Section 5.2B: Limits of Finite Sum
Today we’re going to focus on the Riemann Sums. Remember that Riemann Sums use rectangles to estimate
the area under the curve. As you use more and more rectangles, you will get a better and better approximation
of the total area.
Just as a reminder, the Riemann Sum formulas are as follows. Given a function f (x), an interval [a, b] an integer
n, ∆x = (b − a)/n, and P∆ = {x0 , x1 , x2 , . . . xn−1 , xn } where xk = a + k · ∆x.
Left Riemann Sum =
n−1
X
∆x (f (xk ))
Right Riemann Sum =
k=0
n
X
∆x (f (xk ))
k=1
Let us consider the function f (x) = 16−x2 on the interval [0, 4]. Last class, we used n = 4 rectangles to estimate
the area. For an arbitrary integer n, then our ∆x = 4/n. Thus, our partition of [0, 4] is:
P∆
=
{a + 0∆x, a + 1∆x, a + 2∆x, a + 3∆x, · · · , a + (n − 1)∆x, a + n∆x}
4
4
4
4
,2
,3
, · · · (n − 1)
,4
=
0, 1
n
n
n
n
Remember that for the Riemann Sum rectangles, your width is uniform: ∆x. Your height is determined by
evaluating your partition points into your function. The Left Riemann Sum will use the left endpoint and all the
middle values, while the Right Riemann Sum will use the right endpoint and all the middle values.
RRS
=
n X
4k
4
f
n
n
∆x =
k=1
=
n X
4
k=1
=
n
16 −
4k
n
2 !
4
4k
, and xk =
.
n
n
2
Since f (x) = 16 − x , f
n X
4
16k 2
16 −
n
n2
4k
n
=
16 −
4k
n
2 !
Square the term, then distribute.
k=1
=
n X
64
k=1
64k 2
− 3
n
n
=
64
64
·n− 3
n
n
n(n + 1)(2n + 1)
6
=
64 −
64(2n3 + 3n2 + n)
6n3
=
64 −
128n3 + 192n2 + 64n
6n3
Now use the constant term rule and square summation rule.
Distribute and simplify.
And as you limit this value as n goes to infinity, the Right Riemann Sum converges to 64 - 128/6. Doing the
same exploration will yield that the Left Riemann sum will also converge to the same number. This brings us to
our next vocabulary term.
Example: Compute the limit of the Right Riemann Sum as n → ∞ of f (x) = 9 − 3x2 on the interval [0, 3].
Section 5.3: The Definite Integral
Definition: A function f (x) is Riemann Integrable over an interval [a, b] if its Left and Right Riemann sums
both converge to the same value. The next section brings us to an important theorem, from p. 307:
Theorem 1 - Integrability of Continuous Functions: If a function f (x) is continuous over the interval [a, b]
or if f (x) has at most finitely many jump discontinuities there, then f (x) is integrable over [a, b].
Given a Riemann Integrable function f (x) on an interval [a, b], we define the definite integral of f over [a, b] as
the convergent limit, as n goes to ∞ of the Left and Right Riemann sums of f (x) on [a, b]. The definite integral
is denoted as:
Z
b
f (x)dx
a
where a and b are your limits of integration, f (x) is your integrand, and dx is your variable of integration.
This notation should look familiar. When we get to section 5.4, we’ll tie it altogether.
Properties of Definite Integrals: Given two functions f (x) and g(x) both integrable on [a, b], we have:
(a) Sum and Difference Rule:
Z b
Z b
Z
(f ± g)(x)dx =
f (x)dx ±
a
a
b
g(x)dx
a
(c) Constant Multiple Rule:
Z b
Z b
k · f (x)dx = k ·
f (x)dx
a
(b) Zero Width Interval:
Z a
f (x)dx = 0
a
(d) Order of Integration:
Z b
Z a
f (x)dx = −
f (x)dx
a
(e) Additivity Rule:
Z b
Z
f (x)dx +
a
a
c
c
Z
f (x)dx =
b
b
f (x)dx
a
Examples: Use the properties listed below to do answer the following.
1. Given that f and g are integrable and that:
Z 2
Z 5
f (x)dx = 4,
f (x)dx = 12,
1
2
Use the properties of integration to compute the following:
Z 7
Z 5
(a.)
f (x)dx
(b.)
f (x)dx
7
Z
(d.)
5
g(x)dx = 15,
1
2
Z
g(x)dx
5
5
Z
(f (x) − g(x))dx
(e.)
1
1
Z
(c.)
1
5f (x)dx
1
Z
(f.)
2
(3f (x) + 2g(x))dx
1
2. Graph the integrands and use the areas to evaluate the integrals listed below:
Z 4
Z 0p
Z 1
x
− 3 dx
(b.)
25 − x2 dx
(c.)
(2 − |x|) dx
(a.)
−4
−1
−2 2
3. Use Theorem 1 to give an example of a function f (x) that is not integrable on the interval [−5, 5].