Math 1310
Review session
1. (Optimization and Derivatives)
(a) You are to build a cone (with no base) of a fixed volume V with the minimum
amount of material. What is the ratio of height to radius h/r of your cone? Remember the following formulas:
• the volume of a cone is 31 πr2 h;
√
• the surface area of a cone without base is πr r2 + h2 .
Solution: Since V is fixed, we can see V = 13 πr2 h as a relation between h and
r. Hence we can express h in terms of r as follows
r2 =
3V
.
πh
r
r
Thus we get that the surface is
S(h) = π
3V
πh
3V
+ h2 .
πh
To find the value of h giving the minimum surface, we can minimize S 2 (h) as
well. Thus we get to study
3V
3V
2
2 3V
2
S (h) = π
+ h = 3V π
+h .
πh πh
πh2
Thuse we have
6V
(S (h)) = 3V π − 3 + 1 .
πh
If we set it equal to zero, we get the equation
0
2
6V
= 1,
πh3
which leads to
r
h=
Now, r =
q
3V
,
πh
3
6V
.
π
thus we get
h
=
r
r
3
6V
π
r
πh
=
3V
r
3
v q
u
u 3 6V
6V t π π
.
π
3V
2. (Integrals)
Comupute the following integrals.
R∞
3
(a) 1 3x2 e−3x dx
Solution: We notice that the polynomial 3x2 is the derivative of x3 . Hence if
we set u = x3 , we get du = 3x2 dx. By definition of improper integral we have
Z t
Z ∞
3
2 −3x3
3x2 e−3x dx
(1)
3x e
dx = lim
t→∞
1
1
Rt
3
So we first compute the proper integral 1 3x2 e−3x dx and use substitution. We
see that for x = 1, then u = 1, while for x = t we get u = t3 . Hence we get
t
Z
2 −3x3
3x e
Z
dx =
1
−3u
e−3u du
(2)
1
is an antiderivative of e−3u , so we get
Now, we know that − e 3
t3
Z
t3
3
−e
−3u
1
1
e−3t
e−3u t3
]1 = 3 −
du = [−
3
3e
3
(3)
Hence we get
Z
1
(b)
R∞
0
∞
3
−3x3
3x2 e
dx = lim
t→∞
e−3t
1
−
3e3
3
!
=
1
3e3
(4)
xe−3x dx
Rt
Solution: We first use integration by parts in 0 xe−3x dx and then take the
limit as t goes to ∞ to calculate the improper integral. We see that it is easy
Page 2
to take an antiderivative of e−3x and if we differentiate x we are left with 1.
−3x
Hence in integration by parts we integrate e−3x (the antiderivative is − e 3 )
and differentiate x. I.e, in the u and v notation (see page 383 of the book)
u = x and dv = e−3x :
t Z t
Z t
e−3x
xe−3x
−3x
−
−
dx
(5)
xe dx = −
3
3
0
0
0
Now we study separately the second summand, i.e.
Z
−
0
t
e−3x
1
−
dx =
3
3
Z
t
−3x
e
0
−3x t
1
e
dx =
−
3
3 0
(6)
Hence we get
t
Z
xe−3x dx = −
0
te−3t e−3t 1
−
+
3
9
9
Hence to compute the improper integral we are left with two limits.
We have
e−3t
lim
=0
t→∞ 9
and
t H
1
lim te−3t = 3t = lim 3t = 0
t→∞
t→∞ 3e
e
(c) Find an antiderivative for
(7)
(8)
(9)
√ 1
2 x+3+x
Solution: The problem is well stated just for x > −3, since we are taking the
square root of x + 3. So from now on we will assume x > −3. Now, we want to
calculate the integral
Z
1
√
dx
(10)
2 x+3+x
√
First we do substitution to get rid of the square root. If we set y = x + 3,
then y 2 = x + 3 and hence 2ydy = dx If we do this substitution in the integral,
we get
Z
Z
1
2y
√
dx =
dy
(11)
2y + y 2 − 3
2 x+3+x
We notice that we can factor y 2 + 2y − 3 = (y − 1)(y + 3). Now we want to write
A
2y
B
A(y + 3) + B(y − 1)
=
+
=
2
2y + y − 3
y−1 y+3
(y + 3)(y − 1)
Page 3
(12)
This leads to the system
(
Ay + By = 2
3A − B = 0
(13)
From the second equation we get B = 3A; sustituting in the first one we have
3A + A = 2. Hence we have A = 21 and B = 23 . This tells us that it’s enough to
compute the integrals
Z
Z
1
3
dy,
dy
(14)
2(y − 1)
2(y + 3)
We know that an antiderivative of x1 is ln |x|, and using this idea we have
Z
Z
1
1
1
1
dy =
dy = ln |y − 1|
(15)
2(y − 1)
2
y−1
2
and
Z
3
3
dy =
2(y + 3)
2
Z
3
1
dy = ln |y + 3|
(16)
y+3
2
√
Now in such functions we have to substitute back y = x + 3 and hence we get
that a possible antiderivative of the function we started with is
√
√
3
1
ln | x + 3 − 1| + ln | x + 3 + 3|
2
2
Page 4
(17)
3. (Limits) Compute the following limits
√
√
(a) limx→∞ 3x2 + 8x + 6 − 3x2 + 3x + 1
Solution: We first manipulate the expression
√
√
3x2 + 8x + 6 − 3x2 + 3x + 1
√
√
√
√
3x2 + 8x + 6 + 3x2 + 3x + 1
2
2
√
= ( 3x + 8x + 6 − 3x + 3x + 1) √
3x2 + 8x + 6 + 3x2 + 3x + 1
3x2 + 8x + 6 − 3x2 − 3x − 1
√
=√
3x2 + 8x + 6 + 3x2 + 3x + 1
5x + 5
√
=√
2
3x + 8x + 6 + 3x2 + 3x + 1
5x + 5
√ =√ 6
1
8
3
√
√
√
√
3x 1 + 3x + 3x2 + 3x 1 + 3x + 3x2
=√ 3 1+
√8
3x
5 + x5
√
6
+
+ √3x
3
1+
2
√3
3x
+
√1
3x2
(18)
Now we see
5
=0
x→∞ x
3
1
lim 1 + √ + √
=1
x→∞
3x
3x2
8
6
lim 1 + √ + √
=1
x→∞
3x
3x2
(19)
√
√
5
3x2 + 8x + 6 − 3x2 + 3x + 1 = √
x→∞
2 3
(20)
lim
and thus we get
lim
(b) limx→∞ x sin
π
x
Solution: We have
lim x sin
x→∞
π x
=
= lim
x→∞
Page 5
π
x
sin
1
x
(21)
H
= lim
π
x
cos
− xπ2
− x12
π = lim πcos
=π
x→∞
x
x→∞
(22)
(c) limx→0+ sin x ln x
Solution: We know that if limx→a f (x) = c and limx→a g(x) = d, then we have
limx→a f (x)g(x) = cd (see page 104 for a review of such properties). Then we
can write sin x ln x = sinx x (x ln x). We know that limx→0+ sinx x = 1 (if you do
not remember it, check it with l’Hospital’s Rule). Then it is enough to study
limx→0+ x ln x. To this aim, we use l’Hospital’s Rule
lim+ x ln x = lim+
ln x
1
x
1
H
= lim x1
x→∞ − 2
x
x→0
x→0
= lim+ −x = 0
x→0
Hence limx→0+ sin x ln x = limx→0+
sin x
x
Page 6
(limx→0+ x ln x) = 1 · 0 = 0.
(23)