Math 131.
Riemann Sums and Definite Integrals
Larson Section 4.3
This section formalizes what is meant by a definite integral; much of the process will be similar
to finding areas using the limit definition.
Definition of a Riemann Sum Let f be defined on the closed interval [a, b], and let ∆ be a
partition of [a, b] given by
a = x0 < x1 < x2 < . . . < xn−1 < xn = b
and let ∆xi = xi − xi−1 denote the width of the ith subinterval in the partition, [xi−1 , xi ].
If ci is any point in the ith subinterval, then the sum
n
X
f (ci )∆xi , xi−1 ≤ ci ≤ xi
i=1
is called a Riemann sum of f for the partition ∆.
Note that the sums from the previous section were examples of Riemann sums, but they were
quite specialized, in that the widths of the partitions were always equal, and the ci were usually
chosen quite systematically or for convenience (like the right endpoint of the interval).
Given any partition ∆ as above, the norm of the partition k∆k is the width of the largest
subinterval in the partition.
Example 1. (a) Find the Riemann sum for f (x) = 0.6x2 + 1.1 over the closed interval [1, 3]
where ∆ is the partition of [1, 3] given by
1 < 1.5 < 1.8 < 2.1 < 2.6 < 3
and ci is the midpoint of the ith subinterval for i = 1, 2, 3, 4, 5.
(b) Do a rough sketch of the rectangles that represent the sum you found in (a). Can you
conclusively state that your sum overestimates or underestimates the actual area under the
graph of y = 0.6x2 + 1.1 for 1 ≤ x ≤ 3, or is it difficult to determine?
Solution: (a) The Riemann sum is then
5
X
f (ci )∆xi = f (1.25)(0.5) + f (1.65)(0.3) + f (1.95)(0.3) + f (2.35)(0.5) + f (2.80)(0.4)
i=1
= (2.03750)(0.5) + (2.73350)(0.3) + (3.38150)(0.3) + (4.41350)(0.5) + (5.80400)(0.4)
= 7.381600
(b) A graph of the rectangles from the Riemann sum is as follows:
8 y
6
4
2
x
1
2
3
4
From the graph it is hard to tell whether the sum overestimates or underestimates the true
area because there are corners of the rectangles that lie above the curve, and parts of the
area under the curve that are outside the rectangles. (The actual area to 6 decimal places
is 7.400000, and so the Riemann sum slightly underestimated the area by 0.018400 square
units, but that small amount is not obvious from the graph).
The Definite Integral. Suppose f is defined on the closed interval [a, b] and the limit
lim
n
X
k∆k→0
f (ci )∆xi
i=1
exists, then f is integrable on [a, b] and the limit is denoted by
lim
k∆k→0
Z
n
X
Z
f (ci )∆xi =
b
f (x) dx
a
i=1
b
We call
f (x) dx the definite integral of f from a to b. The number a is called the lower
a
limit of integration and the number b is called the upper limit of integration.
Properties of Integrals. Suppose f is continuous on the closed interval [a, b].
Z b
(a) Then
f (x) dx exists.
a
Z
(b) If, additionally, f (x) ≥ 0 on [a, b], then
b
f (x) dx is the area under the graph of y = f (x)
a
that lies above the x-axis and between the lines x = a and x = b.
Example 2. Express the areas of the shaded regions below the graph of y = 0.3x2 + 0.8 as
integrals.
8 y
8 y
6
6
4
4
2
2
x
−1
−2
1
2
3
4
x
−1
−2
5
1
2
3
4
5
Solution: The area of the shaded region in the graph to the left above is given by
Z
4
[0.3x2 + 0.8] dx.
0
4
Z
[0.3x2 + 0.8] dx
The area of the shaded region in the graph to the right above is given by
2
Integrals can be calculated using their limit definition, much like areas. The following is an
example of this.
Z
Example 3. This problem will go through a step-by-step calculation of
n→∞
(x2 − 2x + 1) dx
0
using the formal definition of the integral as the limit
" n
#
Z b
X
f (x) dx = lim
f (x∗k )∆x
a
9
k=1
where f (x) = x2 − 2x + 1 is integrable because it is continuous.
(a) Divide the interval [0, 9] into n subintervals of equal length ∆x. What is ∆x?
(b) Let x∗k be chosen as the right-hand endpoint of the k-th subinterval. Express x∗k in terms
of k and n.
(c) Express f (x∗k )∆x in terms of k and n.
(d) With the help of appropriate summation formula(s), express
n
X
f (x∗k )∆x in terms of n.
k=1
(e) Take the limit of the expression in (d) to find the integral. That is
" n
#
Z 9
X
(x2 − 2x + 1) dx = lim
f (x∗k )∆x =
0
n→∞
k=1
Solution: (a) ∆x =
9−0
9
=
n
n
9
9k
=
n
n
" # 2
9k
9k
9
729k 2 162k 9
(c) f (x∗k )∆x =
−2
+1
=
− 2 +
n
n
n
n3
n
n
(b) x∗k = 0 + k(∆x) = k ·
(d)
n
X
k=1
f (x∗k )∆x
=
n X
729k 2
k=1
n3
162k 9
− 2 +
n
n
n
n
n
729 X 2 162 X
9X
=
k − 2
k+
1
n3 k=1
n k=1
n k=1
162 n(n + 1)
9
729 n(n + 1)(2n + 1)
− 2
+ (n)
=
3
n
6
n
2
n
729
1
3
1
=
2 + + 2 − 81 1 +
+9
6
n n
n
(e) Using part (d), the integral is then
Z
"
9
(x2 − 2x + 1) dx =
0
lim
n→∞
=
lim
n→∞
#
f (x∗k )∆x
k=1
=
n
X
729
6
1
3
2+ + 2
n n
1
− 81 1 +
+9
n
729
− 81 + 9 = 171
3
As you can see from the above example, if you can avoid using the limit definition to compute
an integral, it will probably save you time and work. The following example does precisely this
by taking advantage of the connection between integrals and area.
Example 4. The integral given below represents the area of a region below a nonnegative
continuous function and above the x-axis.
Z 4
x
8−
dx.
2
1
(a) Describe and sketch the region.
(b) Evaluate the integral by using geometry to find the area of the region.
Solution: (a) The shaded region lies below the function y = 8 − x2 and above the x-axis,
and it is between the vertical lines x = 1 and x = 4. The region is the shaded region given
in the graph below.
12 y
10
8
6
4
2
x
−6−5−4−3−2−1
−2
1 2 3 4 5 6
(b) The shaded region forms a trapezoid, where the base is b = 4−1 = 3 units. The heights
of the sides are h1 = 7.5 and h2 = 6.0 units respectively. Therefore, the area is
h1 + h2
7.5 + 6.0
A=b
= (3)
= 20.25
2
2
Several important and natural properties of integrals are listed next. These properties will be
useful for the rest of your calculus career.
Some Important Properties of Integrals.
a
Z
1. If f is defined at x = a, then
f (x) dx = 0.
a
Z
a
Z
f (x) dx = −
2. If f is integrable on [a, b], then
b
b
f (x) dx.
a
3. (Additive Property) If f is integrable on the three closed intervals determined by a, b,
and c, then
Z b
Z c
Z b
f (x) dx =
f (x) dx +
f (x) dx
a
a
c
4. (Linearity Property) If f and g are integrable on [a, b] and k is a constant, then the
functions kf and f ± g are integrable on [a, b], and
Z b
Z b
Z b
Z b
Z b
kf (x) dx = k
f (x) dx
and
[f (x) ± g(x)] dx =
f (x) dx±
g(x) dx.
a
a
a
a
a
5. (Preservation of Inequality) If f and g are integrable on [a, b] and f (x) ≤ g(x) for each
x in [a, b], then
Z b
Z b
g(x) dx
f (x) dx ≤
a
a
Example 5. The following graphs illustrate the additive propery of integrals above. The
function y = |x − 3| + 4 is used in both graphs below. The area of the shaded pink region in
the graph below to the left is given by the integral
Z 5
(|x − 3| + 4) dx.
−1
Z
In the graph to the right below, the shaded yellow region has area equal to
Z 5
(|x − 3| + 4) dx.
while the shaded green region has area equal to
3
(|x − 3| + 4) dx,
−1
3
12 y
12 y
10
10
8
8
6
6
4
4
2
2
x
−6−5−4−3−2−1
−2
x
−6−5−4−3−2−1
−2
1 2 3 4 5 6
1 2 3 4 5 6
Together the areas of the green and yellow regions add up to the area of the pink region, and
so
Z 5
Z 3
Z 5
(|x − 3| + 4) dx
(|x − 3| + 4) dx +
(|x − 3| + 4) dx =
−1
−1
3
which illustrates the additive property with a = −1, c = 3 and b = 5.
Z
5
(|x − 3| + 4) dx.
Example 6. Use the graphs given above to evaluate the integral
−1
Solution: The integral is the area of the shaded pink region which forms two trapezoids
(see the graph to the right, one yellow and one green).
The yellow trapezoid has b = 4 units. The heights of the sides are h1 = 8 and h2 = 4 units
respectively. Therefore, its area, A1 , is
h1 + h2
8+4
A1 = b
= (4)
= 24.00
2
2
The green trapezoid has b = 2 units. The heights of the sides are h1 = 4 and h2 = 6 units
respectively. Therefore, its area, A2 , is
h1 + h2
4+6
A2 = b
= (2)
= 10.00
2
2
Therefore,
5
Z
(|x − 3| + 4) dx = A1 + A2 = 24.00 + 10.00 = 34.00
−1
The following example provides illustrations of using the various properties of definite integrals
given earlier. The functions are intentionally generic so that you focus on using properties of
integration, rather than on the specific functions, to obtain the answers.
Example 7. ZSuppose f and gZare integrable on the Zclosed intervals determined by −3, 5 and
5
5
5
2. Given that
f (x)dx = 8,
f (x)dx = 11, and
g(x) dx = 7, evaluate:
−3
2
Z
2
Z
(a)
f (x) dx and
8f (x) dx
−3
−3
5
Z
(b)
Z
5
[f (x) − g(x)] dx
[f (x) + g(x)] dx and
−3
−3
−3
Z
−3
2
(c)
−3
Z
f (x) dx and
g(x) dx
5
5
5
Z
Z
5
[8f (x) − 3g(x)] dx and
(d)
−3
[8f (x) + 3g(x)] dx
−3
2
Z
[15f (x) − 2001g(x)] dx
(e )
2
2
Z
(f)
Z
5
[g(x) + f (x)] dx +
−3
g(x) dx
2
Z
5
(g) If α and β are two constants, write
(αf (x) + βg(x)) dx in terms of α and β
−3
Z
(h)
5
2
Z
g(x) dx +
2
g(x) dx
5
Solution: (a)
Z
2
Z
5
f (x) dx −
f (x) dx =
−3
Z
−3
5
f (x) dx = 8 − 11 = −3,
2
2
Z
Z
2
f (x) dx = (8)(−3) = −24
8f (x) dx = 8
and
−3
−3
Z
5
Z
g(x) dx = 8 + 7 = 15, and similarly,
f (x) dx +
−3
−3
−3
5
Z
[f (x) + g(x)] dx =
(b)
Z
5
5
[f (x) − g(x)] dx = 8 − 7 = 1
−3
Z
−3
Z
5
−3
5
g(x) dx = −7
f (x) dx = −8 and similarly
f (x) dx = −
(c)
−3
Z
5
(d)
5
Z
5
Z
5
Z
g(x) dx = (8)(8) − (3)(7) = 43,
f (x) dx − 3
[8f (x) − 3g(x)] dx = 8
−3
−3
−3
and
Z
5
Z
5
[8f (x) + 3g(x)] dx = 8
5
f (x) dx + 3
−3
Z
Z
−3
g(x) dx = (8)(8) + (3)(7) = 85
−3
2
[15f (x) − 2001g(x)] dx = 0 because the upper and lower limits of integration are the
(e)
2
same.
(f)
Z
2
Z
Z
2
Z
2
−3
Z 5
−3
Z 2
2
=
Z
5
g(x) dx
g(x) dx +
f (x) dx +
g(x) dx =
[g(x) + f (x)] dx +
−3
5
2
g(x) dx = −3 + 7 = 4
f (x) dx +
−3
−3
(g)
Z
5
Z
5
5
g(x) dx = 8α + 7β.
f (x) dx + β
[αf (x) + βg(x)] dx = α
−3
Z
−3
−3
(h)
Z
5
Z
g(x) dx +
2
2
Z
Z
g(x) dx −
g(x) dx =
5
5
2
5
g(x) dx = 0
2