Review Problems for Final Exam
April 08, 2015
Review Problems for Chapter 12
1. Let the point P (1, 2) and the points Q(4, 1), compute the distance between P and Q.
√
1.
Answer.
10
The distance betweem P and Q is:
p
p
√
−−→
|AB| = (1 − 4)2 + (2 − 1)2 = 32 + 12 = 10.
2. Let u = h3, −4i and v = h1, 2i, then
(a) Compute u + 2v.
h5, 0i
(a)
Answer.
u + 2v = h3, −4i + 2h1, 2i
= h3, −4i + h2, 4i
= h3 + 2, −4 + 4i
= h5, 0i
(b) Find the angle between u and v?
(b)
Answer.
√ √ cos−1 − 55 or π − cos−1 55
Recall the formula:
cos(θ) =
u·v
.
|u||v|
Notice that
u · v = 3 · 1 + (−4) · 2
= 3−8
= −5
p
32 + (−4)2
|u| =
√
=
9 + 16
√
=
25
= 5
p
|v| =
12 + 2 2
√
=
1+4
√
=
5.
So we get
√
−5
1
5
√ = −√ = −
cos(θ) =
.
5
5· 5
5
Therefor, we have
√ !
5
θ = cos−1 −
5
√ !
5
= π − cos−1
.
5
3. Let P be the plane which passes through the point (3, 0, 1) and is parallel to the plane 3x − 2y − z = 0, then
(a) Write down the equation of the plane P .
3x − 2y − z = 8
(a)
Answer.
Since the plane P is parallel to the plane 3x − 2y − z = 0, then h3, −2, −1i is a normal vector
to the plane P . Since the plane P passes through the point (3, 0, 1), then the equation of P is
3(x − 3) − 2(y − 0) − (z − 1) = 0.
That is,
3x − 2y − z = 8.
(b) Find the value of a such that the point (a, 2, 0) is on the plane P .
4
(b)
Answer.
By the result of part a, since (a, 2, 0) is on the plane P , then
3a − 2 · 2 − 0 = 8.
Then 3a = 12, that is,
a = 4.
(c) Find the value of b such that the plane P is orthogonal to the plane 2x − by = 1.
−3
(c)
Answer.
By the result of part a, we know that h3, −2, −1i is a normal vector to the plane P . For the
plane 2x − by = 1, then h2, −b, 0i is a normal vector to the plane 2x − by = 1. By the assumption, we
have
h3, −2, −1i · h2, −b, 0i = 0.
That is,
3 · 2 − 2 · (−b) − 1 · 0 = 0.
That is, 6 + 2b = 0. So
b = −3.
4. Let f (x, y) = ex and g(x, y) = y sin(xy), then
(a) Compute fx (x, y).
ex
(a)
Answer.
It’s easy to see that
fx (x, y) = ex .
(b) Compute fy (x, y).
0
(b)
Answer.
For fy (x, y), we should think x is a constant, then ex is a constant. So
fy (x, y) = 0.
(c) Compute gy (x, y).
sin(xy) + xy cos(xy)
(c)
Answer.
By the product rule and the chain rule, we have
gy (x, y) = sin(xy) + y cos(xy) · x
= sin(xy) + xy cos(xy)
5. (a) Let f (x, y) = y sin(xy), compute fyx (x, y).
2y cos(xy) − xy 2 sin(xy)
(a)
Answer.
First Approach:
fy (x, y) = sin(xy) + y cos(xy) · x
By the product rule and the chain rule
= sin(xy) + xy cos(xy)
fyx (x, y) = cos(xy) · y + y cos(xy) − xy sin(xy) · y
By the chain rule, the product rule and the chain rule
2
= 2y cos(xy) − xy sin(xy)
Second Approach:
fx (x, y) = y cos(xy) · y
By the chain rule
2
= y cos(xy)
fyx (x, y) = fxy (x, y)
By Clairaut’s theorem
= 2y cos(xy) − y 2 sin(xy) · x
By the product rule and the chain rule
2
= 2y cos(xy) − xy sin(xy)
(b) Can you find a function h(x, y) such that hxx = y 2 cos(xy 2 ) and hxy = xy 2 cos(xy 2 )? (Do not need to
find h(x, y), just put ‘Yes’ or ‘No’)
No
(b)
Answer.
If such a function h(x, y) exists, by the Clairaut’s theorem, we must have
hxxy = hxyx .
Since hxx = y 2 cos(xy 2 ), then
hxxy = 2y cos(xy 2 ) − y 2 sin(xy 2 ) · 2xy
2
3
By the product rule and the chain rule
2
= 2y cos(xy ) − 2xy sin(xy ).
Since hxy = xy 2 cos(xy 2 ), then
hxyx = y 2 cos(xy 2 ) − xy 2 sin(xy 2 ) · y 2
2
2
4
2
= y cos(xy ) − xy sin(xy ).
So we get hxxy 6= hxyx , contradiction.
By the product rule and the chain rule
6. Let g(x, y) = 2x2 + y 4 + 1, find a local minimum point of g(x, y).
(0, 0)
6.
Answer.
It’s easy to see that
g(x, y) = 2x2 + y 4 + 1 ≥ 0 + 0 + 1 = 1 = g(0, 0).
So (0, 0) is a local minimum point of g(x, y). In fact, this is the only one minimum point of g(x, y) in R2 .
7. Let R = {(x, y) : x2 + y 2 ≤ 4} and f (x, y) = 2x2 + 2y 2 + 1, then
(a) Find the absolute maximum value of f (x, y) on R.
9
(a)
Answer.
For all (x, y) in R, we know that
x2 + y 2 ≤ 4.
Notice that
f (x, y) = 2x2 + 2y 2 + 1 = 2(x2 + y 2 ) + 1.
For the first term 2(x2 + y 2 ), it has the largest possible value 2 · 4 = 8 which can be achieved for every
point on the boundary of R. So the absolute maximum value of f (x, y) on R is 8 + 1 = 9.
(b) Find the absolute minimum value of f (x, y) on R.
1
(b)
Answer.
It’s easy to see that (0, 0) is in R and
f (x, y) = 2x2 + 2y 2 + 1 ≥ 0 + 0 + 1 = 1 = f (0, 0),
for all (x, y) on R.
So f (0, 0) = 1 is the absolute minimum value of f (x, y) on R.
8. (a) Find the value of a such that the gradient of f (x, y, z) = aex + xy + sin(z) at point (0, 0, 0) is orthogonal
to a normal vector of the plane 2x − z = 1.
1
2
(a)
Answer.
Since
fx = aex + y,
fy = x,
and fz = cos(z),
then
∇f (0, 0, 0) = ha, 0, 1i.
Notice that a normal vector of the plane 2x − z = 1 can be h2, 0, −1i, by the assumption, we have
0 = ha, 0, 1i · h2, 0, −1i = 2a − 1.
Then
a=
1
2
(b) Can you find a function h(x, y) such that ∇h(x, y) = hcos(y), sin(x)i? (Do not need to find h(x, y), just
put ‘Yes’ or ‘No’)
No
(b)
Answer.
If we can find h(x, y) such that ∇h(x, y) = hcos(y), sin(x)i, then
hx = cos(y),
and hy = sin(x).
By Clairaut’s theorem, we must have hxy = hyx . But we know that
hxy = − sin(y),
and hyx = cos(x),
contradiction.
(c) Solve the system
x3 = xy
.
y−x=2
(0, 2), (−1, 1) and (2, 4)
(c)
Answer.
From the first equation, x3 = xy, we have x(x2 − y) = 0, then either x = 0 or y = x2 .
• When x = 0, plug x = 0 into the second equation y − x = 2, then y = 2. So we have one solution
(0, 0).
• When y = x2 , plug y = x2 into the second equation x2 − x = 2, that is, x2 − x − 2 = 0, then x = 2
or x = −1. So we have two solutions (2, 4) and (−1, 1).
In summary, we have three solutions:
(0, 2),
(−1, 1),
and
(2, 4).
9. Let f (x, y) = x2 + xy 2 − 2x + 1, then
(a) Find all critical points of f (x, y).
√
√
(0, 2), (0, − 2) and (1, 0).
(a)
Answer.
Let’s compute the first order partial derivatives of f (x, y), we have
fx (x, y) = 2x + y 2 − 2,
and fy (x, y) = 2xy.
For critical points, we need to solve the following system:
2x + y 2 − 2 = 0,
(1)
2xy = 0
(2)
The equation (2) is much simpler than equation (1), so let’s solve (2) first. Since 2xy = 0, then either
x = 0 or y = 0.
√
√
2 − 2 = 0, then either y =
When x = 0, plug√x = 0 into (1),
we
get
y
2
or
y
=
−
2. So we have two
√
critical points (0, 2) and (0, − 2).
When y = 0, plug y = 0 into (1), we get 2x − 2 = 0, then x = 1. So we have one critical point (1, 0).
In summary, we have three critical points:
√
√
(0, 2), (0, − 2), and (1, 0).
(b) Compute the Hessian matrix of f (x, y).
(b)
Answer.
2 2y
2y 2x
.
Since fx (x, y) = 2x + y 2 − 2 and fy (x, y) = 2xy, then
fxx (x, y) = 2,
fxy (x, y) = fyx (x, y) = 2y,
and fyy (x, y) = 2x.
So the Hessian matrix of f (x, y) is:
2 2y
2y 2x
.
(c) Compute the discriminant D(x, y) of f (x, y).
4x − 4y 2 .
(c)
Answer.
Since
fxx (x, y) = 2,
fxy (x, y) = fyx (x, y) = 2y,
and fyy (x, y) = 2x,
then
D(x, y) = fxx fyy − (fxy )2 = 2 · (2x) − (2y)2 = 4x − 4y 2 .
(d) Classify all critical points of f (x, y). (Make a table)
Answer.
Since D(x, y) = 4x − 4y 2 , then
√
D(0, 2) = −8,
√
D(0, − 2) = −8,
and D(1, 0) = 4.
Since fxx (x, y) = 2, then
(x,√y)
(0, √2)
(0, − 2)
(1, 0)
10. Sketch two level curves of the surface z =
Answer.
get
D(x, y)
−8
−8
4
fxx (x, y)
2
2
2
Conclusion
saddle point
saddle point
local minimum point
x2
y2
+ .
9
16
For the level curves, we just need to set z = z0 for some z0 . For example, we set z = z0 = 1, we
x2
y2
+
= 1,
9
16
which is an ellipse.
If we set z = z0 = 2, we get
x2
y2
+
= 2,
9
16
that is,
x2
y2
+
= 1,
18 32
which is also an ellipse.
Remark: You must draw these two level curves in the same xy-plane, do not separately.
11. Find and sketch the level curves of the function f (x, y) = y − x2 − 1.
Answer. The level curve with level z0 is described by the equation y − x2 − 1 = z0 , that is, y = x2 + 1 + z0 ,
which is a parabola in the xy-plane. For example,
• With z0 = 0, the level curve is the parabola y = x2 + 1.
• With z0 = −1, the level curve is the parabola y = x2 .
• With z0 = −2, the level curve is the parabola y = x2 − 1.
• With z0 = 1, the level curve is the parabola y = x2 + 2.
12. Use the Lagrange multipliers to find the maximum and minimum values of f (x, y) = xy subject to x2 +y 2 −xy =
9.
Answer.
The objective function is f (x, y) = xy, and the constraint is g(x, y) := x2 + y 2 − xy − 9 = 0, so
we should solve the following system:
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
Notice that
∇f (x, y) = hy, xi,
and ∇g(x, y) = h2x − y, 2y − xi.
So we need to solve the system:
2
2
y = λ(2x − y)
(3)
x = λ(2y − x)
(4)
x + y − xy = 9
(5)
It’s easy to see that λ 6= 0, otherwise, by (3) and (4), we have x = y = 0, which does not satisfy (5). By (3)
and (4), we have
yλ(2y − x) = xλ(2x − y).
Sine λ 6= 0, then
2y 2 − xy = 2x2 − xy.
That is, y 2 = x2 , then y 2 − x2 = 0. So (y − x)(y + x) = 0, then y = x or y = −x.
When y = x, plug y = x into (5), we get x2 + x2 − x2 = 9, that is, x2 = 9. So x = 3 or x = −3. Since
y = x,then y = 3 or y = −3. So when (x, y) = (3, 3), by (3), we get 3 = λ(2 · 3 − 3), then λ = 1. When
(x, y) = (−3, −3), by (3), we get −3 = λ(−2 · 3 + 3), then λ = 1. So we have two solutions:
(x, y, λ) = (3, 3, 1),
and
(x, y, λ) = (−3, −3, 1).
√
√
2 + x2 + x2 = 9, that is, x2 = 3. So x =
When y = −x, plug y = √
−x into (5),
we
get
x
3
or
x
=
−
√
√
√
√
√
√ 3.
Since y = −x,then y = − 3 or
y
=
3.
So
when
(x,
y)
=
(
3,
−
3),
by
(3),
we
get
3
=
λ(2
·
3
+
3),
√ √
√
√
√
then λ = 31 . When (x, y) = (− 3, 3), by (3), we get − 3 = λ(−2 · 3 − 3), then λ = 13 . So we have two
solutions:
√
√ 1
√ √ 1
(x, y, λ) = ( 3, − 3, ), and (x, y, λ) = (− 3, 3, ).
3
3
Notice that
√
√ √
√
and f ( 3, − 3) = f (− 3, 3) = −3.
f (3, 3) = f (−3, −3) = 9,
So the maximum value of f (x, y) = xy subject to x2 + y 2 − xy = 9 is 9, and the minimum value of f (x, y) = xy
subject to x2 + y 2 − xy = 9 is −3.
13. Find the maximum and minimum values of the objective function f (x, y) = 2x2 + y 2 + 2, where x and y lie
on the ellipse C given by g(x, y) = x2 + 4y 2 − 4 = 0.
Answer.
Step 1: Let’s compute the gradients of f and g, we have
∇f (x, y) = h4x, 2yi,
and ∇g(x, y) = h2x, 8yi.
Let’s solve the system:
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
That is,
4x = λ · (2x) = 2λx
(6)
2y = λ · (8y) = 8λy
(7)
2
2
x + 4y − 4 = 0
(8)
It’s easy to see that λ 6= 0, otherwise, by (6) and (7), we have x = y = 0, which does not satisfy (8). By (6)
and (7), we have
4x(8λy) = 2λx(2y).
Since λ 6= 0, then 32xy = 4xy, that is, xy = 0. So either x = 0 or y = 0.
When x = 0, plug x = 0 into (8), we get 4y 2 − 4 = 0, that is, y 2 = 1, so y = 1 or y = −1. By (7), we have
λ = 14 . So we have two solutions
1
1
(x, y, λ) = 0, 1,
, and (x, y, λ) = 0, −1,
.
4
4
When y = 0, plug y = 0 into (8), we get x2 − 4 = 0, then x = 2 or x = −2. By (6), we have λ = 2. So we
have two solutions:
(x, y, λ) = (2, 0, 2) , and (x, y, λ) = (−2, 0, 2) .
In summary, we have four solutions:
1
1
(x, y, λ) = 0, 1,
, (x, y, λ) = 0, −1,
,
4
4
(x, y, λ) = (2, 0, 2) ,
and
(x, y, λ) = (−2, 0, 2) .
Step 2: Compute the values of f (x, y) on the above points, we have
f (0, 1) = 2 · 02 + 12 + 2
= 3
f (0, −1) = 2 · 02 + (−1)2 + 2
= 3
f (2, 0) = 2 · 22 + 02 + 2
= 10
f (−2, 0) = 2 · (−2)2 + 02 + 2
= 10.
In summary, the minimum values of f (x, y) subject to the constraint g(x, y) = 0 is 3 = f (0, 1) = f (0, −1),
and the maximum values of f (x, y) subject to the constraint g(x, y) = 0 is 10 = f (2, 0) = f (−2, 0).
14. Let f (x, y) = 2x2 + 2y 2 − 6x and R := {(x, y) : x2 + y 2 ≤ 9}, then
(a) Use Lagrange multipliers to find the maximum and minimum values of f (x, y) on the boundary of R.
Answer.
Let g(x, y) = x2 + y 2 − 9, then the objective function is f (x, y) = 2x2 + 2y 2 − 6x and the
constraint is g(x, y) = x2 + y 2 − 9 = 0. So we need to solve the system:
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
Notice that
∇f (x, y) = h4x − 6, 4yi,
∇g(x, y) = h2x, 2yi.
So we need to solve the system:
4x − 6 = 2λx
(9)
4y = 2λy
(10)
2
x +y
If λ = 0, by (9) and (10), then x =
(11).
If λ 6= 0, by (9) and (10), we have
3
2
2
= 9
(11)
and y = 0. It’s easy to see that (x, y) = (3/2, 0) is not a solution to
2λy(4x − 6) = 2λx(4y).
Since λ 6= 0, then 2y(4x − 6) = 8xy, that is, 8xy − 12y = 8xy. So y = 0. Plug y = 0 into (11), we get
x2 = 9, then x = 3 or x = −3. If x = 3, by (9), then 6 = 6λ, that is, λ = 1. If x = −3, by (9), then
−12 = −6λ, that is, λ = 2. We get two solutions
(x, y, λ) = (3, 0, 1),
and
(x, y, λ) = (−3, 0, 2).
and
(x, y, λ) = (−3, 0, 2).
In summary, we get two solutions:
(x, y, λ) = (3, 0, 1),
Notice that
f (3, 0) = 2 · 32 − 6 · 3 = 0,
and f (−3, 0) = 2 · (−3)2 + 6 · 3 = 36.
So the maximum value of f (x, y) on the boundary of R is 36, and the minimum value of f (x, y) on the
boundary of R is 0.
(b) Find the absolute maximum and minimum values of f (x, y) on R.
Answer.
Find critical points of f (x, y) inside R. Since ∇f (x, y) = h4x − 6, 4yi, then we have only one
critical point:
3
(x, y) = ( , 0).
2
Notice that
3
f ( , 0) = 2 ·
2
2
3
3
9
9
−6· =2· −9=−
2
2
4
2
By the result of part a, we know that the absolute maximum value of f (x, y) on R is 36, and the absolute
minimum value of f (x, y) on R is − 92 .
Review Problems for Chapter 5 and Chapter 7
2
Z
15. (a) Write down the midpoint Riemann sum with n = 50 to approximate
1
50
X
(a)
k=1
Answer.
Since n = 5, then ∆x =
So the midpoint Riemann sum is: x∗k =
f (x∗k )∆x =
k=1
50
X
k=1
0.02
1 + 0.02(k − 0.5)
2−1
1
=
= 0.02 and
50
50
for all k = 0, 1, 2, · · · , 5.
xk = 1 + k∆x = 1 + 0.02k,
50
X
1
dx using sigma notations.
x
xk−1 + xk
1 + 0.02(k − 1) + 1 + 0.02k
=
= 1 + 0.02(k − 0.5) and
2
2
50
X
0.02
1
· 0.02 =
.
1 + 0.02(k − 0.5)
1 + 0.02(k − 0.5)
k=1
2
Z
(b) Let f and g be two integrable functions on [1, 2], if
Z 2
[2f (x) − 3g(x)] dx.
Z
2
f (x) dx = 2 and
1
g(x) dx = 1, compute
1
1
1
(b)
Answer.
In fact, we have
Z 2
Z
[2f (x) − 3g(x)] dx = 2
1
2
Z
f (x) dx − 3
1
2
g(x) dx
1
= 2·2−3·1
= 1.
√
1 − x2 , if 0 ≤ x ≤ 1,
16. Let f (x) =
, and R be the region bounded by the graph of f (x) and x-axis between
2 − 2x,
if x > 1,
x = 0 and x = 3, then
(a) Find the area of R.
(a)
π
+4
4
Answer.
The region bounded by the graph of f (x) and x-axis between x = 0 and x = 3 consists of a
1
-disc and a triangle.
4
1
1
π
1
π
For the -disc, its area is π · 12 = . For the triangle, its area is · 2 · 4 = 4. So the area of R is + 4.
4
4
4
2
4
(b) Find the net area of R.
(b)
π
−4
4
1
-disc is above the x-axis, it has positive net area, but the triangle is below the x-axis,
4
π
it has negative net area. So the net are of R is − 4.
4
Answer.
The
Z
3
f (x) dx.
(c) Compute
0
π
−4
4
(c)
Z
3
Z
Answer.
By the geometric meaning of
f (x) dx, it’s just the net area of R, that is,
0
π
− 4.
4
1, if x is a rational number
17. Let f (x) =
, then
0, if x is not a rational number
3
f (x) dx =
0
(a) What’s the value for the left Riemann sum for any regular partition of [0, 1]?.
1
(a)
Answer.
For any regular partition with n, we know that ∆x =
xk = k∆x =
It’s easy to see that left end point x∗k =
So the left Riemann sum is:
n
X
for all k = 0, 1, 2, · · · , n.
k−1
is always a rational number, which implies that f (x∗k ) = 1.
n
f (x∗k )∆x
k=1
k
,
n
1
and
n
=
n
X
k=1
n
X
1
1
∆x =
= n · = 1.
n
n
k=1
(b) For any partition of [0, 1], is it true that we can always find a Riemann sum whose value is 0? (Just put
‘Yes’ or ‘No’).
Yes
(b)
Answer.
For any partition a = x0 < x1 < x2 < · · · < xn = n, for any subinterval [xk−1 , xk ], we can
take an irrational number x∗k which belongs to [xk−1 , xk ], which implies that f (x∗k ) = 0. So we get
n
X
k=1
f (x∗k )∆xk
=
n
X
k=1
0 · ∆xk = 0.
(c) Is f integrable on [0, 1]? (Just put ‘Yes’ or ‘No’).
No
(c)
Answer.
If f is integrable on [0, 1], for any regular partition, when n → 0, we should get the same
limit. But by the results of part a and part b, we could get two limits 1 and 0, contradiction. Therefore,
f is not integrable on [0, 1].
18. (a) Find an antiderivative of |x|.
x
Z
|t| dt =
(a)
0
1 2
2x ,
− 12 x2 ,
if x ≥ 0,
if x < 0.
Z
Answer.
0
tiderivative of |t|. Notice that
Z
x
|t| dt =
0
(b) Compute
d
dx
1 2
2x ,
− 12 x2 ,
if x ≥ 0,
if x < 0.
x
Z
sin2 (t) dt.
1
sin2 (x)
(b)
Answer.
x
|t| dt is an an-
By the Fundamental Theorem of Calculus, we know that the function
In fact, we have
Z
d
dx
x
sin2 (t) dt = sin2 (x) ·
1
d
d
(x) − sin2 (1) ·
(1)
dx
dx
= sin2 (x).
(c) Compute
d
dx
Z
e2x
sin(t)
dt.
t
x−1
2 sin(e2x ) −
(c)
Answer.
In fact, we have
d
dx
(d) Compute lim
x→0
1
x
Z
e2x
Z
sin(e2x ) d 2x
sin(x − 1) d
(e ) −
·
(x − 1)
·
2x
e
dx
x−1
dx
sin(e2x )
sin(x − 1)
=
· 2e2x −
2x
e
x−1
sin(x
−
1)
= 2 sin(e2x ) −
.
x−1
sin(t)
dt =
t
x−1
x2
cos(t2 ) dt.
x
-1
(d)
Z
Answer.
sin(x − 1)
x−1
Let f (x) =
x2
Z
2
cos(t ) dt, then f (0) =
x
d
f 0 (x) =
dx
0
cos(t2 ) dt = 0 and
0
Z
x
x2
!
cos(t2 ) dt
= cos(x4 ) · 2x − cos(x2 ) = 2x cos(x4 ) − cos(x2 ).
Approach I: Notice that
1
lim
x→0 x
x2
Z
cos(t2 ) dt = lim
x→0
x
f (x)
f (x) − f (0)
= lim
= f 0 (0) = −1.
x→0
x
x−0
Approach II: By l’Hospital’s rule, we have
1
lim
x→0 x
Z
x2
f (x)
f 0 (0)
=
= f 0 (0) = −1.
x
1
cos(t2 ) dt = lim
x→0
x
Z r q
√
(e) Evaluate
x x x dx.
8 15
x8 +C
15
(e)
Answer.
In fact, since
r q
r q
√
1
x x x =
x x · x2
r q
3
x x2
q
3
=
x · x4
q
7
=
x4
=
7
= x8
So we get
Z r q
Z
√
7
x x x dx = x 8 dx =
Z
(f) Evaluate
7
1
8 15
x 8 +1 = x 8 + C.
15
+1
10x · 32x dx.
1
1
e(ln 10+2 ln 3)x + C or
90x + C
ln 10 + 2 ln 3
ln 90
(f)
Answer.
7
8
Approach I: Notice that
10x · 32x = ex ln 10 · e2x ln 3 = e(ln 10+2 ln 3)x .
So we get
Z
10x · 32x dx =
=
Z
e(ln 10+2 ln 3)x dx
1
e(ln 10+2 ln 3)x + C.
ln 10 + 2 ln 3
Approach II: Notice that
10x · 32x = 10x · 9x = 90x = ex ln 90 .
So we get
Z
x
10 · 3
2x
Z
dx =
ex ln 90 dx =
1 x ln 90
1
e
+C =
10x + C.
ln 90
ln 90
Z
(g) Evaluate
cos(2x)
dx.
cos(x) − sin(x)
sin(x) − cos(x) + C
(g)
Answer.
Notice that
cos(2x) = 2 cos2 (x) − 1 = 2 cos2 (x) − [cos2 (x) + sin2 (x)] = cos2 (x) − sin2 (x).
Then
cos2 (x) − sin2 (x)
cos(2x)
=
= cos(x) + sin(x).
cos(x) − sin(x)
cos(x) − sin(x)
So we get
Z
Z
(h) Evaluate
cos(2x)
dx =
cos(x) − sin(x)
Z
[cos(x) + sin(x)] dx = sin(x) − cos(x) + C.
cos(2x)
dx.
· sin2 (x)
cos2 (x)
− cot(x) − tan(x) + C or −
(h)
Answer.
2
+C
sin(2x)
Approach I: Notice that
cos(2x) = 2 cos2 (x) − 1 = 2 cos2 (x) − [cos2 (x) + sin2 (x)] = cos2 (x) − sin2 (x).
Then
cos2 (x) − sin2 (x)
1
1
cos(2x)
=
=
−
= csc2 (x) − sec2 (x).
2
2
2
2
2
cos (x) sin (x)
cos (x) sin (x)
sin (x) cos2 (x)
So we get
Z
cos(2x)
dx =
2
cos (x) sin2 (x)
Z
[csc2 (x) − sec2 (x)] dx = − cot(x) − tan(x) + C.
Approach II: Since sin(2x) = 2 sin(x) cos(x), then
Z
Z
cos(2x)
cos(2x)
dx
=
1
2 dx
2
2
cos (x) · sin (x)
sin(2x)
Z 2
cos(2x)
dx
= 4
sin2 (2x)
Z
1
= 2
du let u = sin(2x), then du = 2 cos(2x) dx
u2
2
= − +C
u
2
= −
+ C Since u = sin(2x).
sin(2x)
Z
(i) Evaluate
r
1+x
+
1−x
r
1−x
1+x
(i)
!
dx.
2 sin−1 (x) + C
Answer.
Notice that
r
1+x
+
1−x
r
1−x
1+x
=
=
=
=
√
√
1+x
1−x
√
+√
1−x
1+x
√
√
√
√
1+x· 1+x
1−x· 1−x
√
√
√
+√
1−x· 1+x
1+x· 1−x
1−x
1+x
√
+√
1 − x2
1 − x2
2
√
.
1 − x2
So we get
Z r
Z
(j) Evaluate
√
1+x
+
1−x
r
1−x
dx =
1+x
Z
√
2
dx = 2 sin−1 (x) + C.
1 − x2
dx
√ .
x+ 3x
1
1
1 1 1 1
6
2
3
6
6 x − x + x − ln x + 1 + C
3
2
(j)
Answer.
Let u =
√
6
x, then x = u6 and dx = 6u5 du. So we get
Z
Z
dx
6u5
√
√
du
dx
=
u3 + u2
x+ 3x
Z
u3
= 6
du
u+1
By the long division, we have
u3
1
= u2 − u + 1 −
.
u+1
u+1
Then
Z
Z
(k) Evaluate
ex
1
u −u+1−
du
u+1
1 3 1 2
= 6 u − u + u − ln |u + 1| + C
3
2
1
1
1 1 1 1
6
2
3
6
= 6 x − x + x − ln x + 1 + C
3
2
dx
√
√ dx = 6
x+ 3x
Z 2
1
dx.
+ e−x
tan−1 (ex ) + C
(k)
Answer.
In fact, we have
Z
Z
1
ex
dx
=
dx
ex + e−x
e2x + 1
Z
1
=
du Let u = ex , then du = ex dx
2
u +1
= tan−1 (u) + C
= tan−1 (ex ) + C
Since u = ex .
1
Since u = x 6
Z
√
(l) (0.5 points) Evaluate
−x2
dx
.
− 8x − 7
−1
sin
(l)
Answer.
x+4
3
+C
By completing square, we have
−x2 − 8x − 7 = −(x + 4)2 + 16 − 7
= 9 − (x + 4)2 .
Then
Z
Z
(m) Evaluate
dx
√
2
−x − 8x − 7
Z
1
p
dx
9 − (x + 4)2
Z
1
√
du Let u = x + 4
=
2
3 − u2
u
= sin−1
+C
3
−1 x + 4
+ C Since u = x + 4.
= sin
3
=
tan−1 (x) dx.
x tan−1 (x) −
(m)
Answer.
1
ln |x2 + 1| + C
2
By ‘ILATE’ rule, let
1
u = tan−1 (x) =⇒ du =
dx
2
Z1 + x
dv = dx =⇒ v = dx = x.
Then we get
Z
−1
tan
Z
(x) dx =
u dv
Z
= uv − v du
Z
x
dx
+1
Z
1
dw
= x tan−1 (x) −
Let w = x2 + 1
2
w
1
= x tan−1 (x) − ln |w| + C
2
1
= x tan−1 (x) − ln |x2 + 1| + C Since w = x2 + 1.
2
−1
= x tan
Z (n) Evaluate
ln(ln x) +
1
ln x
(x) −
x2
(n)
dx.
x ln(ln x) + C
Answer.
Let
1
1
u = ln(ln x) =⇒ du =
· dx
Z ln x x
dv = dx =⇒ v = dx = x.
Then
Z Z
Z
1
1
ln(ln x) +
dx =
ln(ln x) dx +
dx
ln x
ln x
Z
Z
1
=
u dv +
dx
ln x
Z
Z
1
dx
= uv − v du +
ln x
Z
Z
1
1
1
= x ln(ln x) − x ·
· dx +
dx
ln x x
ln x
Z
Z
1
1
= x ln(ln x) −
dx +
dx
ln x
ln x
= x ln(ln x) + C.
(o) Find the area of the region bounded by the graph of ln x and x-axis between x = e−1 and x = e.
2 − 2e−1
(o)
Answer.
Notice that the question is to find the are not the net area. So the area is:
Z e
Z 1
Z e
| ln x| dx =
| ln x| dx +
| ln x| dx
e−1
e−1
1
Z 1
Z e
= −
ln x dx +
ln x dx.
e−1
1
Notice that
Z
Z
ln x dx = x ln x −
x·
1
dx
x
Z
= x ln x −
1 dx
= x ln x − x + C.
So we get
Z
e
| ln x| dx =
e−1
− [x ln x − x]|1e−1 + [x ln x − x]|e1
= −[1 ln 1 − 1] + [e−1 ln e−1 − e−1 ] + [e ln e − e] − [1 ln 1 − 1]
= 1 − e−1 − e−1 + e − e + 1
= 2 − 2e−1
(p) Compute lim
n→∞
1
3
3
1
+
2
+
·
·
·
+
n
.
n4
(p)
1
4
Answer.
Write
1
3
3
1
+
2
+
·
·
·
+
n
as a Riemann sum:
n4
n
n
n 1
1 X 3
1X k 3
1 X k3
3
3
1 + 2 + ··· + n = 4
=
.
k =
n4
n
n
n3
n
n
k=1
k=1
k=1
So we know that
lim
n→∞
1
1 + 2 3 + · · · + n3 =
4
n
n
n→∞
Z
=
=
1X
n
lim
k=1
3
k
n
1
x3 dx
0
1 4 1
x
4 0
=
(q) Compute lim n
n→∞
1
.
4
1
1
1
+
+ ··· + 2 .
n2 + 1 n2 + 2 2
2n
π
4
(q)
Answer.
Write n
n
1
1
1
+
+ ··· + 2
n2 + 1 n2 + 2 2
2n
1
1
1
+ 2
+ ··· + 2
2
2
n +1 n +2
2n
=n
n
X
k=1
as a Riemann sum:
n
n
k=1
k=1
1
1 X n2
1X
1
=
=
2 .
2
2
2
2
n +k
n
n +k
n
1+ k
So we get
lim n
n→∞
1
1
1
+ 2
+ ··· + 2
2
2
n +1 n +2
2n
n
1X
1
= lim
k 2
n→∞ n
k=1 1 + n
Z 1
1
dx
=
1
+
x2
0
1
= tan−1 (x)0
= tan−1 (1) − tan−1 (0)
π
=
.
4
1
(r) Compute lim
n→∞ n
π
2π
n−1
sin + sin
+ · · · + sin
π .
n
n
n
2
π
(r)
Answer.
Since sin(π) = 0, then
1
n
n−1
n
π
2π
n−1
1X
k
1X
k
sin + sin
+ · · · + sin
π =
sin
π =
sin
π .
n
n
n
n
n
n
n
k=1
k=1
n
Then
1
lim
n→∞ n
π
2π
n−1
sin + sin
+ · · · + sin
π
=
n
n
n
=
=
=
=
=
Z
(s) Evaluate
n
1X
k
lim
sin
π
n→∞ n
n
k=1
Z 1
sin(xπ) dx
0
1
1
− cos(πx)
π
0
1
1
− cos(π) + cos(0)
π
π
1
1
+
π π
2
.
π
sin3 (x) cos2 (x) dx.
1
1
sin4 (x) cos(x) sin2 (x) cos(x) 2 cos(x)
−
−
+C
(s) − cos3 (x) + cos5 (x) + C or
3
5
5
15
15
Answer.
Approach I: In fact, we have
Z
Z
3
2
sin (x) cos (x) dx =
sin2 (x) cos2 (x) · sin(x) dx
Z
=
[1 − cos2 (x)] cos2 (x) · sin(x) dx
Z
= − [cos2 (x) − cos4 (x)] d cos(x)
1
1
= − cos3 (x) + cos5 (x) + C.
3
5
Approach II: Since sin2 (x) + cos2 (x) = 1, then
Z
Z
sin3 (x) cos2 (x) dx =
sin3 (x)[1 − sin2 (x)] dx
Z
Z
=
sin3 (x) dx − sin5 (x) dx.
Recall the reduction formula:
Z
Z
sinn−1 (x) cos(x) n − 1
n
+
sinn−2 (x) dx.
sin (x) dx = −
n
n
Then
Z
sin4 (x) cos(x) 4
sin (x) dx = −
+
5
5
5
Z
sin3 (x) dx.
So we get
Z
sin3 (x) cos2 (x) dx =
=
=
Z
sin4 (x) cos(x) 1
+
sin3 (x) dx
5
5
Z
sin4 (x) cos(x) 1
sin2 (x) cos(x) 2
+
−
+
sin(x) dx + C
5
5
3
3
By the reduction formula again
4
sin (x) cos(x) sin2 (x) cos(x) 2 cos(x)
−
−
+ C.
5
15
15
Z
(t) Evaluate
tan(x) sec2 (x) dx.
1
1
tan2 (x) + C or sec2 (x) + C
2
2
(t)
Approach I: Notice that d tan(x) = sec2 (x) dx, then
Z
Z
tan(x) sec2 (x) dx =
tan(x) d tan(x)
Answer.
1
tan2 (x) + C.
2
Approach II: Notice that d sec(x) = sec(x) tan(x) dx, then
Z
Z
tan(x) sec2 (x) dx =
sec(x) d sec(x)
=
=
1
sec2 (x) + C.
2
Approach III: Notice that sec2 (x) = 1 + tan2 (x), then
Z
Z
2
tan(x) sec (x) dx =
tan(x)[1 + tan2 (x)] dx
Z
Z
=
tan(x) dx + tan3 (x) dx.
Recall the reduction formula:
Z
tann−1 (x)
−
tan (x) dx =
n−1
Z
tan2 (x)
−
2
Z
n
Then
Z
tan3 (x) dx =
tann−2 (x) dx.
tan(x) dx.
So we get
Z
Z
(u) Evaluate
tan(x) sec2 (x) dx =
sec3 (x) dx.
(u)
Z
Answer.
tan2 (x)
+ C.
2
Let I =
Z
I =
sec(x) tan(x) ln | sec(x) + tan(x)|
+
+C
2
2
sec3 (x) dx, then
sec(x) · sec2 (x) dx
Z
=
=
=
=
=
sec(x) d tan(x) Since d tan(x) = sec2 (x) dx
Z
sec(x) tan(x) − tan(x) d sec(x) Use the integration by parts
Z
sec(x) tan(x) − tan2 (x) sec(x) dx Since d sec(x) = sec(x) tan(x) dx
Z
sec(x) tan(x) − [sec2 (x) − 1] sec(x) dx Since tan2 (x) = 1 = sec2 (x)
Z
Z
sec(x) tan(x) − sec3 (x) dx + sec(x) dx
= sec(x) tan(x) − I + ln | sec(x) + tan(x)| + C.
So we get
2I = sec(x) tan(x) + ln | sec(x) + tan(x)| + C.
Hence
Z
I=
Z
(v) Evaluate
sec3 (x) dx =
sec(x) tan(x) ln | sec(x) + tan(x)|
+
+ C.
2
2
tan2 (x) sec3 (x) dx.
sec3 (x) tan(x) sec(x) tan(x) 1
−
− ln | sec(x) + tan(x)| + C
4
8
8
(v)
Notice that 1 + tan2 (x) = sec2 (x), then
Z
Z
Z
Z
2
3
2
3
5
tan (x) sec (x) dx = [sec (x) − 1] sec (x) dx = sec (x) dx − sec3 (x) dx.
Answer.
Z
Recall the reduction formula for
Z
secn (x) dx:
secn−2 (x) tan(x) n − 2
sec (x) dx =
+
n−1
n−1
n
Z
secn−2 (x) dx.
Then we have
Z
Z
sec5 (x) dx =
tan2 (x) sec3 (x) dx =
=
=
Recall
sec3 (x) tan(x)
4
3
sec (x) tan(x)
4
sec3 (x) tan(x)
4
3
sec (x) tan(x)
4
Z
3
sec3 (x) dx
4
Z
Z
3
3
+
sec (x) dx − sec3 (x) dx
4
Z
1
−
sec3 (x) dx
4
Z
1 sec(x) tan(x) 1
−
+
sec(x) dx .
4
2
2
+
Z
sec(x) dx = ln | sec(x) + tan(x)| + C.
Then
19. (a) Evaluate
tan2 (x) sec3 (x) dx =
Z
sec2 (x) ln(tan(x) + 2) dx.
(a)
Answer.
Z
sec3 (x) tan(x) sec(x) tan(x) 1
−
− ln | sec(x) + tan(x)| + C.
4
8
8
Z
(tan(x) + 2) ln(tan(x) + 2) − tan(x) + C
Notice that d tan(x) = sec2 (x) dx, then
Z
2
sec (x) ln(tan(x) + 2) dx =
ln(u + 2) du Let u = tan(x), then d tan(x) = sec2 (x) dx
Z
=
ln v dv Let v = u + 2
Z
1
= v ln v − v · dv Use the integration by parts
v
Z
= v ln v −
dv
= v ln v − v + C
= (u + 2) ln(u + 2) − (u + 2) + C
Since v = u + 2
= (tan(x) + 2) ln(tan(x) + 2) − tan(x) + C
Since u = tan(x).
Z
(b) Evaluate
x2 e3x dx.
2
1 2 3x 2 3x
x e − xe + e3x + C
3
9
27
(b)
Answer.
Use the integration by parts, by ’ILATE’ rule, let
u = x2 =⇒ du =Z2x dx
1
dv = e3x dx =⇒ v = e3x dx = e3x .
3
Then
Z
x2 e3x dx =
Z
u dv
Z
= uv − v du
Z
1 2 3x 2
=
xe3x dx
x e −
3
3
Z
1 2 3x 2
x de3x Since de3x = 3e3x dx
x e −
=
3
9
Z
1 2 3x 2 3x 2
=
x e − xe +
e3x dx Use the integration by parts, by ’ILATE’ rule
3
9
9
2
1 2 3x 2 3x
x e − xe + e3x + C.
=
3
9
27
Z
(c) Evaluate
x sin(x) cos(x) dx.
1
1
− x cos(2x) + sin(2x) + C
4
8
(c)
Answer.
In fact, use the integration by parts, by ’ILATE rule, we have
Z
Z
1
x sin(2x) dx Since sin(2x) = 2 sin(x) cos(x)
x sin(x) cos(x) dx =
2
Z
1
= −
x d cos(2x) Since d cos(2x) = −2 sin(2x) dx
4
Z
1
1
cos(2x) dx
= − x cos(2x) +
4
4
1
1
= − x cos(2x) + sin(2x) + C.
4
8
Z
(d) Evaluate
sin3 (x) dx.
− cos(x) +
(d)
Answer.
1
cos3 (x) + C
3
In fact, we have
Z
Z
sin3 (x) dx =
sin2 (x) · sin(x) dx
Z
=
[1 − cos2 (x)] sin(x) dx Since sin2 (x) + cos2 (x) = 1
Z
= − [1 − cos2 (x)] d cos(x) Since d cos(x) = − sin(x) dx
= − cos(x) +
1
cos3 (x) + C.
3
Z
(e) Evaluate
sin−1 (x) dx.
x sin−1 (x) +
(e)
Answer.
p
1 − x2 + C
Use the integration by parts, by ’ILATE’ rule, we have
u = sin−1 (x) =⇒ du = √
1
dx
1 − x2
dv = dx =⇒ v = x.
Then
Z
−1
sin
Z
(x) dx =
u dv
Z
= uv − v du
−1
= x sin
Z
√
(x) −
x
dx
1 − x2
1 du
√
Let u = 1 − x2 , then du = −2xdx
2 u
1
1
1
u− 2 +1 + C
= x sin−1 (x) + · 1
2 −2 + 1
= x sin−1 (x) +
1
= x sin−1 (x) + u 2 + C
p
= x sin−1 (x) + 1 − x2 + C
Since u = 1 − x2 .
π
(f) Let R be the region bounded by y = sin(x) cos(x), the x-axis, the line x = 0 and the line x = . Find
2
the volume of the solid that is generated when the region R is revolved about the x-axis. .
π2
16
(f)
Answer.
In fact, we have
Z
Volume =
π
2
π[sin(x) cos(x)]2 dx
2
Z π
2
1
π
sin(2x) dx Since sin(2x) = 2 sin(x) cos(x)
2
0
Z π
π 2 1 − cos(4x)
1 − cos(4x)
dx Since
= sin2 (2x)
4 0
2
2
π
2
π x 1
− sin(4x) 4 2 8
0
π π
·
Since sin(2π) = sin(0) = 0.
4 4
π2
.
16
0
=
=
=
=
=
Z
20. (a) Evaluate
(x2
1
dx.
+ 1)2
(a)
1
1
x
tan−1 (x) + · 2
+C
2
2 x +1
Let x = tan(θ), then dx = sec2 (θ) dθ and x2 + 1 = tan2 (θ) + 1 = sec2 (θ). So we get
Z
Z
1
1
dx =
· sec2 (θ) dθ
2
2
(x + 1)
sec4 (θ)
Z
1
=
dθ
sec2 (θ)
Z
=
cos2 (θ) dθ
Z
1 + cos(2θ)
1 + cos(2θ)
=
dθ Since cos2 (θ) =
2
2
1
1
=
θ + sin(2θ) + C
2
4
1
1
=
θ + sin(θ) cos(θ) + C Since sin(2θ) = 2 sin(θ) cos(θ).
2
2
Answer.
Since x = tan(θ), then
θ = tan−1 (x),
sin(θ) = √
x
,
+1
x2
and
cos(θ) = √
1
x2
+1
.
So we get
Z
(x2
Z
(b) Evaluate
1
1
x
1
dx = tan−1 (x) + · 2
.
2
+ 1)
2
2 x +1
tan3 (x) dx.
1
sec2 (x) − ln | sec(x)| + C
2
(b)
Answer.
In fact, we have
Z
Z
tan2 (x)
3
dx
tan (x) dx =
sec(x) tan(x) ·
sec(x)
Z
sec2 (x) − 1
=
sec(x) tan(x) ·
dx Since 1 + tan2 (x) = sec2 (x)
sec(x)
Z
sec2 (x) − 1
d sec(x) Since d sec(x) = sec(x) tan(x) dx
=
sec(x)
1
=
sec2 (x) − ln | sec(x)| + C.
2
Z
(c) Evaluate
dx
3
(4 − 4x2 ) 2
.
1
x
·√
+C
8
1 − x2
(c)
Answer.
Notice that
Z
dx
1
3 =
2
8
(4 − 4x ) 2
Z
dx
3
(1 − x2 ) 2
.
Let x = sin(θ), then 1 − x2 = cos2 (θ) and dx = cos(θ)dθ. So we get
Z
Z
dx
1
cos(θ)
=
dθ
3
3 (θ)
2
8
cos
(4 − 4x ) 2
Z
1
=
sec2 (θ) dθ
8
1
=
tan(θ) + C.
8
x
Since x = sin(θ), then tan(θ) = √
. So we get
1 − x2
Z
dx
1
x
·√
+ C.
3 =
2
8
1 − x2
(4 − 4x ) 2
Z
dx
(d) Evaluate
3 .
(4 + 4x2 ) 2
1
x
+C
·√
8
1 + x2
(d)
Answer.
Notice that
Z
dx
1
3 =
8
(4 + 4x2 ) 2
Z
dx
3
(1 + x2 ) 2
.
Let x = tan(θ), then 1 + x2 = sec2 (θ) and dx = sec2 (θ)dθ. So we get
Z
Z
sec2 (θ)
1
dx
dθ
=
3
8
sec3 (θ)
(4 + 4x2 ) 2
Z
1
1
=
dθ
8
sec(θ)
Z
1
=
cos(θ) dθ
8
1
sin(θ) + C.
=
8
x
. So we get
Since x = tan(θ), then sin(θ) = √
1 + x2
Z
dx
1
x
·√
+ C.
3 =
2
8
1 + x2
(4 + 4x ) 2
Z
dx
(e) Evaluate
3 .
2
(4x − 4) 2
x
− √
+C
8 x2 − 1
(e)
Answer.
Notice that
Z
dx
1
3 =
8
(4x2 − 4) 2
Z
dx
3
(x2 − 1) 2
.
Let x = sec(θ), then x2 − 1 = tan2 (θ) and dx = sec(θ) tan(θ)dθ. So we get
Z
Z
dx
1
sec(θ) tan(θ)
=
dθ
3
2
8
tan3 (θ)
(4x − 4) 2
Z
1
sec(θ)
=
dθ
8
tan2 (θ)
Z
1
1
cos2 (θ)
=
·
dθ
8
cos(θ) sin2 (θ)
Z
1
cos(θ)
=
dθ
8
sin2 (θ)
Z
1
d sin(θ)
=
8
sin2 (θ)
1
+ C.
= −
8 sin(θ)
√
1
x2 − 1
. So we get
Since x = sec(θ), that is, cos(θ) = , then sin(θ) =
x
x
Z
x
dx
+ C.
3 = − √
2
8 x2 − 1
(4x − 4) 2
(f) Find the partial fraction decomposition of
x2
.
x3 − 16x
1
1
+
2(x + 4) 2(x − 4)
(f)
Answer.
Since x3 − 16x = x(x2 − 16) = x(x + 4)(x − 4), then the partial fractional decomposition of
2
x
is:
3
x − 16x
x2
A
B
C
= +
+
.
x3 − 16x
x
x+4 x−4
So we get
x2 = A(x + 4)(x − 4) + Bx(x − 4) + Cx(x + 4)
= Ax2 − 16A + Bx2 − 4Bx + Cx2 + 4Cx
= (A + B + C)x2 + (−4B + 4C)x − 16A
So we get
A+B+C = 1
(12)
−4B + 4C = 0
(13)
−16A = 0
(14)
By (14), then A = 0. So we get B + C = 1, plug B = 1 − C into (13), we get −4(1 − C) + 4C = 0, that
1
1
is, 8C = 4, which implies that C = . Since B = 1 − C, then B = . Therefore, we get
2
2
x2
1
1
=
+
.
3
x − 16x
2(x + 4) 2(x − 4)
Z
(g) Evaluate
x3 + 3x2 − 3x + 2
dx.
x3 − 2x2
x + ln |x| +
(g)
1
+ 4 ln |x − 2| + C
x
x3 + 3x2 − 3x + 2
5x2 − 3x + 2
=
1+
. Since x3 −2x2 =
x3 − 2x2
x3 − 2x2
5x2 − 3x + 2
x2 (x − 2), then the partial fraction decomposition of
has the form:
x3 − 2x2
Answer.
Using the long division, we know that
B
5x2 − 3x + 2
A
C
= + 2+
.
3
2
x − 2x
x
x
x−2
Then
5x2 − 3x + 2 = Ax(x − 2) + B(x − 2) + Cx2
= A(x2 − 2x) + B(x − 2) + Cx2
= (A + C)x2 + (−2A + B)x − 2B.
So we have
A+C = 5
(15)
−2A + B = −3
(16)
−2B = 2
(17)
By (17), then B = −1. Plug B = −1 into (16), then −2A − 1 = −3, which implies that A = 1. Plug
A = 1 into (15), then C = 5 − A = 4. So we get
A = 1,
B = −1,
That is, the partial fraction decomposition of
and C = 4.
5x2 − 3x + 2
is:
x3 − 2x2
5x2 − 3x + 2
1
4
1
= − 2+
.
x3 − 2x2
x x
x−2
Therefore, we have
Z
Z
Z
Z
Z
Z 3
5x2 − 3x + 2
1
1
4
x + 3x2 − 3x + 2
dx =
1 dx +
dx = x +
dx −
dx +
dx
x3 − 2x2
x3 − 2x2
x
x2
x−2
1
= x + ln |x| + + 4 ln |x − 2| + C.
x
Z (h) Evaluate
sec2 (2x) +
3
x
dx.
1
tan(2x) + 3 ln |x| + C
2
(h)
Answer.
Z 3
sec (2x) +
x
2
Z
dx =
=
Z (i) Evaluate
3e
2x
x2
−
2
2
sec (2x) dx + 3
Z
1
dx
x
1
tan(2x) + 3 ln |x| + C.
2
dx.
(i)
3 2x 1 3
e − x +C
2
6
Answer.
Z Z
Z
x2
1
2x
2x
3e −
dx = 3 e dx −
x2 dx
2
2
1
1
1
= 3 · e2x − ·
x2+1 + C
2
2 2+1
3 2x 1 3
=
e − x + C.
2
6
Z (j) Evaluate
1
1
sin(4x) + cos(3x)
4
3
dx.
−
(j)
1
1
cos(4x) + sin(3x) + C
16
9
Answer.
Z Z (k) Evaluate
1
1
sin(4x) + cos(3x) dx =
4
3
Z
Z
1
1
sin(4x) dx +
cos(3x) dx
4
3
1
1 1
1
· − cos(4x) + · sin(3x) + C
=
4
4
3 3
1
1
= − cos(4x) + sin(3x) + C.
16
9
2
1
2
tan(x) − csc (5x) dx.
3
5
2
1
− ln | cos(x)| +
cot(5x) + C
3
25
(k)
Answer.
Z Z (l) Evaluate
1
2
2
tan(x) − csc (5x) dx =
3
5
Z
Z
2
1
tan(x) dx −
csc2 (5x) dx
3
5
2
1
1
=
· (− ln | cos(x)|) − · − cot(5x) + C
3
5
5
2
1
= − ln | cos(x)| +
cot(5x) + C.
3
25
5
2
+
−√
2
3 + x2
2−x
dx.
−1
−2 sin
(l)
x
√
2
5
+ √ tan−1
3
x
√
3
+C
Answer.
Z Z (m) Evaluate
5
2
−√
+
2 − x2 3 + x2
sec(x)
2 sec(x) tan(x) +
2
Z
Z
1
1
dx = −2 √
dx + 5
dx
3 + x2
2 − x2
x
5
x
−1
−1
√
√
= −2 sin
+ √ tan
+ C.
2
3
3
dx.
2 sec(x) +
(m)
1
ln | sec(x) + tan(x)| + C
2
Answer.
Z sec(x)
2 sec(x) tan(x) +
2
Z
Z
1
dx = 2 sec(x) tan(x) dx +
sec(x) dx
2
1
= 2 sec(x) + ln | sec(x) + tan(x)| + C.
2
Z
21. Let f be a differentiable function on [2, 4] such that f (2) = 1 and f (4) = 5, compute
2
f (2x)f 0 (2x) dx.
1
6
21.
Answer.
In fact, we have
Z 2
Z
f (2x)f 0 (2x) dx =
1
4
f (u)f 0 (u) ·
2
Z
=
1
2
Z
=
1
2
Z
=
1
2
4
1
du Let u = 2x, then du = 2dx
2
f (u)f 0 (u) du
2
f (4)
w dw
Let w = f (u), then dw = f 0 (u)du
f (2)
5
w dw
1 1 2 5
· w
2 2 1
1
· [52 − 12 ]
4
1
· 24
4
6.
=
=
=
=
Z
22. Use the right Riemann sum to compute
1
1
(2x + 1) dx. (Hint:
0
n
X
k=
k=1
n(n + 1)
.)
2
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
right Riemann sum to approximate
(2x + 1) dx. Now for any n, the regular partition using n tells us that
0
1−0
1
∆x =
= and
n
n
xk = 0 + k∆x =
So the right Riemann sum is: x∗k = xk =
n
X
k
n
k
,
n
for all k = 0, 1, 2, · · · , n.
and
f (x∗k )∆x
=
k=1
=
n
X
k
1
f
·
n
n
k=1
n 1 X 2k
·
+1
n
n
k=1
!
n
n
X
2X
=
k+
1
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
Z
23. Use the integration by parts to get a reduction formula of the integral
Z
Answer.
Let In =
ex sinn (x) dx.
ex sinn (x) dx, by ‘ILATE’ rule, we have
Z
sinn (x) dex
Z
= ex sinn (x) − nex sinn−1 (x) cos(x) dx
Z
= ex sinn (x) − n sinn−1 (x) cos(x) dex
Z
x
n
x
n−1
x
n−2
2
n−1
= e sin (x) − n e sin
(x) cos(x) − e (n − 1) sin
cos (x) − sin
(x) sin(x) dx
Z
Z
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1) ex sinn−2 cos2 (x) dx − n ex sinn (x) dx
Z
x
n
x
n−1
= e sin (x) − ne sin
(x) cos(x) + n(n − 1) ex sinn−2 [1 − sin2 (x)] dx − nIn
In =
Since sin2 (x) + cos2 (x) = 1
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1)
Z
ex sinn−2 dx − n(n − 1)
Z
ex sinn (x) dx − nIn
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1)In−2 − n(n − 1)In − nIn
= ex sinn (x) − nex sinn−1 (x) cos(x) + n(n − 1)In−2 − n2 In .
So we get
In =
Z
24. Evaluate
Answer.
n2
1 x n
e sin (x) − nex sinn−1 (x) cos(x) + n(n − 1)In−2 .
+1
2x3
dx.
x3 − 4x
Notice that
2x3
2x3
2x2
=
=
.
x3 − 4x
x(x2 − 4)
x2 − 4
By Long Division, we have
2x2
2(x2 − 4) + 8
8
=
=2+ 2
.
2
2
x −4
x −4
x −4
For
x2
8
, since x2 − 4 = (x + 2)(x − 2), then
−4
x2
8
A
B
=
+
.
−4
x+2 x−2
Then
8 = A(x − 2) + B(x + 2) = Ax − 2A + Bx + 2B = (A + B)x − 2A + 2B.
So we get
A+B = 0
(18)
−2A + 2B = 8
(19)
By (18), then A = −B. Plug A = −B into (19), then 4B = 8, that is, B = 2. Since A = −B, then A = −2.
So we get
8
−2
2
=
+
.
2
x −4
x+2 x−2
Hence we get
2x3
2
2
=2−
+
.
3
x − 4x
x+2 x−2
Therefore, we have
Z
Z
Z
Z
2
2
2x3
dx
=
2
dx
−
dx
+
dx
x3 − 4x
x+2
x−2
= 2x − 2 ln |x + 2| + 2 ln |x − 2| + C.
Z
25. Evaluate the improper integral
0
Answer.
0
Z
√
3
√
3
1
√ dx.
x(x + 3 x)
In fact, we have
Z
For
∞
∞
Z
1
√
√ dx =
3
x(x + 3 x)
lim
a→0+
a
1
1
√
√ dx + lim
3
b→∞
x(x + 3 x)
Z
b
1
√
3
1
√ dx.
x(x + 3 x)
√
1
√ dx, let u = 3 x, then x = u3 and dx = 3u2 du. So we get
x(x + 3 x)
Z
Z
√
1
1
√
√
dx =
· 3u2 du Let u = 3 x
3
3
3
u(u + u)
x(x + x)
Z
1
= 3
du
2
u +1
= 3 tan−1 (u) + C
√
√
= 3 tan−1 ( 3 x) + C Since u = 3 x.
So we get
Z
0
∞
√
3
1
√ dx =
x(x + 3 x)
=
√ 1
√ b
lim 3 tan−1 ( 3 x)a + lim 3 tan−1 ( 3 x)1
a→0+
b→∞
√
√
3
lim [3 tan−1 (1) − 3 tan−1 ( 3 x)] + lim [3 tan−1 ( b) − 3 tan−1 (1)]
a→0+
b→∞
−1
= 3 tan
=
π
(1) − 3 · 0 + 3 · − 3 tan−1 (1)
2
3π
.
2
y
,
2+1
x
26. Solve the initial value problem
.

y(0) = 1.

 y0 =
Answer.
Since y 0 =
y
y0
1
,
then
= 2
, that is,
2
x +1
y
x +1
1
1
dy = 2
dx.
y
x +1
So we get
Z
1
dy =
y
Z
x2
1
dx.
+1
So we get
ln |y| = tan−1 (x) + C.
Since y(1) = 1, then y(x) > 0 for all x and
π
+ C.
4
ln 1 = tan−1 (1) + C =
π
π
Then C = − . So ln y = tan−1 (x) − , that is,
4
4
−1 (x)− π
4
y = etan
.
1
Z
27. Find the function f (x) such that f (x) = 1 +
tf (t) dt.
x
1
Z
Answer.
Plug x = 1 into the equation f (x) = 1 +
tf (t) dt, then
x
Z
1
f (1) = 1 +
tf (t) dt = 1.
1
Z
Differentiate both sides of f (x) = 1 +
1
tf (t) dt, then
x
f 0 (x) = −xf (x).
Let y = f (x), then y 0 = −xy, that is,
y0
= −x, which implies that
y
1
dy = −x dx.
y
So
Then
Z
1
dy = −
y
Z
x dx.
1
ln |y| = − x2 + C.
2
Since y(1) = 1, then y(x) > 0 for all x and
1
ln 1 = − + C.
2
1
Then C = , that is,
2
1
1
ln y = − x2 + .
2
2
So we get
y(x) = f (x) = e−
x2
+ 12
2
.
Review Problems for Probability Theory
28. The length of time X, needed by students in a course to complete a 1 hour exam is a random variable X with
k(x2 + x), if 0 ≤ x ≤ 1,
PDE given by f (x) =
, then
0,
otherwise.
(a) Find the value k.
Answer.
Since f (x) is a PDF, then
∞
Z
f (x) dx
1 =
−∞
Z 1
k(x2 + x) dx
1 3 1 2 1
= k x + x 3
2
0
1 1
= k·
+
3 2
5
=
k.
6
=
0
So
6
k= .
5
(b) Find the CDF.
x
Z
Answer.
For the CDF , we have F (x) =
f (t) dt. Since f (x) =
−∞
• For x ≤ 0, we have
Z
x
x
Z
F (x) =
k(x2 + x), if 0 ≤ x ≤ 1,
, then
0,
otherwise.
f (t) dt =
0 dt = 0.
−∞
−∞
• For x ≥ 1, we have
x
Z
F (x) =
f (t) dt
−∞
Z 0
=
Z
1
f (t) dt +
−∞
Z 0
=
Z
−∞
Z 1
x
f (t) dt +
0
0 dt +
=
Z
f (t) dt
1
1
Z
f (t) dt +
0
x
0 dt
1
f (t) dt
0
= 1
By the computation in part a.
• For any 0 ≤ x ≤ 1, we have
Z
x
F (x) =
f (t) dt
−∞
Z 0
=
Z
f (t) dt +
−∞
Z 0
=
=
Z
0
x
6 2
(t + t) dt
5
6 1 3 1 2 x
t + t 5 3
2
0
6 1 3 1 2
x + x .
5 3
2
=
f (t) dt
0
0 dt +
−∞
x
In summary, we have

0,


if x ≤ 0,
 6 1
1
F (x) =
x3 + x2 , if 0 ≤ x ≤ 1,
 5 3
2


1,
if x ≥ 1.
(c) Find the probability that a randomly selected student will finish the exam in less that half an hour.
Answer.
By the result of part (b), we have
" 2 #
1
1
6
6
6 4
1
1 3 1
1
1
1
1
Pr X <
=F
= ·
+ ·
= ·
= ·
·
+
= .
2
2
5
3
2
2
2
5 24 8
5 24
5
(d) Find the mean time needed to complete in an 1 hour exam.
Answer.
In fact, we have
∞
Z
xf (x) dx
E(X) =
−∞
Z 1
6
x · (x2 + x) dx
5
=
0
=
=
=
=
=
6
5
Z
1
(x3 + x2 ) dx
0
6 1 4 1 3 1
x + x 5 4
3
0
6 1 1
·
+
5 4 3
6 7
·
5 12
7
.
10
ax2 + b, if 0 ≤ x ≤ 1,
for some constants a and b such that f (x) is a probability density
0,
otherwise
function for some continuous random variable X.
29. Let f (x) =
(a) Find conditions for a and b.
Answer.
Since f (x) is a PDF for some random variable, then f (x) ≥ 0 for all x. For 0 ≤ x ≤ 1,
since f (x) = ax2 + b is a non-negative quadratic polynomial and symmetric
with respect to x = 0, then
Z
∞
f (1) = a + b ≥ 0 and f (0) = b ≥ 0. On the other hand, we must have
f (x) dx = 1, then
∞
Z
∞
1 =
f (x) dx
−∞
Z 1
=
(ax2 + b) dx
ha
i1
x3 + bx 3
0
a
+ b.
3
0
=
=
So we get
a
+ b = 1, that is, a + 3b = 3. In summary, we must have the following conditions:
3
a + b ≥ 0,
b ≥ 0,
and a + 3b = 3.
(b) Compute a and b such that E(X) = 1/2.
Answer.
Since E(X) = 1, then
1
2
= E(X)
Z ∞
=
xf (x) dx
−∞
1
Z
x(ax2 + b) dx
=
0
Z
=
1
(ax3 + bx) dx
a 4 b 2 1
x + x 4
2
0
a b
+ .
4 2
0
=
=
So we get
1
a b
+ = , that is, a + 2b = 2. By the result of part a, we have a + 3b = 3, then
4 2
2
a = 0,
and b = 1.
Review Problems for Chapter 8 and Chapter 9
30. For a sequence
{an }∞
n=1
Z
1
such that an =
√
0
dx
.
n2 + x2
(a) Find the explicit formula of an .
√
n2 + 1 1 ln + n
n
(a)
Z
Answer.
For
√
dx
, let x = n tan(θ), then
+ x2
Z
Z
dx
√
=
n2 + x2
Z
=
n2
dx = n sec2 (θ) dθ and n2 + x2 = n2 sec2 (θ). So
n sec2 (θ)
dθ
n sec(θ)
sec(θ) dθ
= ln | sec(θ) + tan(θ)| + C.
Since tan(θ) =
x
, then
n
√
sec(θ) =
So we get
Z
n2 + x2
.
n
√
n2 + x2
dx
x
√
= ln + + C.
n
n
n2 + x2
Hence we have
Z
1
√
an =
0
=
=
=
=
dx
+ x2
n2
√
1
n2 + x2
x ln + n
n 0 √
√
n2
n2 + 1 1 0
+ − ln + ln n
n
n
n
√
n2 + 1 1 + − ln 1
ln n
n
√
n2 + 1 1 ln + .
n
n
(b) If {an }∞
n=1 has a limit, find this limit.
0
(b)
Answer.
In fact, since
"√
lim
n→∞
#
n2 + 1 1
+
= 1.
n
n
Then
lim an = ln 1 = 0.
n→∞
31. For a sequence {an }∞
k=1 such that a1 = 1 and an+1 = 4an for all n ≥ 1. Find the explicit formula of an .
4n−1
31.
Answer.
that
Since an+1 = 4an for all n ≥ 1, then {an } is a geometric sequence with ratio r = 4, which implies
an = a1 rn−1 = 1 · 4n−1 = 4n−1 .
32. (a) Compute lim
n→∞
√
cos( n) 3 tan−1 (n)
.
2· √
+
n3 + 3
n
0
(a)
√
π
π
Notice that −1 ≤ cos( n) ≤ 1 and − ≤ tan−1 (n) ≤ , then
2
2
√
cos( n)
π
π
tan−1 (n)
π
1
√
√
≤
≤ √ , and −
≤
≤
,
−
2(n3 + 3)
n3 + 3
2(n3 + 3)
n
n
n
Answer.
for all n ≥ 1.
π
π
= 0, by the squeeze theorem, we have
Since lim √ = lim
n→∞
n n→∞ 2(n3 + 3)
√
cos( n)
tan−1 (n)
√
= 0.
lim
= lim
n→∞
n→∞
n3 + 3
n
Hence we get
√
cos( n) 3 tan−1 (n)
lim 2 · √
+
= 0.
n→∞
n3 + 3
n
n3 + 2n
23n−1
−1
(b) Compute lim tan
+ n−10 .
n→∞
n2 + 2n + 1
9
π
2
(b)
n3 + 2n
Answer.
Since lim
= ∞, then lim tan−1
n→∞ n2 + 2n + 1
n→∞
geometric sequence with ratio r:
r =
=
=
=
Then
lim
n→∞
n3 + 2n
n2 + 2n + 1
23·2−1 91−10
·
92−10 23·1−1
25 1
·
22 9
23
9
8
< 1.
9
23n−1
= 0.
9n−10
Therefore, we get
lim
n→∞
−1
tan
n3 + 2n
n2 + 2n + 1
23n−1
π
+ n−10 = .
9
2
π
= . For
2
23n−1
, it’s a
9n−10
(c) Is the series
∞
X
k=100
√
k+1
convergent or divergent?
k
divergent
(c)
√
Answer.
Approach I: Let f (x) =
x+1
=
x
r
x+1
=
x2
r
1
1
+ , then f (x) is positive and decreasx x2
ing. Notice that
∞
Z
Z
f (x) dx =
1
1
Z √
For
∞
√
x+1
dx.
x
x+1
dx, then
x
Z
Z √
√
u
x+1
dx =
· 2u du let u = x + 1, then x = u2 − 1 and dx = 2udu
2
x
u −1
Z
u2
= 2
du
u2 − 1
Z 1
du
= 2
1+ 2
u −1
Z
1
= 2u + 2
du.
2
u −1
By the partial fraction decomposition, we have
1
1
1
=
−
.
u2 − 1
2(u − 1) 2(u + 2)
Then
Z √
Z 1
1
x+1
dx = 2u + 2 ·
−
du
x
2(u − 1) 2(u + 2)
Z 1
1
= 2u +
−
du
u−1 u+1
= 2u + ln |u − 1| − ln |u + 1| + C
√
√
√
= 2 x + 1 + ln | x + 1 − 1| − ln | x + 1 + 1| + C
√
x + 1 − 1
√
+ C.
= 2 x + 1 + ln √
x + 1 + 1
Notice that
√
lim 2 x + 1 = ∞,
x→∞
and
√
x + 1 − 1
= ln 1 = 0.
lim ln √
x→∞
x + 1 + 1
Then
√
x + 1 − 1 ∞
√
= ∞.
f (x) dx = 2 x + 1 + ln √
x
+
1
+
1
1
1
√
∞
X
k+1
So by the integral test, we know that the series
is divergent.
k
Z
∞
k=100
Approach II: Notice that
√
k+1
1
≥ ,
k
k
for all k ≥ 1.
∞
X1
X
Since
diverges, by the comparison test, then the series
k
k=100
√
k+1
is divergent.
k
(d) Is the series
∞
X
4
convergent or divergent?
2
k
ln
k
k=9
convergent
(d)
Answer.
Z
Let f (x) =
4
, then f (x) is positive and decreasing for x ≥ 9. Notice that
x ln2 x
Z ∞
Z ∞
4
f (x) dx =
dx
x ln2 x
9
9
Z ∞
1
= 4
dx.
x ln2 x
9
1
dx, then
x ln2 x
For
Z
1
dx =
x ln2 x
Z
du
1
Let u = ln x, then du = dx
u2
x
1
= − +C
u
1
= −
+ C Since u = ln x.
ln x
So we get
Z
∞
f (x) dx =
9
=
So by the integral test, we know that the series
33. Is the series
∞
X
k=1
We get
√
3 4
√ k +1
k5 +9
1
∞
X
1
7
6
k=1 k
∞ √
3
X
k4 + 1
√
k=1
4
is convergent.
2
k
ln
k
k=9
convergent
√
√
3
3
k4 + 1
k4
1
1
1
For the term √
, it behaves like √ = 5 4 = 7 as k → ∞, so let’s compare it with 7 .
k5 + 9
k5
k2−3
k6
k6
7
k6
Since
∞
X
√
3
k4 + 1
√
convergent or divergent?
k5 + 9
33.
Answer.
1 ∞
4 −
ln x 9
4
.
ln 9
k5 + 9
p
√
3
6
(k 4 + 1)2 · k 7
k4 + 1 7
· k6 = p
= √
→ 1,
6
k5 + 9
(k 5 + 9)3
is a p-series with p = 7/6, then
∞
X
1
7
k=1
as k → ∞.
converges. By the limit comparison test, then the series
k6
converges.
∞
X
(k!)2
34. Is the series
convergent or divergent?
(2k)!
k=1
34.
convergent
Answer.
Let ak =
(k!)2
, then
(2k)!
ak+1
ak
=
[(k+1)!]2
[2(k+1)]!
(k!)2
(2k)!
=
(k + 1)!
k!
2
·
(2k)!
(2k + 2)!
1
(2k + 2)(2k + 1)
1
k2
→ , as k → ∞.
4k 2 + 6k + 2
4
= k2 ·
=
By the ratio test, then the series
∞
X
k=1
35. Evaluate the series
∞ h
X
cos
π n
3
n=2
√
3
k4 + 1
√
is convergent.
k5 + 9
π n i
− 2 · − tan
.
6
35.
Answer.
Since cos
∞ h
X
π 3
cos
36. Is the series
π √3
1
= and tan
=
, then
2
6
3
π n
n=2
3
∞
X
k 10 10k (k!)2
k=1
(2k)!
√
3
1
− +
2
3
∞ ∞ π n
π n
π n i
X
X
cos
− tan
− 2 · − tan
=
−2
6
3
6
n=2
n=2
!
√ n
∞ n
∞
X
X
1
3
=
−2
−−
2
3
n=2
n=2
√ 2
1 2
− 33
2
√
=
−2·
1 − 12
1 + 33
1
1
√
=
−2·
2
3+ 3
√
1
3− 3
−2·
=
2
√6
1
3
=
−1+
2
√ 3
1
3
= − +
.
2
3
convergent or divergent?
36.
divergent
Answer.
Let ak =
k 10 10k (k!)2
, then
(2k)!
ak+1
ak
=
=
(k+1)10 10k+1 [(k+1)!]2
[2(k+1)]!
k10 10k (k!)2
(2k)!
(k + 1)10 10k+1 [(k +
[2(k + 1)]!
1)!]2
·
(2k)!
k 10 10k (k!)2
(k +
10k+1
(k + 1)! 2
(2k)!
=
·
·
·
10
k
k
k!
(2k + 2)!
10
10
1
1
· 10 · (k + 1)2 ·
=
1+
k
(2k + 2)(2k + 1)
10
1
(k + 1)2
= 10 1 +
·
k
(2k + 2)(2k + 1)
1
5
→ 10 · 1 · = , as k → ∞.
4
2
1)10
By the ratio test, we know that the series
∞
X
k 10 10k (k!)2
(2k)!
k=1
37. Consider the power series
∞
X
2k (x − 3)k
k+1
k=1
diverges.
.
(a) Find the radius of convergence of this series.
1
2
(a)
Answer.
Let ck =
2k
, then
k+1
ck+1
2k+1 k + 1
2k+1 k + 1
k+1
=
· k = k ·
=2·
→ 2,
ck
k+2
k+2
k+2
2
2
as k → ∞.
1
So the radius of convergence is R = .
2
(b) Find the interval of convergence of this series.
(5/2, 7/2)
(b)
1
and center is a = 3, then the interval of convergence
2
is (a − R, a + R) = (3 − 1/2, 3 + 1/2) = (5/2, 7/2).
Answer.
Since the radius of convergence is R =
38. Find the value of x such that
∞
X
ekx = 1.
k=1
− ln 2
38.
Answer.
∞
∞
X
X
x k
Since
(e ) =
ekx = 1, then |ex | < 1 and
k=1
k=1
∞
X
k=1
ekx =
ex
= 1.
1 − ex
1
So we get ex = 1 − ex , that is, 2ex = 1. So we get ex = , that is,
2
x = ln
39. Find the power series of f (x) =
1
= − ln 2.
2
4
centered at 0.
4 + x2
∞ X
1 k 2k
−
x
4
39.
k=0
Answer.
In fact, we have
f (x) =
=
4
4 + x2
4
1
·
4 1 + x2
4
=
1
1+
x2
4
1
2
1 − − x4
2
∞ X
x2
=
−
if
4
k=0
∞ X
1 k 2k
x .
−
=
4
=
2
x − < 1, that is, |x| < 2
4
k=0
40. Find the power series of f (x) =
1
centered at 1.
4 − 2x
∞
X
(x − 1)k
40.
k=0
Answer.
2
In fact, we have
f (x) =
=
=
=
=
1
4 − 2x
1
4 − 2(x − 1) − 2
1
2 − 2(x − 1)
1
1
·
2 1 − (x − 1)
∞
1X
(x − 1)k if |x − 1| < 1
2
k=0
=
∞
X
(x − 1)k
k=0
2
.
k
∞ 2
X
x −1
41. Find the function represented by the series
3
k=0
.
3
4 − x2
41.
Answer.
In fact, we have
k
∞ 2
X
x −1
k=0
3
=
=
=
2
x − 1
< 1, that is, |x2 − 1| < 3, then |x| < 2
if 3 1
1−
x2 −1
3
3
3 − x2 + 1
3
.
4 − x2
42. Find the function represented by the series
∞
X
x2k
k=1
4k
.
1
− ln(1 − x2 )
4
42.
Answer.
Approach I: Let f (x) =
∞
X
x2k
k=1
4k
, then
f 0 (x) =
=
d
dx
∞
X
x2k
k=1
∞
X x2k−1
k=1
x
2
!
4k
2
if |x2 | < 1
1 − x2
x
1
.
= − · 2
2 x −1
=
So we get
1
f (x) = −
2
Z
x2
x
1
dx = − ln |1 − x2 | + C.
−1
4
Taking x = 0 in the above identity, we get C = 0. So we get
∞
X
x2k
k=1
1
= − ln(1 − x2 ).
4k
4
Approach II: In fact, we have
∞
X
x2k
k=1
4k
∞
=
=
=
=
=
=
1 X yk
Let y = x2
4
k
k=1
∞ Z
X
1
y k−1 dy
4
k=1
!
Z X
∞
1
k−1
y
dy
4
Z k=1
1
1
dy
4
1−y
1
− ln |1 − y| + C if |y| < 1, that is, |x| < 1
4
1
− ln(1 − x2 ) + C.
4
Taking x = 0 in the above identity, we get C = 0. So we get
∞
X
x2k
k=1
43. Consider the power series
1
= − ln(1 − x2 ).
4k
4
∞
X
(−1)k 32k+1 x3k+1 .
k=0
(a) Find the radius of convergence of this series.
2
3− 3
(a)
Answer.
Notice that
∞
X
k 2k+1 3k+1
(−1) 3
x
k=0
∞
∞
X
X
k 2k+1 3k
=
(−1) 3
x ·x=x
(−1)k 32k+1 x3k .
k=0
Let y = x3 , then
k=0
∞
∞
X
X
(−1)k 32k+1 x3k+1 = x
(−1)k 32k+1 y k .
k=0
k=0
Let ck = (−1)k 32k+1 , then
32k+3
|ck+1 |
= 2k+1 = 9,
|ck |
3
Then the radius of convergence of the series
∞
X
(−1)k 32k+1 y k is
k=0
r
∞
X
convergence of the series
(−1)k 32k+1 x3k+1 is
k=0
for all k ≥ 0.
3
1
. Since y = x3 , then the radius of
9
2
1
= 3− 3 .
9
(b) Evaluate the sum of this series.
(b)
3x
1 + 9x3
Answer.
In fact, we have
∞
X
∞
X
=
(−1)k 32k · 3 · x3k · x
k 2k+1 3k+1
(−1) 3
x
k=0
k=0
= 3x
∞
X
(−1 · 32 · x3 )k
k=0
= 3x
∞
X
(−9x3 )k
k=0
3x
1 + 9x3
=
44. Is the series
∞
X
sin(k 2 + 1)
√
convergent or divergent?
k3 − 1
k=1
convergent
44.
Answer.
For
∞
X
k=1
Notice that
√
if | − 9x3 | < 1.
1
k3 − 1
sin(k 2 + 1) ≤ √ 1
√
.
k3 − 1 k3 − 1
, since
√ 1
k3 −1
1
3
k2
√
k3
=√
=
k3 − 1
r
k3
→ 1,
−1
k3
as k → ∞.
∞
X
1
∞
X
3
1
Since
By the limit comparison test, then
> 1, then
3 is a p-series with p =
3 converges.
2
2
2
k=1 k
k=1 k
∞
∞ ∞
X
X
X
sin(k 2 + 1) 1
sin(k 2 + 1)
converges. So
√
√
√
converges. By the comparison test, then
converges.
k3 − 1
k3 − 1 k3 − 1
k=1
k=1
k=1
45. Find the Taylor series of e2x centered at 1.
∞ 2 k
X
e 2
45.
k=0
Answer.
In fact, we have
e2x = e2(x−1)+2)
= e2 · e2(x−1)
∞
X
[2(x − 1)]k
= e2
k!
k=0
= e2
=
∞
X
2k
k!
k=0
∞
X e2 2k
k=0
k!
(x − 1)k
(x − 1)k .
k!
(x − 1)k
46. Find the Maclaurin series of log3 (x + 1).
∞
X
(−1)k
46.
k=1
Answer.
Notice that
Since ln(1 − x) = −
k=1
k
xk
ln(x + 1)
.
ln 3
log3 (x + 1) =
∞
X
xk
k ln 3
, then
ln(x + 1) = −
∞
X
(−x)k
k
k=1
So we get
.
∞
∞
k=1
k=1
X (−1)k
1 X (−x)k
log3 (x + 1) = −
=
xk .
ln 3
k
k ln 3
47. Identify the functions represented by the power series
∞
X
xk
k=1
k
.
− ln |1 − x|
47.
Answer.
Approach I: In fact, we have
∞
X
xk
k=1
k
∞ Z
X
=
x
k−1
Z
dx =
k=1
∞
X
!
x
k−1
Z
dx =
k=1
∞
X
!
x
k=0
Z
1
dx if |x| < 1
1−x
= − ln |1 − x| + C.
=
Plug x = 0 into the above identities, then C = 0. So we get
∞
X
xk
k=1
Approach II: Let f (x) =
∞
X
xk
k=1
k
= − ln |1 − x|.
, then
d
f (x) =
dx
0
k
∞
X
xk
k=1
So we have
Z
f (x) =
!
k
∞
X
k=1
xk−1 =
1
,
1−x
if |x| < 1.
1
dx = − ln |1 − x| + C.
1−x
Since f (0) = 0, then C = 0. Hence we get
∞
X
xk
k=1
k
= − ln |1 − x|.
k
dx
48. Identify the functions represented by the power series
∞
X
k(k − 1)xk
3k
k=2
6x2
(3 − x)3
48.
Answer.
.
In fact, we have
∞
X
k(k − 1)xk
k=2
3k
=
=
∞
X
k=2
∞
X
x k
k(k − 1)
3
k(k − 1)y k
Let y =
k=2
=
∞
X
x
3
y 2 · k(k − 1)y k−2
k=2
= y2
∞
X
d2 k
(y )
dy 2
k=2
2
∞
X
2
k=2
2
d2
y
d2
= y
dy 2
!
y
k
if |y| < 1, that is, |x| < 3
dy 2 1 − y
2y(1 − y) + y 2
2 d
= y
dy
(1 − y)2
= y
d
2y − y 2
dy (1 − y)2
(2 − 2y)(1 − y)2 + (2y − y 2 ) · 2(1 − y)
y2 ·
(1 − y)4
(2 − 2y)(1 − y) + 2(2y − y 2 )
y2 ·
(1 − y)3
2 − 2y − 2y + 2y 2 + 4y − 2y 2
y2 ·
(1 − y)3
2
y2 ·
(1 − y)3
2y 2
(1 − y)3
2
2 · x3
x
Since y =
x 3
3
1−
= y2
=
=
=
=
=
=
3
=
6x2
.
(3 − x)3
49. Identify the functions represented by the power series
∞
X
k=2
49.
xk
.
k(k − 1)
(1 − x) ln(1 − x) + x
Answer.
Let f (x) =
∞
X
k=2
xk
, then
k(k − 1)
∞
X
d
dx
f 0 (x) =
k=2
!
xk
k(k − 1)
∞
X
xk−1
=
k−1
k=2
∞
X
xk−1
k−1
d
dx
00
f (x) =
k=2
∞
X
=
!
xk−2
k=2
1
1−x
=
Then
Z
0
f (x) =
if |x| < 1.
1
dx = − ln |1 − x| + C.
1−x
Since f 0 (0) = 0, then C = 0. So we get
f 0 (x) = − ln |1 − x|.
Then
Z
f (x) = − ln(1 − x) dx
Z
=
ln u du Let u = 1 − x
Z
1
= u ln u − u · du Use the integration by parts
u
= u ln u − u + C
= (1 − x) ln(1 − x) − (1 − x) + C
Since u = 1 − x
= (1 − x) ln(1 − x) + x + C.
Since f (0) = 0, then C = 0. So we get
∞
X
k=2
xk
= (1 − x) ln(1 − x) + x.
k(k − 1)
50. Consider differential equation y 0 (x) = 2xy with initial condition y(0) = 1, find the power series solution to the
this differential equation.
∞
X
x2k
50.
k=0
Answer.
Let y(x) =
∞
X
k!
ck xk , since y(0) = 1, then c0 = 1. Since y 0 (x) = 2xy, then
k=0
0
y (x) =
d
dx
= 2x ·
∞
X
!
k
ck x
k=0
∞
X
k=0
ck xk =
=
∞
X
kck xk−1 =
k=0
∞
X
k=0
(2ck )xk+1
∞
X
(k + 1)ck+1 xk
k=0
∞
X
=
(2ck−1 )xk .
k=1
So we have c1 = 0 and (k + 1)ck+1 = 2ck−1 for all k ≥ 1, that is,
ck =
2
ck−2 ,
k
for all k ≥ 2,
and
c0 = 1,
c1 = 0.
So we know that ck = 0 if k is odd. If k = 2m is even, then
2
2
2
2
2 2
1
2
c2m−2 =
·
c2m−4 =
·
· · · · c0 =
.
c2m =
2m
2m 2m − 2
2m 2m − 2
4 2
m!
So we have
y(x) =
∞
X
m=0
m=0
51. (a) Use the integral test to show that the series
∞
X
k=2
Answer.
Let f (x) =
∞
X
x2m
2
= ex .
m!
c2m x2m =
1
is convergent.
k 2 (k + 1)
1
, then f (x) is positive and decreasing. By the partial fractional decomx2 (x + 1)
position, we have
A
B
C
1
= + 2+
.
x2 (x + 1)
x
x
x+1
Both sides of the above identity multiplied by x2 (x + 1), then
1 = Ax(x + 1) + B(x + 1) + Cx2
= Ax2 + Ax + Bx + B + Cx2
= (A + C)x2 + (A + B)x + B.
Then we get
A + C = 0,
A + B = 0,
and B = 1.
So we get
A = −1,
B = 1,
and C = 1.
So we get
1
1
1
1
=− + 2 +
.
+ 1)
x x
x+1
x2 (x
Then
Z
Z
Z
Z
1
1
1
f (x) dx = −
dx +
|dx +
dx
2
x
x
x+1
1
= − ln |x| − + ln |x + 1| + C
x
x + 1
1
+ C.
= − + ln x
x Hence we have
Z
∞
Z
f (x) dx =
1
b
lim
f (x) dx
1
1 b
= lim − + ln 1 + b→∞
x
x 1
1
1
= lim
+ ln 1 + + 1 − ln 2
b→∞
b
b
= 0 + ln 1 + 1 − ln 2
b→∞
1
= 1 − ln 2.
So by the integral test, we have the series
∞
X
k=2
1
is convergent.
+ 1)
k 2 (k
(b) Use the comparison test to show that the series
∞
X
k=2
Answer.
Notice that k 2 (k + 1) = k 3 + k 2 ≥ k 3 , then
0<
Since
∞
X
k=1
1
is convergent.
k 2 (k + 1)
1
1
≤ 3,
k 2 (k + 1)
k
for all k ≥ 1.
∞
∞
X
X
1
1
is
a
p-series
with
p
=
3,
then
converges. By the comparison test, then the series
3
k
k3
k=1
k=1
1
converges.
k 2 (k + 1)
(c) Evaluate the series
∞
X
k=2
1
.
+ 1)
k 2 (k
∞
X
π2
1
=
Hint:
and the result of the partial fraction decomposition in Problem (a).
k2
6
k=1
Answer.
By the computation in part (a), we have
1
1
1
1
=− + 2 +
.
+ 1)
k k
k+1
k 2 (k
Notice that
∞
X
1
π2
=
,
k2
6
and
k=1
∞ X
1
k=1
1
−
= 1.
k k+1
Then we have
∞
X
k=2
1
2
k (k + 1)
∞ X
1
1
1
=
+
−
k2 k + 1 k
k=2
∞
∞ X
X
1
1
1
=
−
−
k2
k k+1
=
=
52. Assume that the series
∞
X
n=1
∞
X
nan − 2n + 1
n=1
ln
an
an+1
n+1
k=2
π2
k=2
−1−
6
π2 3
− .
6
2
1
2
converges, where an > 0 for all n ≥ 1. Is the series − ln a1 +
convergent or divergent. If it’s convergent, evaluate the sum of the series − ln a1 +
∞
X
n=1
ln
an
.
an+1
Answer.
For any m ≥ 1, let Sm = − ln a1 +
m
X
an
, then
an+1
ln
n=1
Sm = − ln a1 +
= − ln a1 +
m
X
n=1
m
X
ln
an
an+1
[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→∞
we must have lim
n→∞
lim
Since lim
n→∞
n+1
n=1
nan − 2n + 1
= 0. Hence we get
n+1
n→∞
∞
X
nan − 2n + 1
converges, by the divergence test,
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→∞
In summary, we know that the series − ln a1 +
m→∞
∞
X
ln
n=1
− ln a1 +
∞
X
n=1
an
an+1
ln
an
an+1
is convergent and
= − ln 2.