n
2
1
=
[n + 1 + n]
n
2n + 1
=
n
→ 2, as n → ∞.
1
·
n
So we get
Z
1
(2x + 1) dx = 2.
0