Calculus I (part 4): Introduction to Integration
(by Evan Dummit, 2012, v. 1.35)
Contents
4
Introduction to Integration
1
4.1
Denite Integrals and Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
4.2
The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4.3
Indenite Integrals, Evaluating Denite Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
4.4
Dierentiating Integrals
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
4.5
Areas
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
4.6
Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
4.7
Arclength, Surface Area, Volume, Moments
4
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
4.7.1
Arclength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
4.7.2
Surface Area, Surfaces of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
4.7.3
Volume, Solids of Revolution
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
4.7.4
Center of Mass, Moments
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
Introduction to Integration
We introduce the problem of nding the area under a curve, dene denite integrals using Riemann sums, prove
the Fundamental Theorem of Calculus and introduce indenite integrals of basic functions, and discuss substitution
techniques for integrating other functions. We also cover some basic applications of integration to computing areas,
arclengths, volumes, and moments and masses.
4.1 Denite Integrals and Riemann Sums
•
The fundamental problem motivating our development of integration is: given a continuous function
an interval
◦
what is the area under the curve
f (x)
Since
f (x) = x
f (x) = x
between
x=0
2.0
1.5
1.0
0.5
1.0
and
and below the graph of
y = f (x),
assuming
x = 2.
1
·2·2= 2
2
Here is a graph of the area in question:
0.5
x-axis
is linear and passes through the origin, the area forms a right triangle, with base and
height both equal to 2. The area is therefore
◦
on
is positive.
Example: Find the area under
◦
f (x)
y = f (x)?
By area under the curve we mean area above the
that
•
[a, b],
1.5
2.0
1
.
◦
More generally, the area underneath
a,
height both equal to
•
and hence is
f (x) = x
1 2
a .
2
x=0
between
and
x=a
is a right triangle with base and
To evaluate areas more complicated than those which have formulas from basic geometry, we need a more
general method.
•
[a, b] into many small pieces,
x-axis with one vertex on the graph of y = f (x).
The idea is the following: divide the interval
and then in each interval draw a
rectangle with base on the
Then as the number of rectangles
grows, the collective area of the rectangles will ll the area under the graph.
◦
f (x) = x
Here are a pair of pictures illustrating this idea, for
on
[0, 2],
dividing the interval into 4 and
20 equally-sized subintervals respectively:
2.0
2.0
1.5
1.5
1.0
1.0
0.5
0.5
0.5
•
1.0
1.5
2.0
0.5
1.0
1.5
2.0
[a, b], a partition of [a, b] into n subintervals is a list of x-coordinates x0 , x1 , x2 , · · · , xn−1 , xn
x0 = a, xn = b, and x0 < x1 < x2 < · · · < xn−1 < xn dividing [a, b] into the subintervals [x0 , x1 ], [x1 , x2 ],
. . . , [xn−1 , xn ]. The norm of the partition P is the width of the largest subinterval.
Denition: For an interval
with
◦
The only partitions we will be interested in are partitions of
that case, for
◦
•
0≤i≤n
we have
b−a
xi = a + i ·
n
[a, b]
n
into
equally-sized subintervals. In
.
However, in some applications it may be useful to use partitions where the intervals have dierent sizes.
Denition: For
f (x) a continuous function and P
the (left) Riemann sum of
f (x)
on
[a, b]
[a, b] into n subintervals, we dene
n
X
f (xi−1 ) [xi − xi−1 ].
RSP (f ) =
a partition of the interval
corresponding to
P
to be
k=1
◦
Reminder: If
g(x)
is a function, then the notation
n
X
g(k)
means
g(1) + g(2) + g(3) + · · · + g(n).
k=1
◦
Note: There are other types of Riemann sums than the left-endpoint sum.
function
f
Instead of evaluating the
at the left endpoint of each subinterval, we could just as well dene a Riemann sum using the
right endpoint instead: we would use
n
X
f (xi ) [xi − xi−1 ]
instead of the denition above.
k=1
∗
◦
Or we could use the midpoint of each interval, or (in general) any desired point in each subinterval.
In the case where
n
X
f (xi−1 ) ∆x,
k=1
•
Denition: For
f (x)
P
is the partition with
where
b−a
∆x =
n
n
equally-sized subintervals, we can write the Riemann sum as
is the common width of the subintervals.
a continuous function, we dene the denite integral
L such that, for every > 0, there exists
norm(P ) < δ , we have |RSP (f ) − L| < .
the value of
P
with
◦
a
δ>0
Remark: It can be proven (with signicant eort) that, if
the hypotheses actually does exist.
2
´b
f (x) dx of f (x) on [a, b] to be
) such that for every partition
a
(depending on
f (x) is continuous,
then a value of
L satisfying
◦
Notation: All of the parts of the denite integral notation are needed when writing an integral. The
part labels the variable of integration (and behaves exactly as a dierential), and the
function being integrated. The values
[a, b]
◦
b
and
dx
indicates the
are called the limits of integration, and specify the range
where the function is to be integrated.
Essentially, what this denition means is: the value of the denite integral is the limit of the Riemann
sums of
◦
a
f (x)
f,
as the size of the subintervals in the partition becomes small.
Observe the similarity between the notation
´b
a
n
X
f (xi−1 ) ∆x
for a Riemann sum, and the notation
k=1
f (x) dx
for the denite integral.
This similarity is deliberate: the idea of Leibniz (who developed
the notation) is that in the limit of
∆x → 0,
∆x
the term
becomes the dierential
dx,
and the sum
turns into an integral sign.
•
Note that our geometric motivation for integration involved nding the area under the graph of a function
y = f (x),
where we implicitly assumed that
f (x)
require that
◦
−1
times the result for
−f (x)
where
f (x)
is positive, we
+f (x).
So we can interpret the denite integral of a negative function as giving a negative area: that is, if we
interpret the area as being negative if
•
However, the denition via Riemann sums did not
If we follow the denition through and evaluate Riemann sums for
obtain
◦
f (x) ≥ 0.
be nonnegative: it makes perfectly good sense for negative-valued functions as well.
f (x) < 0,
the denite integral makes sense for all functions.
Riemann sums provide us with a way to compute areas under curves formally. Unfortunately, working with
Riemann sums is fairly cumbersome, as an example will demonstrate:
•
Example: Use Riemann sums to nd the area under
◦
Let us take a partition of
[0, 2]
into
n
f (x) = x
between
x=0
and
equal subintervals each of width
Riemann sum.
2
· i,
n
◦
We have
◦
Applying the Riemann sum formula gives
xi =
0 ≤ i ≤ n,
for
and
∆x =
◦
We can pull out the constant
Sn =
n
X
f (xi−1 ) ∆x =
terms to obtain
n X
2
k=1
Sn =
One can show by induction that the sum of the rst
shows
•
2
n
4
·
n2
n
n
X
Now as
◦
Therefore, the area is 2.
n
2
· (i − 1) · .
n
n
[i − 1] =
k=1
4
· (0 + 1 + 2 + · · · + (n − 1)).
n2
positive integers is
4 (n − 1)n
(n − 1)
1
Sn = 2 ·
=2·
=2· 1−
.
n
2
n
n
◦
and compute the resulting
2
.
n
k=1
◦
2
,
n
x = 2.
n(n + 1)
;
2
becomes large, we see that the value of the Riemann sums approach
2,
using this formula
since
1
n
approaches 0.
As with limits and derivatives, it is much easier to work with integrals after we have proven some basic
results on manipulating them. Here are some properties of denite integrals which are more or less immediate
consequences of the Riemann sum denition:
◦
Let
a < b < c be arbitrary, C
be an arbitrary constant, and
f (x) and g(x) be continuous functions.
the following properties hold:
ˆ
◦
b
C dx = C · (b − a)
Integral of constant:
.
a
ˆ
◦
ˆ
b
C · f (x) dx = C ·
Constant multiple of a function:
a
f (x) dx
a
3
b
.
Then
ˆ
◦
ˆ
b
◦
ˆ
b
f (x) dx −
Subtraction of functions:
f (x) ≥ 0,
.
b
[f (x) − g(x)] dx
a
ˆ
Nonnegativity: if
ˆ
b
g(x) dx =
a
◦
[f (x) + g(x)] dx
a
a
a
ˆ
b
g(x) dx =
f (x) dx +
Addition of functions:
ˆ
b
.
a
b
f (x) dx ≥ 0
then
.
a
∗
f (x) ≥ g(x),
As a corollary, if
subtraction property, we see that
∗
ˆ
b
a
f (x) dx ≥
ˆ
c
f (x) dx +
Union of intervals:
a
´b
a
and using the
g(x) dx.
f (x) dx
ˆ
f (x) dx = 0
.
a
b
f (x) dx = −
f (x) dx
b
a
. In particular,
a
a
Backwards interval:
ˆ
c
f (x) dx =
b
ˆ
◦
´b
In other words, we can integrate inequalities of functions.
ˆ
◦
f (x) − g(x) ≥ 0
then by applying this property to
. [This is actually a denition.]
a
4.2 The Fundamental Theorem of Calculus
•
Roughly speaking, the Fundamental Theorem of Calculus says that dierentiation and integration are inverses:
integrating the derivative of a function or dierentiating the integral of a function will (essentially) give back
the original function.
•
•
We will prove several other important theorems on the way to the Fundamental Theorem.
Min-Max inequality for integrals:
´b
a
If
f (x)
is a continuous function on
[a, b],
then
(b − a) · min[a,b] (f ) ≤
f (t)dt ≤ (b − a) · max[a,b] (f ).
◦
Proof: By denition,
g(x) ≤ h(x),
min(f )[a,b] ≤ f (x) ≤ max(f )[a,b] .
∗
If
∗
Applying this result shows that
∗
Since the rst and last integrals are just integrals of constants, we get
then integrating both sides from
´b
a
a
min[a,b] (f )dx ≤
to
b
´b
a
shows that is true
f (x)dx ≤
´b
a
´b
a
g(x) dx ≤
´b
a
max[a,b] (f )dx.
(b−a)·min[a,b] (f ) ≤
(b − a) · max[a,b] (f ).
•
Mean Value Theorem (for integrals): If
(a, b)
◦
for which
The value
1 ´b
f (c) =
f (t)dt.
b−a a
f (x)
is a continuous function on
1 ´b
f (t)dt is called the average
b−a a
value (or mean value) of
[a, b],
f
h(x) dx.
´b
a
f (t)dt ≤
then there exists some
on the interval
[a, b].
c
in
Intuitively,
the Mean Value Theorem says that there is a point in the interval where the function is equal to its average
value.
◦
Proof: Dividing through by
(b−a) everywhere in the Min-Max inequality gives min[a,b] (f ) ≤
1 ´b
f (t)dt ≤
b−a a
max[a,b] (f ).
∗
Since
f
is continuous, it attains its minimum and maximum values, and so by the Intermediate Value
Theorem it takes every value in between.
∗
But the average value
1 ´b
f (t)dt is between the min and max,
b−a a
and therefore is one of the values
attained.
• ´Fundamental Theorem of Calculus (part 1): For any continuous function f on [a, b], the
x
f (t)dt is continuous, dierentiable, and has the property that F 0 (x) = f (x) on [a, b].
a
4
function
F (x) =
◦
Proof: We look at the dierence quotient
"ˆ
F (x + h) − F (x)
1
lim
= lim
h→0
h→0 h
h
ˆ
x+h
x
f (t)dt −
a
a
∗
The quantity inside the limit is the mean value of
∗
Applying the Mean Value Theorem gives a value
f
ch
#
1
f (t)dt = lim
h→0 h
on the interval
in
[x, x + h]
for
"ˆ
#
x+h
f (t)dt .
x
[x, x + h].
i
1 h´ x+h
which
f
(t)dt
= f (ch ).
h x
F (x + h) − F (x)
= lim f (ch ), and this last limit is just f (x) because f is continuous and
h→0
h→0
h
the points ch approach x as h → 0.
F (x + h) − F (x)
0
exists and is equal to f (x), so F is dierentiable and F (x) = f (x).
∗ Therefore lim
h→0
h
Finally, since F is dierentiable, it is continuous.
∗
•
Then
lim
Fundamental Theorem of Calculus (part 2):
´b
a
If
F
is any antiderivative of the continuous function
f,
then
f (t)dt = F (b) − F (a).
◦
Proof: By the rst part of the Fundamental Theorem, we know that
of
f,
∗
since
G(x) =
G0 (x) = f (x).
´x
a
f (t)dt is an antiderivative
But we also know that any two antiderivatives of a function dier by a constant, because the only
functions with derivative zero are constants.
∗
Therefore,
∗
We also can see easily that
∗
Therefore,
G(x) = F (x) + C
´b
a
for some constant
G(a) = 0
and that
C.
G(b) =
´b
a
f (t)dt.
f (t)dt = G(b) − G(a) = [F (b) + C] − [F (a) + C] = F (b) − F (a),
as desired.
4.3 Indenite Integrals, Evaluating Denite Integrals
•
The Fundamental Theorem of Calculus shows that we can evaluate denite integrals using antiderivatives,
rather than the complicated process of computing limits of Riemann sums.
◦
If
◦
Note:
F (x)
is an antiderivative of
The notation
f (x),
F (x) bx=a
then
´b
a
f (t)dt = F (x) bx=a = F (b) − F (a).
means to evaluate the function
f
from
x = a
to
x = b,
meaning
F (b) − F (a).
•
Using antiderivatives to evaluate integrals is, by a wide margin, far easier than using Riemann sums. As such,
we will need some notation for general antiderivatives.
•
Denition: The indenite integral of
of
f (x) with respect to x, denoted
´
f (x) dx, is the set of all antiderivatives
f (x).
◦
By the results on antiderivatives, if
arbitrary constant
C,
if
f (x)
F (x)
is one antiderivative, then all the others are
is dened for all real
F (x) + C
for an
x.
◦
Extremely Important Note: When writing an indenite integral, the
◦
Computing indenite integrals is in general very dicult: unlike with derivatives, there is no straight-
+C
forward procedure for computing antiderivatives of arbitrary functions.
5
must always be included!
•
From the derivatives we have calculated, we can write a list of simple indenite integrals:
ˆ
xn dx
=
xn+1
+ C, n 6= −1
n+1
1
dx
x
=
ln(x) + C
ex dx
= ex + C
ˆ
ˆ
ˆ
sin(x) dx
= − cos(x) + C
cos(x) dx
=
sin(x) + C
sec2 (x) dx
=
tan(x) + C
sec(x) tan(x) dx
ˆ
1
√
dx
1 − x2
ˆ
1
dx
1 + x2
ˆ
1
√
dx
x x2 − 1
=
sec(x) + C
=
sin−1 (x) + C
=
tan−1 (x) + C
=
sec−1 (x) + C
ˆ
ˆ
ˆ
◦
´ 1
dx = ln |x| + C , with
x
This has the advantage of being dened for negative values of x (which ln(x) is not), but
Remark: For the indenite integral of
absolute values.
1
,
x
there are many sources which write
the downside is that this formula may appear to give nite values for denite integrals that are actually
undened.
If one takes the viewpoint of dening logarithms of negative numbers as having non-real
´ 1
dx = ln(x) + C
x
values, then the two formulas
minus sign gets absorbed into the constant of
systems declare that
absolute values, but
•
´ 1
dx = ln(x) + C
x
by using ln(−x) + C
´ 1
dx = ln |x| + C actually are equivalent: the
x
integration when x is negative. (Many computer algebra
and
for this reason.) We will take the convention of avoiding the
for denite integrals where
x
is negative.
Here are some other antiderivatives of basic functions that crop up occasionally:
ˆ
ln(x) dx
= x ln(x) − x + C
ˆ
tan(x) dx
= − ln(cos(x)) + C
sec(x) dx
=
csc(x) dx
= − ln(csc(x) + cot(x)) + C
cot(x) dx
=
ˆ
ln(sec(x) + tan(x)) + C
ˆ
ˆ
•
Example: Find
´π
0
ln(sin(x)) + C
sin(x) + x2 dx.
´
´π
1
◦ We have
sin(x) + x dx = − cos(x) + x3 + C , so 0 sin(x) + x2 dx =
3
1
1
1
− cos(π) + π 3 − − cos(0) + (0)3 = 2 + π 3 .
3
3
3
•
Example: Find
2
´ π/3
π/4
tan(x) dx.
6
1 3 π
− cos(x) + x x=0 =
3
◦
We have
´
tan(x) dx = − ln(cos(x)) + C ,
√
1
(− ln(cos(π/4))) = ln( 2) = ln(2)
2
so
´ π/3
π/4
π/3
tan(x) dx = (− ln(cos(x))) x=π/4 = (− ln(cos(π/3))) −
.
4.4 Dierentiating Integrals
•
Using integration, we can construct new functions whose derivatives we can nd using the Fundamental
Theorem of Calculus.
•
Example: If the function
◦
erf(x)
is dened via
Note: The integration variable is
t
2 ´x
2
erf(x) = √ 0 e−t dt,
π
and not
x:
x for
t.)
erf(x).
this is necessary because the limits of integration cannot
contain the variable of integration the integral
cannot use
nd the derivative of
2 ´x
2
√ 0 e−x dx
π
does not make sense. (Thus, since we
the variable of integration, we replace it with a dummy variable instead in this case
we called it
◦
To nd the derivative, we can just use the Fundamental Theorem of Calculus:
2
2 ´ x −t2
2
d
√ 0 e
dt = √ e−x
dx
π
π
we have
erf 0 (x) =
◦
directly from the rst part of the Fundamental Theorem.
Remark: This function is called the error function and shows up very often in statistics due to its
close connection to the normal distribution.
erf(x)
way to write the error function
It can be proven (though it is hard!)
that there is no
in terms of the elementary functions (i.e., as a sum, product,
or composition of any number of polynomials, exponentials, logarithms, and trigonometric or inverse
trigonometric functions of
x).
Thus, the description above as an integral is, in some sense, the simplest
way of describing the error function. So we have actually constructed a new function that we could not
have described without using integration.
•
Example: If
◦
g(x) =
´ x2 sin(t)
dt,
x
t
nd
The idea here is to rearrange
g 0 (x).
g(x)
into simpler pieces to which we can apply the Fundamental Theorem
of Calculus.
◦
Specically, the Fundamental Theorem tells us how to nd the derivative of the function
by
´ x sin(t)
h(x) = 0
dt:
t
we have
sin(x)
h (x) =
.
x
◦
Now, the critical observation to make is that
◦
Taking the derivative and applying the Chain Rule
sin(x)
2 sin(x2 ) − sin(x)
=
x
x
h(x)
dened
0
´ x sin(t)
´ x2 sin(t)
dt − 0
dt = h(x2 ) − h(x).
0
t
t
sin(x2 )
0
0 2
0
then yields g (x) = 2x h (x ) − h (x) = 2x ·
−
x2
g(x) =
.
4.5 Areas
•
Using integrals, we can compute areas under curves: the area under
x=a
•
x=b
is
´b
a
As an integral, this area is
´2
0
y=x
and above the
x dx =
2
x
2
As an integral, this area is
x-axis,
between
2
2
2
x=0 = 2 − 0 = 2
2
2
y = 2ex + sin(x) and above
´ 2π x
given by
[2e + sin(x)] dx.
0
Example: Find the area underneath
◦
y = f (x)
and above the
x-axis
between
f (x) dx.
Example: Find the area underneath
◦
•
and
7
the
x=0
and
x = 2.
.
x-axis,
between
x=0
and
x = 2π .
•
´
´
[2ex + sin(x)] dx = 2 ex dx + sin(x) dx = 2ex − cos(x) + C .
2π
´ 2π x
[2e + sin(x)] dx = (2ex − cos(x)) 2π
− 1 − 2e0 − 1 = 2e2π − 2 .
x=0 = 2e
0
◦
As an indenite integral, we have
◦
Then
To nd the area between two curves
´
y = f (x)
and
y = g(x)
for
x
in an interval
[a, b],
we need only subtract
the area under the lower curve from the area under the upper curve. Note that if the curves cross, the region
of integration needs to be broken into multiple pieces. To ensure nothing is missed, follow these steps:
◦
Step 1: Graph both curves (on the region of integration).
◦
Step 2: Find any points of intersection, and identify which function is larger on each interval.
◦
Step 3: Integrate [larger]
◦
− [smaller]
on each interval, and sum the results.
Note: If we want the (nite) area enclosed between the two curves, the region of integration is the interval
between the leftmost and rightmost intersection points.
Otherwise, the region of integration must be
given explicitly.
•
Example: Find the area between
◦
y=x
and
y = x2
between
x=0
and
x = 2.
Step 1: Here is a plot of both functions. We can see that they cross once on
[0, 2]:
4
3
2
1
0.5
◦
1.0
1.5
2.0
x = x2
x = 0 and x = 1, so the curves cross at x = 1. On [0, 1] we see that y = x
[1, 2] we see that y = x2 is the upper curve.
´1
´2
◦ Step 3: The desired area is A = 0 x − x2 dx + 1 x2 − x dx. Evaluating each integral yields A =
2
x
x3 1
x2 2
8
1 1
x3
1 1
−
−
−
+
−2 −
−
= 1.
x=0 +
x=1 =
2
3
3
2
2 3
3
3 2
Step 2: Solving
gives
is the upper curve, and on
•
Example: Find the nite area enclosed between
◦
y = x3 − 3x
and
y = x.
Step 1: Here is a plot of both functions. We see that they intersect three times.
5
-2
-1
1
2
-5
x = x3 − 3x gives x3 − 4x = 0 or x(x − 2)(x + 2) = 0, so they cross at x = −2, 0, 2. The
3
interval of interest is [−2, 2]. We see from the plot that on [−2, 0] that y = x − 3x is the upper curve,
and on [0, 2] that y = x is the upper curve.
´0 ´2
◦ Step 3: The desired area is A = −2 (x3 − 3x) − x dx + 0 x − (x3 − 3x) dx. Evaluating each integral
4
x
x4 2
yields A =
− 2x2 0x=−2 + 2x2 −
x=0 = (0 − (−4)) + (4 − 0) = 8 .
4
4
◦
Step 2: Solving
4.6 Substitution
ˆ
•
The general substitution formula states that
f 0 (g(x)) · g 0 (x) dx = f (g(x)) + C
. It is just the Chain Rule,
written in terms of integration via the Fundamental Theorem of Calculus. However, we generally don't use
the formula written this way. To do a substitution, follow this procedure:
◦
Step 1: Choose a substitution
u = g(x).
8
∗
◦
Step 2: Compute the dierential
∗
◦
If a function has a 'complicated' argument, a good rst guess is the complicated thing.
du = g 0 (x) dx.
Dierentials, it turns out, automatically keep track of everything during substitutions.
Step 3: Rewrite the original integral in terms of
∗
∗
∗
u:
3a: Rewrite the integral to peel o what will become the new dierential
3b: Write the remaining portion of the integrand in terms of
3c: Find the new limits of integration in terms of
limits are
x=a
and
x = b,
the new ones will be
du.
u.
u, if the integral is a denite integral. [If the old
u = g(a) and u = g(b). In a very concrete sense,
these are the same points.]
∗
3d: Write down the new integral.
If the integral is indenite, substitute back in for the original
variable.
•
Substitution is best learned by example.
•
Example: Evaluate
•
´3
0
2
2x ex dx.
◦
Step 1: The exponential has a 'complicated' argument
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
Step 3d: Putting it all together gives
Example: Evaluate
◦
du = 2x dx.
x=0
´3
0
´3
0
2
ex
so we try setting
u = x2 .
2
ex · (2x dx).
u = 02
corresponds to
x2 ,
ex
2
, which is just
eu .
x = 3 corresponds to u = 32 .
´9
· (2x dx) = 0 eu du = eu 9u=0 = e9 − 1
and
.
´ e (ln(x))2
dx.
1
x
Step 1: It might not look like any function has a 'complicated' argument, but if we think carefully we
can see that the numerator is what we get if we plug in
ln(x)
to the squaring function. So we try setting
u = ln(x).
◦
du =
1
dx.
x
´3
2
◦
Step 3a: We can rearrange the integral as
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
•
Step 2: The dierential is
corresponds to
Step 3d: Putting it all together gives
Example: Evaluate
◦
x=1
0
[ln(x)] ·
1
dx .
x
2
[ln(x)]
u = ln(1) = 0
, which is just
and
x=e
u2 .
corresponds to
´1
´ e (ln(x))2
1 1
dx = 0 u2 du = u3 1u=0 =
1
x
3
3
u = ln(e) = 1.
.
´1 √
x 3x2 + 1 dx.
0
Step 1: Here we see that the square root function has the 'complicated' argument
3x2 + 1
so we try
u = 3x2 + 1.
◦
Step 2: The dierential is
◦
Step 3a: We can rearrange the integral as
6·
1
,
6
du = 6x dx.
´1√
0
3x2 + 1 ·
1
· (6x dx).
6
Note that we introduced a factor of
but that is okay since it's just multiplication by 1.
√
◦
Step 3b: The remaining portion of the integrand is
◦
Step 3c: We see that
◦
Step 3d: Putting it all together gives
x=0
3x2 + 1 ·
1
,
6
which is just
1 1/2
u .
6
u = 1 and x = 1 corresponds to u = 4.
´1 √
´4 1
1 2
7
x 3x2 + 1 dx = 1 u1/2 du = · u3/2 4u=1 =
0
6
6 3
9
corresponds to
9
.
ˆ
•
3
Example: Evaluate
2
2x
dx.
x2 − 1
u = x2 − 1. The dierential is du = 2x dx, and x = 2 corresponds to u = 3 and x = 3
corresponds to u = 8.
´3 1
´3 1
◦ Rearranging yields 2 2
· (2x dx), and putting everything together gives 2 2
· (2x dx) =
x
−
1
x −1
8
´ 8 −1
u du = ln(u) u=3 = ln 8 − ln 3 .
3
´ 3 2x
2x
1
1
◦ Remark: It's possible to do this one dierently, by observing 2
=
+
. Then
dx =
2 x2 − 1
x −1
x+1 x−1
3
[ln(x + 1) + ln(x − 1)] x=2 = ln(4) + ln(2) − ln(3) = ln(8) − ln(3) as before.
◦
We substitute
4.7 Arclength, Surface Area, Volume, Moments
•
The general principle in many applications of integration (measuring area, surface area, volume, mass, etc.)
is: slice the desired object up into many small pieces, evaluate the desired quantity on the small pieces, and
then add up the results. This will be a Riemann sum for the integral we wish to nd.
•
Using dierentials, we can simplify many of the calculations. The procedure is to determine the dierential
of the quantity we are interested in (in terms of dierentials of quantities we know), and then integrate.
4.7.1
•
Arclength
Arclength Formula: If
x(t)
y(t)
and
are dierentiable and their derivatives are continuous, then the arclength
ˆ
s
of the curve parametrized by
(x(t), y(t))
from
t=a
to
t=b
b
p
s=
is
x0 (t)2 + y 0 (t)2 dt
.
a
◦
Note: The letter
s
is traditionally used to denote arclength. The arclength of a curve (being a length) is
always positive.
◦
If the curve is given explicitly as
ˆ
y = f (x),
one can take
x = t
and
y = f (t)
in the formula to get
b
p
s=
1 + f 0 (t)2 dt
.
a
◦
The idea behind the proof is to slice up the curve into small pieces, approximate each piece with a line
segment, and add up the lengths. As we shrink the size of the pieces, the sum turns into an integral.
◦
Proof:
∗
Over a small time interval
to
∗
The length
2
2
2
2
2
∆s of this line segment is [∆s] = [x(t + ∆t) − x(t)] +[y(t + ∆t) − y(t)] = [∆x] +[∆y]
∗ Taking
the limit
p
(dx)2 + (dy)2 .
•
∆t, the curve is closely approximated by a line segment joining (x(t), y(t))
(x(t + ∆t), y(t + ∆t)).
as
∆t → 0
∗
Dividing both sides by
∗
Integrating
ds
dt
from
dt
t=a
Example: Find the arclength of
turns the dierences into dierentials:
gives
to
ds
=
dt
t=b
s
dx
dt
2
+
dy
dt
2
=
p
shows that the arclength is
x = cos(t), y = sin(t)
x0 (t) = − sin(t) and y 0 (t) = cos(t),
´ 2π
the arclength s =
1 dt = 2π .
0
from
t=0
to
2
[ds] = [dx] + [dy]
´bp
a
x0 (t)2 + y 0 (t)2 dt,
t = 2π .
x0 (t)2 + y 0 (t)2 = sin2 (t) + cos2 (t) = 1.
We have
◦
Then
◦
Note: Of course, we already know that the circumference of the unit circle is
10
2
, so
2π .
.
ds =
x0 (t)2 + y 0 (t)2 .
◦
so
2
as desired.
•
◦
We have
x0 (t) = 3t2 − 3
(3t2 + 3)2 .
◦
4.7.2
•
x = t3 − 3t, y = 3t2
Example: Find the arclength of
Then the arclength
s=
and
´2
1
y 0 (t) = 6t,
so
t=1
from
to
t = 2.
x0 (t)2 + y 0 (t)2 = 9t4 − 18t2 + 9 + 36t2 = 9t4 + 18t2 + 9 =
(3t2 + 3) dt = t3 + 3t 2t=1 = 14 − 4 = 10
.
Surface Area, Surfaces of Revolution
Surfaces of Revolution: The graph of the dierentiable function
about the
x-axis,
y = f (x)
between
x=a
and
x=b
is rotated
creating a surface. We wish to nd the area of this surface. Using integration, we can show
ˆ
that the result is
b
p
1 + f 0 (x)2 dx
2πf (x) ·
A=
.
a
◦
The idea behind the proof is to slice the surface into horizontal strips, and add up their areas.
◦
Proof: Over a small interval
[x, x + ∆x],
each strip is a thin conical slice whose radius of revolution is
the distance to the axis and whose outer length is a piece of the arclength of the curve.
∗
∗
So the dierential area of each strip is
The radius is
f (x),
dA = 2π · [radius] · [arclength].
and for the arclength, we approximate the curve with a line segment as in the
∗
The dierential of the surface area is therefore
p
∗
•
•
Integrating from
x=1
so
dA
= 2πf (x) ·
dx
1 + (f 0 (x))2 .
x=a
to
x=b
then yields the formula.
Example: Find the area of the surface formed by revolving the portion of the graph of
and
4.7.3
p
(dx)2 + (dy)2 .
p
dA = 2πf (x) · (dx)2 + (dy)2 ,
calculation for the arclength of a curve to obtain
about the
between
x=0
x-axis.
◦
We have
◦
Now we substitute
◦
So we get
a = 0, b = 1,
A=
y = x3
´ 10
1
and
f (x) = x3
with
f 0 (x) = 3x2 ,
so the formula gives
A=
´1
0
√
2πx3 1 + 9x4 dx.
u = 1 + 9x4 : we have du = 36x3 dx; also, x = 0 gives u = 1, and x = 1 gives u = 10.
π 2 3/2 10
π
π
π√
u du =
· u
· 103/2 − 13/2 =
(103/2 − 1) .
u=1 =
18
18 3
27
27
Volume, Solids of Revolution
General Volume Formula: If a solid running from
ˆ
total volume of the solid is given by
V =
x=a
to
x=b
has cross-sectional area
A(x)
at
x,
then the
b
A(x)dx
.
a
◦
The justication of this formula is similar to that used for the area underneath a curve: the idea is to
approximate a volume using Riemann sums, where instead of rectangles, we use the cross-sections to
form boxes.
•
Example: A cut log running from
square with side length
•
1+
√
x,
x = 0
to
x = 1
has square cross-sections.
If the cross-section at
x
is a
nd the volume of the log.
√ 2
√
A(x) = (1 + x) = 1 + 2 x + x.
´1
√
2
1
4
1
17
V = 0 (1 + 2 x + x) dx = x + 2 · x3/2 + x2 1x=0 = 1 + · 13/2 + · 12 − 0 =
3
2
3
2
6
◦
Since the cross-sections are squares, we have
◦
Then
Solids of Revolution: The area between the graph of
y = f (x)
and the
x-axis
between
x=a
and
.
x=b
is
rotated about an axis, and we wish to nd the volume of the resulting solid. Via the general volume formula,
all that is needed is to choose an axis of integration, nd the area of each cross-section, and then integrate.
11
◦
Disc method: If the axis of integration is the same as the axis of rotation, then the cross-sections are
discs with radius equal to the distance of the function to the axis. When the axis of rotation is the
ˆ
the radius is just the value of the function, and the volume is given by
x-axis,
b
π · f (x)2 dx
V =
.
a
∗
If instead of the area under a curve we are interested in the area between two curves
y = g(x),
generated by the larger function. If
f (x)2 − g(x)2 dx.
◦
y = f (x)
and
we need only subtract the volume generated by the smaller function from the volume
f (x) ≥ g(x)
on
[a, b],
then we obtain the formula
V =
´b
a
π·
This formula is sometimes referred to as the washer method.
Shell method: If the axis of integration is perpendicular to the axis of rotation, the cross-sections are
cylindrical shells, instead, and the shell area will be equal to
2π
times the distance to the axis times
the height of the shell. In the particular case where the axis of rotation is the
ˆ
then the volume
b
2πx · f (x)dx
V =
will be given by
y -axis,
.
a
◦
Note that, by rewriting the function and (if necessary) choosing to integrate along the
the
•
x-axis,
y -axis
instead of
it is possible to use either the disc method or the shell method for any given problem.
Example: The area under the curve
y=
√
3x2 + 1
between
x=1
and
x=4
is rotated about the
x-axis.
Find
the volume of the resulting solid.
◦
x-axis, since the function is a function of x. Since the √
axis of rotation
a = 1, b = 4, and f (x) = 3x2 + 1.
2
√
´4
´4
◦ Then V = 1 π · 3x2 + 1 dx = 1 π · (3x2 + 1) dx = π · x3 + x 4x=1 = 68π − 2π = 66π .
√
• Example: The area under the curve y = 3x2 + 1 between x = 1 and x = 4 is rotated about the y -axis. Find
The easiest axis of integration is the
is the same as the axis of integration, we use the disc method:
the volume of the resulting solid.
◦
The easiest axis of integration is the
x-axis, since the function is a function of x. Since the axis of
√rotation
a = 1, b = 4, and f (x) = 3x2 + 1.
is perpendicular to the axis of integration, we use the shell method:
•
√
3x2 + 1 dx.
This gives
◦
To evaluate this integral we substitute
gives
4.7.4
´4
◦
V =
1
π·x·
u = 49, we see that V =
´ 13
4
u = 3x2 + 1 so that du = 6x dx. Since x = 1 gives u = 4 and x = 4
π
π 3/2
π 2
335π
π √
· u du = · u3/2 49
· 49 − 43/2 = ·[343 − 8] =
.
x=4 =
6
6 3
9
9
9
Center of Mass, Moments
The center of mass of a physical object is its balancing point, where, if the object is supported only at that
point, gravity will not cause it to tip over.
◦
The center of mass is also called the centroid of an object.
◦
Physically, an object behaves as if all its mass is concentrated at its center of mass (hence the name):
the forces exerted by an object are the same as an equivalent point mass concentrated at the object's
center of mass.
•
Using calculus, if we are given formulas for the shape of an object, we can nd the location of its center of
mass.
◦
We would, in real life, be most interested in the physical properties of 3-dimensional objects with variable
densities.
◦
However, the calculations to nd centers of masses for such objects are extremely dicult to do with a
single integration in one variable.
∗
These types of problems are a common application of multivariable calculus: they are far easier to
work out using integration in several variables.
12
◦
Since we are using single-variable integration, we can only discuss the problems of a 1-dimensional rod
with variable density, and a 2-dimensional plate with density depending only on one coordinate.
•
Center of Mass and Moment Formulas (rod): We are given a 1-dimensional rod of variable density
tween
x=a
The total mass
◦
The rst moment
◦
The center of mass
is given by
M0
x̄
M=
is given by
x
is
ρ(x) =
√
be-
ρ(x)dx.
´b
M0 = a xρ(x)dx.
a
is given by the ratio of the rst moment to the total mass:
x̄
ρ(x)
´b
is the average value of the
x-coordinate
x̄ =
´b
xρ(x)dx
M0
= ´a b
.
M
ρ(x)dx
a
over the rod.
Example: Find the mass and center of mass of the rod whose endpoints are at
x=1
and
x=4
and whose
x.
´4√
2
2 3/2 4
14
2 3/2
.
x
· 4 − 13/2 = · 7 =
x=1 =
3
3
3
3
2
´4
√
2
2 62
rst moment is M0 =
x · x dx = x5/2 4x=1 = · 45/2 − 15/2 = · 31 =
.
1
5
5
5
5
62/5
M0
91
=
=
center of mass is located at x̄ =
.
M
14/3
35
◦
The mass is given by
◦
The
◦
The
M=
1
x dx =
ρ(x)
y , whose upper boundary has equation y = g(x) and whose lower boundary has
equation y = f (x), between x = a and x = b.
´b
◦ The total mass is given by M = a ρ(x) · [g(x) − f (x)] dx.
´b
´b 1
◦ The rst moments are given by My = a x·ρ(x)·[g(x) − f (x)] dx and Mx = a ·ρ(x)· g(x)2 − f (x)2 dx.
2
∗ The moment My is the average value of x over the plate, and the moment Mx is the average value
of y over the plate. (The traditional notation for this is, unfortunately, very confusing.)
My
Mx
◦ The center of mass (x̄, ȳ) has coordinates x̄ =
and ȳ =
.
M
M
Center of Mass and Moment Formulas (plate): We are given a 2-dimensional plate with density function
which depends on
•
M
The center of mass
density at
•
x = b.
◦
∗
•
and
x
but not
Example: Find the center of mass of the thin plate with uniform density 1 whose shape is the quarter-disc
x2 + y 2 ≤ 1
◦
with
x≥0
and
y ≥ 0.
The quarter-disc runs from
x = 0 to √
x = 1, the
y = 1 − x2 .
lower boundary is the
x-axis
with equation
y = 0,
and
the upper boundary is the curve
◦
Since the density is 1, the mass of the disc is simply its area, which is
∗
As an integral the mass would be
´1
´1√
0
1 − x2 dx,
π
.
4
but this integral is tricky to evaluate.
´1 1
◦ The moments are My = 0 x · 1 − x2 dx and Mx = 0 · (1 − x2 ) dx.
2
∗ For My we substitute u = 1 − x2 , with
du
=
−2x
. We see x = 0 gives u = 1 and x = 1
´0
´1 1
1
1 √
1 2
so the substitution gives My =
u du = 0 u1/2 du = · · u3/2 1u=0 = .
−
1
2
2
2 3
3
3
´1 1
1
x 1
1 2
1
∗ For Mx we have Mx = 0 · (1 − x2 ) dx = · x −
· −0= .
x=0 =
2
2
3
2 3
3
1/3
4
1/3
4
4 4
◦ Then x̄ =
=
and ȳ =
=
, so the center of mass is
,
.
π/4
3π
π/4
3π
3π 3π
√
gives
u = 0,
Well, you're at the end of my handout. Hope it was helpful.
Copyright notice: This material is copyright Evan Dummit, 2012. You may not reproduce or distribute this material
without my express permission.
13