2
=
cos2(z)dz
0
π
1
2
=
(1 + cos(2z))dz
2 0
π
π
1
2
1
2
=
z
+
sin(2z)
2
4
0
0
π
=
,
4
as we expected.
Example 2.6.18. Let C be the circle with equation x2 + y2 = 1 and let L
be the length of the shorter arc of C between
− 1
√ , 1
√
and
1
√ , 1
√
(see
2
2
2
2
Figure 2.6.2). Since the circumference of C is 2π and this arc is one-fourth of the circumference of C, we should have L = π . We will now show that this
2
agrees with (2.5.28), the formula we derived for computing arc length. Now
√
y =
1 − x2, so
dy
1
x
=
(1 − x2)− 12 (−2x) = − √
.
dx
2
1 − x2
106
CHAPTER 2. INTEGRALS
Hence
2
dy
x2
1 − x2 + x2
1
1 +
=
1 +
=
= √
.
dx
1 − x2
1 − x2
1 − x2
Hence, by (2.5.28),
1
√2
1
L =
√
dx.
− 1
√
1 − x2
2
If we let
x = sin(z)
dx = cos(z)dz,
then
π
4
1
L =
cos(z)dz
− π
4
1 − sin2(z)
π
4
cos(z)
=
dz
− π
cos2(z)
4
π
4
cos(z)
=
dz
− π cos(z)
4
π
4
=
dz
− π
4
π
=
.
2
Exercise 2.6.26.
Use the change of variable x = 2 sin(z) to evaluate
2
4 − x2dx.
−2
1
2
Exercise 2.6.27.
Evaluate
√
dx.
−2
16 − x2
Example 2.6.19. In Example 2.5.8 we saw that the arc length L of the parabola y = x2 over the interval [0, 1] is
1
L =
1 + 4x2dx.
0
However, at that point we did not have the means to evaluate this integral. We
now have most, although not all, of the necessary tools. To begin, we will first
2.6. SOME TECHNIQUES FOR EVALUATING INTEGRALS
107
make the change of variable
u = 2x
du = 2dx,
which gives us
1
2
L =
1 + u2du.
2 0
Next, we recall the trigonometric identity
1 + tan2(t) = sec2(t)
(2.6.26)
(a consequence of dividing each term of the identity cos2(t) + sin2(t) = 1 by
cos2(t)), which is a hint that the change of variable
x = tan(z)
dx = sec2(z)
might be of use. If we let α be the angle for which tan(α) = 2, with 0 < α < π ,
2
and note that tan(0) = 0 and
1 + tan2(z) =
sec2(z) = | sec(z)| = sec(z)
(note that sec(z) > 0 since 0 ≤ z ≤ π ), then
2
1
α
1
α
L =
sec(z) sec2(z)dz =
sec3(z)dz.
2 0
2 0
We may reduce the integral on the right using an integration by parts: Letting
u = sec(z)dz
dv = sec2(z)dx
du = sec(z) tan(z)dz
v = tan(z),
we have
α
α
sec3(z)dz = sec(z) tan(z)|α −
sec(z) tan2(z)dz
0
0
0
α
= sec(α) tan(α) −
sec(z)(sec2(z) − 1)dz
0
√
α
α
= 2 5 −
sec3(z)dz +
sec(z)dz,
0
0
where we have used the fact that tan(α) = 2 and 1 + tan2(α) = sec2(z) to find
√
that sec(α) =
5. It now follows that
α
√
α
2
sec3(z)dz = 2 5 +
sec(z)dz,
0
0
108
CHAPTER 2. INTEGRALS
and so
α
√
1
α
sec3(z)dz =
5 +
sec(z)dz.
0
2 0
Hence
√5 1 α
L =
+
sec(z)dz.
2
4 0
For this reduced integral, we notice that
α
α
sec(z) + tan(z)
α sec2(z) + sec(z) tan(z)
sec(z)dz =
sec(z)
dz =
dz,
0
0
sec(z) + tan(z)
0
sec(z) + tan(z)
and so the change of variable
w = sec(z) + tan(z)
dw = (sec(z) tan(z) + sec2(z))dz
gives us
√
α
2+
5 1
sec(z)dz =
dw.
0
1
w
Thus we now have
√
√
5
1
2+
5 1
L =
+
dw.
2
4 1
w
Although greatly simplified from the integral with which we started, neverthe-
less we cannot evaluate the remaining integral with our current tools. Indeed,
we may use the fundamental theorem of calculus to evaluate, for any rational
number n, any definite integral involving wn, except in the very case we are
facing now, that is, when n = −1. We will fill in this gap in the next section,
and finish this example at that time (see Example 2.7.9).
Example 2.6.20. For a simpler example of the change of variable used in the
previous example, consider the integral
1
1
dx,
−1 1 + x2
the area under the curve
1
y = 1 + x2
over the interval [−1, 1] (see Figure 2.6.3). If we let
x = tan(z)
dx = sec2(z)dz,
2.6. SOME TECHNIQUES FOR EVALUATING INTEGRALS
109
y
1
1
y = 1 + x2
0.75
0.5
0.25
0
1
0
0
0
−
− .75
− .5
− .25
0
0.25
0.5
0.75
1
x
Figure 2.6.3: Region beneath y =
1
over the interval [−1, 1]
1+x2
and note that tan − π = −1 and tan π = 1, then
4
4
1
π
1
4
1
dx =
sec2(z)dz
−1 1 + x2
− π 1 + tan2(z)
4
π
4
sec2(z)
=
dz
− π sec2(z)
4
π
4
=
dz
− π
4
π
=
.
2
You should compare this with the simple approximation we saw in Example
3
6
Exercise 2.6.28.
Evaluate
dx.
−3 9 + x2
1
2
1
Exercise 2.6.29.
Evaluate
dx.
− 1 1 + 4x2
2
Exercise 2.6.30.
Show that for any positive integer n > 2,
1
n − 2
secn(x)dx =
secn−2(x) tan(x) +
secn−2(x)dx.
n − 1
n − 1
110
CHAPTER 2. INTEGRALS
2.7
The exponential and logarithm functions
There are many applications in which it is necessary to find a function y of a
variable t which has the property that
dy = ky
(2.7.1)
dt
for some constant real number k. Examples include modeling the growth of
certain animal populations, where y is the size of the population at time t and
k > 0 depends on the rate at which the population is growing, and describing
the decay of a radioactive substance, where y is the amount of a radioactive
material present at time t and k < 0 depends on the rate at which the element
decays. We will first consider that case k = 1; that is, we will look for a function
y = f (t) with the property that f (t) = f (t).
2.7.1
The exponential function
Suppose f is a differential function on (−∞, ∞) with the property that f (t) =
f (t) for all t (one may show that such a function does indeed exist, although we
will not go into the details here). Now f (t) = f (t) implies, by the fundamental
theorem of calculus, that
t
t
f (x)dx =
f (x)dx = f (t) − f (0)
(2.7.2)
0
0
for all t. The value of f (0 is arbitrary; we will find it convenient to take f (0) = 1.
That is, we are now looking for a function f which satisfies
t
f (t) = 1 +
f (x)dx.
(2.7.3)
0
Suppose we divide [0, t] into N subintervals of equal length ∆x = t , where N is
N
a positive integer, and let x0, x1, x2, . . . , xN be the endpoints of these intervals.
Now for any i = 1, 2, . . . , N , using (2.7.3),
xi
f (xi) = 1 +
f (x)dx
0
xi−1
xi
= 1 +
f (x)dx +
f (x)dx
0
x1−1
xi
= f (xi−1) +
f (x)dx.
(2.7.4)
x1−1
Moreover, for small ∆x,
xi
f (x)dx ≈ f (xi−1)∆x,
(2.7.5)
x1−1
2.7. THE EXPONENTIAL AND LOGARITHM FUNCTIONS
111
and so we have
f (xi) ≈ f (xi−1) + f (xi−1)∆x = f (xi−1)(1 + ∆x).
(2.7.6)
Hence we have
f (x0) = f (0) = 1,
(2.7.7)
f (x1) ≈ f (x0)(1 + ∆x) = 1 + ∆x,
(2.7.8)
f (x2) ≈ f (x1)(1 + ∆x) ≈ (1 + ∆x)2,
(2.7.9)
f (x3) ≈ f (x2)(1 + ∆x) ≈ (1 + ∆x)3,
(2.7.10)
f (x4) ≈ f (x3)(1 + ∆x) ≈ (1 + ∆x)4,
(2.7.11)
..
.
.
.
..
..
(2.7.12)
f (xN ) ≈ f (xN−1)(1 + ∆x) ≈ (1 + ∆x)N
(2.7.13)
Now xN = t and ∆x = t , so we have
N
N
t
f (t) ≈
1 +
.
(2.7.14)
N
Moreover, if we followed the same procedure with N infinite and dx = t , we
N
should expect (although we have not proved)
N
t
f (t)
1 +
.
(2.7.15)
N
We will let e = f (1). That is,
N
1
e = sh
1 +
,
(2.7.16)
N
where N is any positive infinite integer. We call e Euler’s number. Now if t is
any real number, then
N t
N
t
t
1
f (t) =
1 +
= 1 +
= et,
(2.7.17)
N
N
t
where we have used the fact that N is infinite since N is infinite and t is finite.
t
(Note, however, that N is not an integer, as required in (2.7.16). The statement t
is nevertheless true, but this is a detail which we will not pursue here.) Thus
the function
f (t) = et
(2.7.18)
has the property that f (t) = f (t), that is,
d et = et.
(2.7.19)
dt
112
CHAPTER 2. INTEGRALS
In fact, one may show that f (t) = et is the only function for which f (0) = 1
and f (t) = f (t). We call this function the exponential function, and sometimes
write exp(t) for et.
It has been shown that e is an irrational number. Although, like π, we
may not express e exactly in decimal notation, we may use (2.7.16) to find approximations, replacing the infinite N with a large finite value for N . For
example, with N = 200, 000, we find that
200000
1
e ≈
1 +
≈ 2.71828,
(2.7.20)
200000
which is correct to 5 decimal places.
Example 2.7.1. If f (t) = e5t, then f is the composition of h(t) = 5t and
g(u) = eu. Hence, using the chain rule,
f (t) = g (h(t))h (t) = e5t · 5 = 5e5t.
In general, if h(t) is differentiable, then, by the chain rule,
d eh(t) = h (t)eh(t).
(2.7.21)
dt
Example 2.7.2. If f (x) = 6e−x2 , then
f (x) = −12xe−x2 .
Exercise 2.7.1.
Find the derivative of g(x) = 12e−7x.
Exercise 2.7.2.
Find the derivative of f (t) = 3t2e−t.
Example 2.7.3. Let f (t) = et and g(t) = e−t. Since et > 0 for all t, we
have f (t) = et > 0 and f (t) = et > 0 for all t, and so f is increasing on
(−∞, ∞) and the graph of f is concave upward on (−∞, ∞). On the other
hand, g (t) = −e−t < 0 and g (t) = e−t > 0 for all t, so g is decreasing on
(−∞, ∞) and the graph of g is concave upward on (−∞, ∞). See Figure 2.7.1.
Of course, it follows from (2.7.19) that
etdt = et + c.
(2.7.22)
Example 2.7.4. From what we have seen with the examples of derivatives
above, we have
1
e−tdt = −e−t 1 = −e−1 + e0 = 1 − e−1 ≈ 0.6321.
0
0
2.7. THE EXPONENTIAL AND LOGARITHM FUNCTIONS
113
y 15
10
t
5
y = et
y = e−
0
3
2
1
0
1
2
3
−
−
−
t
Figure 2.7.1: Graphs of y = et and y = e−t
Example 2.7.5. To evaluate
xe−x2 dx,
we will use the change of variable
u = −x2
du = −2xdx.
Then
1
1
1
xe−2 dx = −
eudu = − eu + c = − e−x2 + c.
2
2
2
Example 2.7.6. To evaluate
xe−2xdx,
we will use integration by parts:
u = x
dv = e−2xdx
1
du = dx
v = − e−2x.
2
Then
1
1
1
1
xe−2xdx = − xe−2x +
e−2xdx = − xe−2x −
e−2x + c.
2
2
2
4
114
CHAPTER 2. INTEGRALS
4
Exercise 2.7.3.
Evaluate
5e−2xdx.
0
1
Exercise 2.7.4.
Evaluate
x2e−x3 dx.
0
Exercise 2.7.5.
Evaluate
x2e−xdx.
Note that if y = aekt, where a and k are any real constants, then
dy = kaekt = ky.
(2.7.23)
dt
That is, y satisfies the differential equation (2.7.1) with which we began this section. We will consider example applications of this equation after a discussion
of the logarithm function, the inverse of the exponential function.
2.7.2
The logarithm function
The logarithm function is the inverse of the exponential function. That is, for
a positive real number x, y = log(x), read y is the logarithm of x, if and only if
ey = x. In particular, note that for any positive real number x,
elog(x) = x,
(2.7.24)
and for any real number x,
log (ex) = x.
(2.7.25)
Also, since e0 = 1, it follows that log(1) = 0.
Since log(x) is the power to which one must raise e in order to obtain x,
logarithms inherit their basic properties from the properties of exponents. For
example, for any positive real numbers x and y,
log(xy) = log(x) + log(y)
(2.7.26)
since
elog(x)+log(y) = elog(x)elog(y) = xy.
(2.7.27)
Similarly, for any positive real number x and any real number a,
log (xa) = a log(x)
(2.7.28)
since
a
ea log(x) = elog(x)
= xa.
(2.7.29)
2.7. THE EXPONENTIAL AND LOGARITHM FUNCTIONS
115
Exercise 2.7.6.
Verify that for any positive real numbers x and y,
x
log
= log(x) − log(y).
(2.7.30)
y
Note in particular that this implies that
1
log
= − log(y).
(2.7.31)
y
To find the derivative of the logarithm function, we first note that if y =
log(x), then ey = x, and so
d
d
ey =
x.
(2.7.32)
dx
dx
Applying the chain rule, it follows that
dy
ey
= 1.
(2.7.33)
dx
Hence
dy
1
1
=
=
.
(2.7.34)
dx
ey
x
Theorem 2.7.1. For any real number x > 0,
d
1
log(x) =
.
(2.7.35)
dx
x
Example 2.7.7. Since, for all x > 0
d
1
log(x) =
> 0
dx
x
and
d2
1
log(x) = −
< 0,
dx2
x2
the function y = log(x) is increasing on (0, ∞) and its graph is concave down-
ward on (0, ∞). See Figure 2.7.2.
Example 2.7.8. If f (x) = log(x2 + 1), then, using the chain rule,
1
d
2x
f (x) =
(x2 + 1) =
.
x2 + 1 dx
x2 + 1
Exercise 2.7.7.
Find the derivative of f (x) = log(3x + 4).
Exercise 2.7.8.
Find the derivative of y = (x + 1) log(x + 1).
116
CHAPTER 2. INTEGRALS
y
y = log(x)
2
1
0
10
x
−1
−2
Figure 2.7.2: Graph of y = log(x)
Using the fundamental theorem of calculus, it now follows that, for any
x > 0,
x 1 dt = log(t)|x = log(x) − log(1) = log(x).
(2.7.36)
1
1
t
This provides a geometric interpretation of log(x) as the area under the graph
of y = 1 from 1 to x. For example, log(10) is the area under the graph of y = 1
t
t
from 1 to 10 (see Figure 2.7.3).
Example 2.7.9. We may now complete the example, discussed in Example
2.5.8 and continued in Example 2.6.19, of finding the length L of the graph of y = x2 over the interval [0, 1]. In those examples we found that
√
√
1
5
1
2+
5 1
L =
1 + 4x2dx =
+
dw.
0
2
4 1
w
Now we see that
√
2+
5 1
√
√
dw = log(w)|2+ 5 = log(2 +
5),
1
1
w
and so
√5 1
√
L =
+
log(2 +
5).
2
4
Rounding to four decimal places, this gives us L ≈ 1.4789, the same approxi-